Chemistry notes Important terms *Mass of element in a sample

Chemistry notes
Important terms
*Mass of element in a sample
Periods- horizontal rows on a periodic table
Groups- vertical rows on a periodic table
Ionic compound – electron transfer from a metal to a non-metal
Covalent compound- electron sharing between two non-metals
Hydrates- have a specific number of water molecules associated with each formula
unit ( shown with a *#H2O)
Aqueous solutions are solutions in water
Combustion analysis- add O2  H2O +CO2
Limiting reactant- substance that stops the reactions from proceeding
Molarity- concentration of a solution (mole/liters)
Polar molecule : a molecule that is electronegativity charged
Precipitate reaction- when two soluble ionic compounds combined to form an
insoluble product
Acid-base reaction- acid reacts with a base to form a neutral substance
Titration- one solution of known concentration is used to determine the
concentration of another solution through a monitored reaction
End point- tiny excess of OH- ion changes
Oxidation-reduction reaction(redox)- net movement of electrons from one reactant
to the other, goes from less electronegative to more electronegative
rules
1. For atoms in their elemental form, the oxidation number is 0
2. For ions, the oxidation number is equal to their charge
3. For single hydrogen, the number is usually +1 but in some cases it is -1
4. For oxygen, the number is usually -2
5. The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal
to its total charge.
Standard temperature and pressure(STP)- 273.15K and 1 atm
Standard molar volume = 22.4 L
Universal gas constant= .082058 atm*L
Mol*K
Mole fraction(X) the fraction of each element in a compound
Kinetic-molecular theory- describes behavior of gas at macroscopic levels
Root mean squared speed (rms) a molecule moving at this speed has the average
kinetic energy
Effusion- gas escaping through a tiny hole
Diffusion- gas moving through another gas
Equations
Mass of compound in sample * mass of element in compound
Mass of compound
Molecular mass= sum of atomic masses
Avogadro’s number= 6.022*1023= 1 mole
Mass % of element x = moles of X in formula * molar mass of X (g/mole)
Mass(g) of 1 mole of compound
%Yield=actual yield *100
theoretical yield
PV=nRT
n= m=PV
M RT
Pa=Xa * Ptotal
Urms= √3RT/M
R= 8.314 (J/k*mol)
Rate of effusion Rate a =√Mb
Rateb √Ma
Vander Waals equation = (Pn2a) (V-nb) =nRT
V2
a and b are Vander Waals constants
A relates to the number of electrons
B relates to molecular volume
Laws
*law of multiple proportions
if element A and B react to form tow compounds, the different masses of B than
combine with a fixed mass of A can be expressed as a ratio of small whole numbers
*law of mass conservation
the total mass of a substance does not change during a chemical reaction. The
number of substance may change. The total amount of substance is constant
*law of definite composition
no matter what its source a particular compound is composed of the same elements
in the same parts by mass.
*coulombs law
the energy of attraction is directly proportional to the product of the charges and
inversely proportionate to the distance between them
*ideal gas law
PV=nRT
*Boyles law
at constant temperature the volume occupied by a fixed amount of gas is inversely
proportionate to the applied pressure
*Charles law
at constant pressure, the volume occupied by a fixed amount of gas is directly
proportionate to its absolute temperature
*Dalton’s law of partial pressure
in a mixture of unreacting gas the total pressure I the sum of the partial pressure of
the individual gases Ptotal=P1+P2+P3+P4…
*Grahams law of effusion
the rate of effusion of a gas is inversely proportional to the square root of molar
mass
Chapter 16 kinetics
Factors that influence reaction rate
1. Concentration – molecules must collide to react
a. Reaction rate is proportional to concentration of reaction
2. physical state – molecules must mix to collide
a. the more finely divided a solid or liquid reaction , the greater its
surface area per unit volume, the more contact it makes with the
other reactant, the faster it occurs
3. Temperature- molecules must collide with enough energy to react
a. At higher temperature more collisions occur in a given time.
b. Raising the temperature raises the reaction rate
Express your reaction
Rate of motion = Change in position = x2-x1 = Δx
Change in time t2-t1 Δt
Reaction rate- change in concentration; reactant decreases while product
increase at set rate
Rate = - Δ[A]
Δt
[ ] represent concentration in moles per liter
Average , instantaneous, and initial reaction rates
Rate itself varies with time as the reaction proceeds
The instantaneous rate decreases as during the course of the reaction
The slope of straight line from two points on an x/y graph is the average
Initial rate- the instantaneous rate when the reaction occurs
aA+ bB cC+ dD
Rate= - 1 Δ[A] = - 1 Δ[B] = 1 Δ[C] = 1 Δ[D]
c Δt
b Δt
c Δt d Δt
16.3
Rate law and its components
rate law expresses rate as a function of reactant concentrations, product
concentrations, and temperature.
Rate depends only on reactant concentration and temperature
Rate= k[A]m[B]n
k is reaction constant
M and n are the reaction orders aka how the rate is affected by reactant
concentration
The component of the rate law- rate, reaction, order and rate constant must be
found through experiment
Reaction order terms
First order
Rate = k[A]
Second order
Rate = k[A]2
Zero order- doesn‘t depend on A
Rate = k{A}0
Example 2NO + 2H2  N2 + O2
Rate = k[NO]2[H2]
Second order in NO and first order in H2 overall it‘s a third order
Another example: CHCl3 + Cl2  CCl4+HCl
Rate= k[CHCl3][Cl2]1/2
Reaction orders experimentally
Run a series of experiments staring each with a different set of reactant
concentration and obtaining an initial rate for each case
Integrated rate law
Rate -Δ{A} =k[A]m
Δt
First order is Ln[A]0 =kt
[A]t
second order is 1 -1 =kt
[A]t [A]0
zero order is [A}t -[A]0 = -kt
half-life of rate law
half-life of first order is constant
ln 2
k
= half life
second order half life
t1/2= 1 .
k[A]0
zero order half life
t1/2= [A]0
2k
Effects of temperature on reaction
K increases as temperature increase
Use Arrhenius equation
K=Ae=Ea/RT
A is a constant
Ea is activation energy- minimum energy needed for a reaction
R is universal gas constant
T in temperature in K
Higher T larger k increase in rate
Rewrite the equation
Ln k2 = Ea(1 - 1 )
K1
R(T2 T1)
16.6
Explanation of effect of concentration and temperature
There are two theories that explain
1. collision theory-views the reaction rates as the result of parties colliding with
certain frequencies and minimal energy.
2. transition state theory- offers close up view of how energy of collision converts
from reactant to product
collision theory
measured in collisions per time
explains why reactant concentrations are multiplied together in the rate law
Why concentrations are multiplied
Because the different paths that a particle can take to get to the end result
Temperature affects the rate because collisions must have enough energy to
overcome the minimum amount needed in a reaction called activation energy
Two types of activation energy
1. Ea forward- energy difference between activated state and reactant.
2. Ea reverse- difference between activated energy and product
the smaller the Ea the larger the value of k, and the faster a reaction
larger Ea (or lower T)  smaller k decrease rate
effective collision- the molecules must collide so that the reacting atoms make
contact. Aka must have enough energy and particular molecular orientation
TRANSITION STATE THEORY
Explains why activation energy is needed and what the molecules look like.
- if the potential energy is less than the activation energy, molecules recoil,
like billiard balls.
- The kinetic energy pushes them together with enough force to overcome
repulsions and react
Transition state aka atiatied complex: a state that is not product or reactant; very
frail; highest potential energy
16.7 Reaction mechanisms: steps in the overall reaction
reaction mechanism- a sequence of single reaction steps that sum to an overall
reaction
A+B E +D
Reaction intermediate- a substance that is formed and used up during the overall
reaction
Elementary steps- a single molecular event, molecules decompose or collide
Unimolecular reaction
A  B +C
Bimolecular reaction
A +B  C
We use the equations coefficient as the reaction order in rate law for elementary
steps
Rate determining step: a step in a reaction slower than every other that affects
overall rate
Slow step determines rate
We can‘t prose from just data that particular mechanism represents actual
chemical change
16.8 Catalysis: speeding up a chemical reaction
catalysis: a substance that increases the rate without being consumed in the
reaction
catalysis  lower activation energyrate constant higher
homogeneous catalysis: exist in solution with reaction mixture
heterogeneous catalysis: speed up reaction that occur in separate stages
enzyme- a catalyst found in the human body, it is a protein
active site- a small region whose shape results from those of side chains of
amino acids, every enzyme has one
2 models of enzymes
1. lock and key – key(substance) lock is active site
2. induced fit – hand entering a glove
all enzymes function by binding to the reactions transition state and thus
stabilizing it
ozone reactions
O2uv 2O
O + O2O3 (ozone)
O + O3 2O2 (ozone breakdown)
Chapter 17 Equilibrium: the extent of chemical reactions
Focuses on
17.1 Equilibrium state & equilibrium constant
summary:
 kinetics and equilibrium are distinct aspects of a chemical reaction, thus
the rate extent of a reaction are related.
 When the forward and reverse reactions occur at the same rate, the
system has reached dynamic equilibrium.
 The equilibrium constant (K) is a number based on a particular ratio of
products and reactant concentrations: K is small for reaction that reach
equilibrium with a high concentration of reactant(s) and large for reactions
that reach equilibrium with a low concentration of reaction
17.2 reaction quotient & equilibrium constant
17.3 expressing equilibrium w/ pressure
 The reaction quotient and the equilibrium constant are most often
expressed in terms of concentration(Qc & Kc). for gases, they can also be
expressed in terms of partial pressure(Qp & Kp)
 The value of Kp and Kc are related by the derivation that relies on the
ideal gas law Kp=Kc(RT)Δngas
17.4 reaction direction: comparing Q and K
 We compare the value of Q and K to determine the direction I which a
reaction will proceed towards equilibrium
 If Qc<Kc, more product forms
 If Qc> Kc more reactant is formed
 If Qc=Kc there is no net charge
17.5 how to solve equilibrium problem
 In equilibrium problems, we typically use quantities( concentrations or
pressure) of reactants and products to find K, or we use K to find
quantities
 Reaction tables summarize the initial quantities, how they change and the
equilibrium quantities
 To simplify calculation, we assume that if K is small and the initial quantity
of reactant is large, the unknown change in reactant (x) can be neglected.
If this assumption is not justified( that is, if the error that results is greater
than 5%), we use the quadratic formula to find x
 To determine reaction direction, we compare the values of Q and K
17.6 reaction conditions : Le Chatelier‘s principle
 Le Chatelier principle states that if a system at equilibrium is disturbed, the
system undergoes a net reaction that reduces the disturbance and allows
equilibrium to be retained





Changes in concentration cause a net reaction away from the added
component or toward the removed component
For a reaction that involves a change in the number of moles of gas, an
increase in pressure(decrease in volume) causes a net reaction toward
fewer moles of gas, an increase in pressure(decrease in volume) cause a
net reaction towards fewer moles of gas and a decrease in pressure
causes the opposite change.
Although the equilibrium concentration of components changes as a result
of concentration and volume, K does not change. A temperature change,
however, does change K: higher T increase K for an endothermic reaction
(positive ΔHrxn) and decreases K for an exothermic reaction ( negative
ΔHrxn)
A catalyst causes a system to reach the equilibrium point more quickly
Ammonia production is favored by high pressure, low temperature, and
continual removal product. To make the process economical, an
intermediate temperature and a catalyst are used
17.1 equilibrium state and the equilibrium constant
equilibrium: reactant and product concentrations stop changing because the
forward reverse rates have become equal
ratefwd = raterev
no further change is observed at equilibrium because changes in one direction
are equal to changing in the opposite direction
Kfwd[N2O4]eq=krev[NO2]2eq
K =Kfwd = [NO2]2eq = product
Krev = [N2O4]eq reactant
K is the equilibrium constant- number equal to particular ratio of equilibrium
concentration of product and reactant at a particular temperature
1. small K – reaction yields very little product ; it looks like there is ‗no
reaction‘
2. Large K – reaction reaches equilibrium with very little reactant remaining; ―
goes to completion‖
3. Intermediate K – significant amount of both reactant and product are
present
17.2 reaction quotient and equilibrium constant
History: In 1864 two Norwegian chemist, Cato Goldberg & Peter Waage found
that at a given temperature a chemical system reaches a state in which a
particular ratio of reactant to product concentration has a constant value Aka law
of chemical equilibrium or law of mass action
Reaction quotient (Q) –aka mass-action expression
Q= [product]
[reactant]
@ equilibrium Q=K
note
1. ratio of initial concentration varies widely but always gives the same ratio
of equilibrium concentration
2. individual equilibrium concentrations are different in each case, but the
ratio of these concentrations is constant
reaction quotient (Q) is a ratio made up from product concentration terms
multiplied together and divided by reactant concentration terms multiplied
together.
aA +bB cC + dD
a , b , c, d are stoichiometry coefficients
Qc=[C]c[D]d
[A]a[B]b
To construct the reaction quotient you must write the balance equation first
Q & K are unitless numbers because it is a ratio
If the overall reaction is the sum of two or more reactions, the overall reaction
quotient is the product of the reaction quotients
Qoverall = Q1 *Q2 * Q3 *….
Koverall = K1 *K2 *K3 *…
The form of the reaction depends on the direction the balanced equation is
written.
Qc(fwd] = 1/Qc(rev) Kc(fwd) = 1/ Kc(rev)
If you multiply a reaction by a coefficient the Q must be raises to that power
If all coefficients of the balanced equation are multiplied by some factor, that
factor becomes the exponent for relating the reaction quotients and the
equilibrium constant.
N (aA +bB cC +dD)
Q‘=Qn=([C]c[D]d)n K‘=Kn
([A]a[B]b)n
heterogeneous equilibrium- components are in different phases
a pure solid and liquid always has the same concentration, same density
We eliminate the terms for pure liquids and solids from reaction quotient… they
do not change
17.3 expressing equilibria with pressure terms: relation between Kc & Kp
we start with the ideal gas law PV=nRT
if T is constant then pressure is directly proportional to molar concentration
reaction quotient based on partial pressures (Qp)
2NO(g) + O2(g)2NO2(g)
Qp = P2NO2
.
P2NO*PO2
Kp= equilibrium constant based upon pressure
Kp =/= Kc
But you can use change in moles of gasΔn to find
Kp = Kc(RT)Δn
17.4 reaction direction: comparing Q &K
more products make Q larger
more reactant makes Q smaller
3 types of relative size of Q and K
1. if Q< K reactant products
2. if Q>K, reactant  product
3. if Q=K, reactantproducts
17.5 How to solve equilibrium problems
Review:
-reactant and product concentrations are constant over time
-the forward reaction rate equals the reverse reaction rate
-the reaction quotient equals the equilibrium constant Q=K
Principles
- we are given equilibrium quantities and solve for K
- we are given K and initial quantities and some for equilibrium quantities
USE QUANTITES TO FIND EQUILBRUM CONSTANT
Example:
flask 1.5 Liters
H2(g) + I2(g) 2HI(g)
We have
At equilibrium 1.8 mol H2 1.8 mol I2 and .520 mol HI.
We find Kc through concentrations and substituting them into reaction quotient.
Qc = [HI]2
.
[H2][I2]
We first convert amount (mol) to concentration (mol/Liters) using the 1.5 liter
flask.
[H2]=1.8 mol = 1.2 M
1.5 L
We get [I2] =1.2 mol, & [HI] =.347M
Substitute values for Qc
Kc = (0.347)2 . = 8.36 *10-2
(1.2)(1.2)
Using a reaction table
CO2 + C(graphite)  2CO(g)
Qp= P2CO
PCO2
X atm CO22x atm CO
Pressure at equilibrium
PCO2(initial) –X =PCO2
Setup a table
Pressure (atm)
Initial
Change
Equilibrium
CO2 +
0.458
-x
0.458-x
C(graphite)
---------------------

0
+2x
2x
2CO(g)
Use the quadratic formula to solve if in doubt
If a reaction has a relatively small K and a relatively large initial reactant
concentration, the concentration change (x) can often be neglected, x does not
equal 0
[A]initial – [A]reacting = [A}eq =[A]initial
17.6 reaction conditions and the equilibrium state: Le Chatelier‘s principle
Le Chatelier Principle: when a chemical system at equilibrium is disturbed, it
retains equilibrium by undergoing a net reaction that reduces the effects of the
disturbance
Q does not equal K
― Disturbance‖
Three common are change in
1. concentration of component
a. if the concentration goes up, system reacts to consume some
i. equilibrium position moves right when reactant added
b. if the concentration does down, systems reacts to reduce some
i. equilibrium position move to the left when reactant reduced
The equilibrium system reacts to consume some of added substance or
produce some of the removed substance
New table
Concentration(M)
PCl3(g)
+
Cl2(g)
 PCl5(g)
Original equilibrium .2
.125
.6
Disturbance
0
+.075
0
New initial
.2
.2
.6
Change
-x
-x
+x
New equilibrium
.2-x
.2-x
.6+x
Kc stays the same
2. change in pressure(volume)
a. change in concentration of gas
b. adding an inert gas (gas does not take part)
i. no effect
c. changing the volume of vessel
i. decrease volume raises concentration
ii. increase volume decreases concentration
iii. does not alter Kc
iv. more volume = more product
v. more pressure leads to side with less moles gas
3. change in temperature
a. only way to alter K
b. temperature rises, K rises and ΔHrxnx is positive
c. temperature down, K down ΔHrxn negative
A catalyst shortens the time it takes to reach equilibrium but has no effect on the
equilibrium position.
Haber Process
N2(g) + 3H2(g)  2NH3(g)
ΔHrxn =-91.8Kj
1. Decrease concentration of ammonia. NH3 is the product, so removing it will
shift the equilibrium position towards producing more
2. Decreasing Volume(increase in pressure). Because 4 moles of gas react to
form two moles of gas, decrease volume will shift the equilibrium position
toward fewer moles of gas.
3. Decrease temperature. Because the formation of ammonia is exothermic,
decreasing the temperature will shift the equilibrium position towards
formation of product, thereby increasing Kc
Chapter 18 Acid base equilibria
18.1 acid bases in water
 In aqueous solution, water binds the proton released from an acid to form
the hydrated species represented by H3O+(aq)
 In Arrhenius definition , acids contain H and yield H3O+ in water, bases
contain OH and yield OH- in water, and acid-base reaction(neutralization)
is the reaction of H+ and OH- to form H2O
 Acid strength depends on [H3O] related to [HA] in aqueous solution.
Strong acids dissociate completely and weak acids slightly
 The extent of dissociation is expressed by the acid-dissociation constant
Ka. weak acids range from 10-1 to 10-12
 Many acids and bases can be classified qualitatively as strong or weak
based on their formulas
18.2 auto ionization of water and pH scale
 Pure water has a low conductivity because it auto ionizes to a small
extent. This process is described by an equilibrium reaction whose
equilibrium constant is the ion-product for water, Kw(10-14). Thus [H3O]
and [OH] are inversely related: in acidic solution, [H3O] is greater
than[OH]; the reverse is true in basic solutions; and the two are equal in a
neutral solution
 To express small values of [H3O]more simply, we use the pH scale (pH =
-log[H3O]). A high pH represents a low [H3O]. In acidic solution, pH <7.00;
I basic solutions, pH>7.00; and in neutral solutions pH=7.00. Similarly
pOH= -log[OH], and pK = -logK. The sum of pH and pOH equals pKw(14)
18.3 proton transfer and he Bronsted-Lowry acid- base definition
 The bronsted-lowry acid-base definition does not require that bases
contain OH or that acid-base reaction occur I aqueous solutions
 An acid is a species that donates a proton and a base is one that accepts
it


An acid and a base act together in proton transfer. When an acid donates
a proton, it becomes the conjugate base; when a base accepts a proton it
becomes the conjugate acid. I an acid-base reaction, acids and bases
form their conjugates. A stronger acid has a weaker conjugate base and
vice versa
An acid-base reaction proceeds in the net direction I which a stronger
acid and base form a weaker base and acid
18.4 solving problems involving weak-acid equilibra
 Two common types of weak- acid equilibrium problems involve Ka from a
concentration and find a concentration from Ka
 We summarize the information in a reaction table (ICE) and we simplify
the arithmetic by assuming [H3O ]from H2O is so small relative to [H3O]from HA
that it can be neglected and weak acids dissociate so little that
[HA]init=[HA]equilibrium
 The fraction of weak acid molecules that dissociates is greater in a more
dilute solution, even though the total [H3O] is less
 Polyprotic acids have more than one ionizable proton, but we assume
that the first dissociation provides virtually all H3O
18.5 weak bases and their relation to weak acids
 The extent a weak base accepts a proton from water to from OH is
expressed by base dissociation constant Kb
 Bronsted-lowry base include NH3 and amines and the anions of weak
acids. All produce basic solution by accepting H+ from water, which yields
OH and makes [H3O]<[OH}
 A solution of HA is acidic because [HA] >>[A], so [H3O]>>[OH].
 A solution of A is basic because [A]>>[HA] so [OH]>>[H3O]
 By multiplying the expression for Ka of HA and Kb of A we obtain Kw. this
relationship allows us to calculate either Ka of BH, the cation conjugate
acid of weak base B, or Kb of A, the anion conjugate base for weak acid
HA
18.6 molecular properties and acid strength
 The strength of an acid depends on the ease with which the ionizable
proton is released
 For nonmetal hydrides, acid strength increases across the period, with
the electronegativity of the nonmetal (E, and don a group, with the length
of E—H bond
 For oxoacids with the same number of O atoms, acid strength increases
with electronegativity of E; for oxoacids with the same E, acid strength
increases number of O
 Small, highly charged metal ions are acidic in water because they
withdraw electron density from the O-H bond of bound H2O molecules,
releasing an H ion to the solution
18.7 acid-base properties of salt solutions
 Salts that always yield a neutral solution consist of ions that do not react
with water
 Salt that always yield an acidic solution contain unreactive anion and a
cation that releases a proton to water
 Salts that always yield a basic solution contain an unreactive cation and
an anion that accepts protons from water
 If both cation and anion react with water, the ion that reacts to the greater
extent (higher K) determines the acidity or basicity of the salt solution
 If the anion is amphiprotic(first anion a polyprotic acid) the strength of the
anion as an acid (Ka) or a base (Kb) determines the acidity of the salt
solution
18.8 electron-pair donation and the Lewis acid-base definition
 The Lewis acid base definition focuses on the donation or acceptance of
an electron pair to form a new covalent bond in an adduct, the product of
an acid-base reaction. Lewis bases donate the electron pair, the Lewis
acids accept it. Many species that do not contain are Lewis acids.
 Molecules with polar double bonds act as Lewis acids, as do those with
electron deficient atoms.
 Metal ions act as Lewis Acids when they dissolve in water, which acts as
a Lewis base to from an adduct, a hydrated cation
 Many metal ions function as Lewis acids in biomolecules
18.1 acids and bases in water
Acids and bases react, each cancels the properties of the other in a process
called neutralization.
Water is a product of all reactions between strong acids and strong bases:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
An acid dissociates in water to make a hydronium molecule
HA + H2O A- + H3O
Arrhenius acid-bade definition- acids and bases are classified in terms of their
formula and their behavior in water
-
an acid is a substance that has H in its formula and dissociates into water
to yield H3O+
a Base is a substance that has OH in its formula and dissociates into
water to yield OH-
Acids contain covalently bonded H atoms that ionize when their molecules
dissolve in water.
The H+ ion and OH- ion form the base combine h2O
The heat of reaction is abut -57Kj per mole of water formed
Strong acids dissociate completely into water.
Weak acids dissociate slightly into water
Specific equilibrium (Ka)- weak acid in water
Kc[H2O]= Ka=[H3O+][A-]
[HA]
stronger acid higher[H3O+]larger Ka
smaller Ka lower dissociation of HA weaker acid
Acids
Strong acids
Hydrohalic acids
HCl, HBr, HI
Oxoacids- 2 more O‘s than H‘s
HNO3, H2SO4, HClO4
Weak acids- a lot
Hydrohalic acid
HF
H is not bonded to O or Halogens
HCN, H2S
Oxoacids- number of O‘s exceeds by 1 over H‘s
HClO, HNO2, H3PO4
Carboxylic acids- (RCOOH)
CH3COOH, C6H5COOH
Bases
Strong bases- water soluble compound containing O2- or OHGroup 1A(Li, Na, K, Rb, Cs)
M2O, MOH
Group 2A(Ca, Sr, Ba)
MO, M(OH)2
Weak bases- nitrogen with lone pair
Ammonia
NH3
Amines (RNH2, R2NH, R3N)
CH3CH2NH2, (CH3)2NH, (C3H7)3N
18.2 autoionization of water and the pH Scale
Water is an extremely weak electrolyte.
Water dissociates in equilibrium process called autoionization (self- ionizing)
Ion-product constant for water (Kwx)
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0*10-14 (at 25degrees C)
Pure waters concentration 55.5M
One H3O and one OH ion appear for each H2O molecule that dissociates
1. A change in{H3O] causes an inverse change in [OH]
Higher [H3O] lower [OH]
2.Both ions are present in all aqueous systems
In an acidic solution [H3O]>[OH]
In a basic solution
[H3O]<[OH]
In a neutral system
[H3O] = [OH]
[H3O]=Kw .
[OH]
pH scale
pH = -log[H3O+]
[H3O] can varies from 10M to 10-15M
[H3O] = 5.4*10-4M find pH
pH=-log[H3O+]= (-1) (log5.4 +log10-4)=3.27
the number of the significant figures in the concentration equals the number of
digits to the right of the decimal point in the pH value
The higher the pH, the lower the [H3O]
pH of an acidic solution <7.00
pH of a neutral solution = 7.00
pH of a basic solution >7.00
hydroxide ion concentration can be expressed as pOH
pOH = -log[OH]
acidic solutions have a higher pOH than basic solution
pK= -logK
a lower pK corresponds to a higher K
relationship between pH, pOH, pKw
Kw=[H3O][OH]=1.0*10-14
-logKw = (-log[H3O]) + (-log [OH]) = 1*10-14
pKw = pH + pOH= 14 @25degrees C
table 18.1 Relationship between Ka and pKa
Acid Name (formula)
Ka@25C
Hydrogen sulfate ion (HSO4 )
1.0*10-2
Nitrous acid (HNO2)
7.1*10-4
Acetic acid (CH3COOH)
1.8*10-5
Hypobromous acid (HBrO) 2.3*10-9
Phenol (C6H5OH)
1.0*10-10
pKa
1.99
3.15
4.74
8.64
10.00
In the lab we measure acidity with a pH indicator or a pH meter.
18.3 proton transfer and the Bronsted-lowry acid base definition
The Bronsted- Lowry acid-base definition took out the limitations of Arrhenius
definition
1. an acid is a proton donor, and species that donates a H+ ion
a. there must be a H in the formula
2. a base is proton acceptor, any species that accepts a H+ ion
a. a base must contains a lone pair of electron to bind the H
b.
the only requirement for an acid-base reaction is that one species donates a
proton another species accepts it; an acid-base reaction is a proton-transfer
process.
Typical acidic and basic solutions
1. acid donates a proton to watera. HCl + H2O  Cl- +H3O+
2. base accept a proton from water
a. NH3 + H2O  NH4+ + OHH2S+NH3  HS + NH4
Conjugate acid-base pair- HS is conjugate base of acid H2S
NH4 is the conjugate acid of the base NH3
Every acid has a conjugate base, every base has a conjugate acid
Table 18.2 the conjugate pairs in some acid-base reactions
_____________________conjugate pair__
|
|
acid +
Base

Base
|
------------------conjugate pair-------reaction1
reaction 2
reaction 3
reaction 4
reaction 5
reaction 6
HF
+
HCOOH+
NH4+ +
H2PO4- +
H2SO4 +
HPO42- +
H2O
CNCO32OHN2H5+
SO32-






FHCOONH3
HPO4-2
HSO4PO33-
+
Acid
|
+
+
+
+
+
+
H3O+
HCN+
HCOH2O
N2H6+2
HSO3-
A reaction proceeds to the greater extent in the direction in which a stronger acid
and stronger base forma weaker acid and weaker base
Strong  weak
Kc>1 if strong is on left
A weaker acid has a stronger conjugate base
An acid-base reaction proceeds to the right if the acid reacts with a base that is
lower on the list
18.4 solving problems involving weak-acid equilibria
2 general problems
1. given equilibrium concentration, find Ka
2. given Ka and some concentration information, find the other equilibrium
constants
- problem solving approach
o start with what is given in the problem, then apply steps
1. write the balanced equation an Ka expression; these will tell you
what to find
2. define x as the unknown change in concentration during reaction.
Through the use of certain assumptions, also equals [H3O+] and [A-]
at equilibrium
3. construct a reaction table that incorporates the unknown
4. make any assumptions that simplify calculation
a. [H3O] from autoionization is so much smaller [H3O] of
dissociation that it can be ignored
b. weak acid has a small Ka
5. substitute the value of Ka and solve for x
6. check the assumption are justified, test it
percent HA dissociated =[HA]dissoc *100
[HA]init
as the original acid concentration decreases, the percent dissociation of the acid
increases.
do not confuse the concentration of HA dissociated with eh percent of HA
dissociated
polyprotic acid- acids with more than one ionizable proton. Each time a proton
is taken away it gets a new Ka. Ka decreases as protons go
Ka1>Ka2>Ka3
18.5 Weak bases and their relation to weak acids
a base must have a lone pair
base-dissociation constant (Kb)-no base is dissociated during the process
B(aq) + H2O  BH (aq) +OH(aq)
Kb = [BH+][OH-]
[B]
ammonia is the simplest nitrogen-containing compound that acts as a weak base
in water
NH3(aq) + H2O  NH4(aq) +OH(aq)
kb=1.76*10-5
Kb=[NH4][OH]
[NH3]
Amine group: was a NH3 but H atom gets replaced by N group
RNH2, R2NH, R3N
Has a lone pair of electrons to bind proton
Rules
1. the acidity of HA(aq)
a. the equilibrium position moves to the left.
2. basicity of A-(aq)
a. the relative concentration of HA and A determine the acidity or
basicity of the solution
3. In a HA solution, [HA]>>[A-] & [H3O]from HA>>[OH]from H2O, so the solution is
acidic
4. In a A solution, [A]>>[HA]&[OH]from A>> [H3O]from the H2O, so the solution is
basic
HA + H2O  H3O +A
A + H2O  Ha +OH
2H2O  H3O + OH
the sum of the two dissociated reactions is the autoionization of water.
Ka * K b = K w
18.6 Molecular properties and acid strength
The strength of an acid depends on its ability to donate a proton.
Factors for nonmetal hydrides
1. cross a period, nonmetal hydride acid strength increase, because
electronegativity increases.
2. Down a group, nonmetal hydride acid strength increases, as E becomes
longer the E-H bond increases, decreasing the bong strength
a. H2O<H2S<H2Se<H2Te
Trends in acid strength in oxoacids
-All oxoacids have the acidic H atom bonded to an O atom
-2 factors determine acid strength
a. Electronegativity of central non metal
b. the number of O atoms
1. for oxoacids with the same number of O
around the e, acid strength increases across with electronegativity of E
2.for oxoacids with different number of oxygen around a given E, acid strength
increase with the number of O atoms.
a. HNO3> HNO2
acids of hydrated metal ions
hydrated ions transfer H+ to water.
M(NO3)n(s) +xH2O(l) M(H2O)xn+(aq) +nNO3(aq)
If M is small and highly charged
18.7 Acid-base properties of salt solutions
Salts that yield neutral solutions
A salt consist of the anion of a strong acid and the cation of a strong base yields
a neutral solution because the ions do not react with water
HNO3(l) +H2O NO3- (aq) + H3O(aq)
The anion of a strong acid is a weak base in water, a strong acid anion is
hydrated but nothing more
NaOH(s)h2o Na + Oh
All strong bases behave this way
Salt containing ions such as NaCl or Ba(NO3)2 yield neutral solutions because
no reaction takes place between the ions and water.
Salts the yield acidic solutions
-A salt consisting of the anion of a strong acid and the cation of a weak base
yield an acidic solution because the cation acts as a weak acid and the anion
does not react.
NH4Cl(s)H2O NH4 +Cl
{dissolution and hydration}
NH4 (aq) + H2O(l)  NH3 + H3O { dissociation of weak acid}
Small highly charged metal ions make up another group of cations that yield H3O
in solution
Fe(NO3)3 + 6H2O H2O Fe(H2O)6(aq) +3NO3 (dissociates and hydrates)
FE(H2O)6(AQ) + H2O  Fe(H2O)5OH2++H3O (dissociate of weak acid)
Salts that yield basic solutions
-a salt consist of the anion of a weak acid and the cation of a strong base yields a
basic solution in water because the anion acts as a weak base and the cation
does not react.
Salts of weak acid and weak base ions
Acidity depends on Ka and Kb
NH4 +H2O  NH3 + H3O
CN+H2O  HCN +OH
Ka of NH4 = Kw . = 1.0*10-14 = 5.7*10-10
Kbof HCN 1.76*10-5
Kb of CN = Kw . = 1.0*10-14 = 1.6*10-5
Ka of HCN 6.2*10-10
Solution is basics because Kb>Ka
Table 18.3 behavior of salts in water
Salt solution
pH
examples
Neutral:
7.0
Nacl,Kbr,Ba(NO3)2
Acidic:
NH4Cl, NH4NO3,
CH3NH3Br
<7.0
Acidic:
AL(NO3)3, CrBr3,
FeCl3
<7.0
Acidic/ basic
NH4ClO2, NH4CN,
Pb(CH3COO)2
<7.0 if
Ka(cation>Kb(anion)
>7.0 If
Kb(anion)> Ka(cation)
Acidic/ basic:
<7.0 if
NaH2PO4, KHCO3, Ka(anion)>Kb(cation)
NaHSO3
>7.0 if
Kb(cation)>Ka(anion)
Nature of ions
Cation of strong
acid
Anion of strong
base
Cation of weak
base
Anion of strong
acid
Small, highly
charged cation
Anion of strong
acid
Cation of weak
base
Anion of weak
acid
Cation of strong
base
Anion of polyprotic
acid
Ion that react with
water ex:
None
Cation:
NH4 + H2O
NH3 + H3O
Cation:
Al(H2O)6 +H2O
 Al(H2O)5OH
NH4 +H2O
NH3 +H3O
CN+ H2O 
HCN + OH
Anion:
HSO3 +H2O
SO3 + H3O
HSO3+ H2O
H2SO3+OH
18.8 electron-pair donation and the Lewis acid-base definition
definition:
- a base is any species that donates an electron pair
- an acid is any species that accepts an electron pair
compared to Bronsted- Lowry Lewis expands on classes of acid
the product of a Lewis acid-base reaction is called Adduct – a single species
that contains a new covalent bond
B: + H+ B-H+
\ covalent bond
Lewis acid has electron deficiency
- one surrounded by less than 8 electrons
Chapter 19 ionic equilibria in aqueous systems
19.1 equilibria of acid- base buffer systems
 a buffered solution exhibits a much smaller change in pH when H3O
or OH is added that does an unbuffered solution
 a buffer consists of relatively high concentrations of the
components of a conjugate weak acid-base pair. The buffer
component concentration ratio determines the pH and the ratio and
pH are related by the Henderson- hasselbalch equation. As H3O or
OH is added, one buffer component concentration ratio, and
consequently the free[H3O] (and pH), change only slightly
 a concentration buffer undergoes small changes in pH than a dilute
buffer. When the buffer pH equals the pKa of the acid component,
the buffer has its highest capacity
 a buffer has an effective range of pKa+- pH unit.
 To prepare a buffer choose the conjugate acid-base pair, calculate
the ratio of buffer components, determine buffer concentration, and
adjust the final solution to the desires pH
19.2 acid-base titration curves
 An acid-base (pH) indicator is a weak acid that has differently colored
acidic and basic forms and changes color over about 2pH unit
 In a strong acid-strong base titration, the pH starts out low, rises
slowly, then shoots up near the equivalence point (pH=7)
 In a weak acid- strong base titration, the pH starts out higher, than the
strong acid-strong base titration, rises slowly in the buffer region

(pH=pKa at the midpoint), then rises more quickly at the equivalence
point (pH>7)
A weak base-strong acid titration curve has a shape that is the inverse
of the weak acid-strong base curve, with the pH decreasing to the
equivalence point(pH<7)
19.3 equilibria of slightly soluble ionic compounds
 As an approximation, the dissolved portion of a slightly soluble salt
dissociates completely into ions
 In a saturated solution, the ions are in equilibrium with the solid and
the product of the ion concentrations, each raise to the power of its
subscript in the compound‘s formula, has a constant value (Qsp=Ksp)
 The value of Ksp can be obtained from the solubility , and vice versa
 Adding a common ion lowers an ionic compound‘s solubility
 Adding H3O (lowering pH) increases a compounds solubility if the
anion of the compound is that of a weak acid
 If Qsp> Ksp for an ionic compound, a precipitate forms when two
solution, each contains one of the compounds ions, are mixed
 Lakes bounded by limestone-rich soil form buffer systems that
prevent harmful acidification by acid rain
19.4 equilibria involving complex ions
 A complex ion consists of a central metal ion covalently bonded to
two or more negatively charges or neutral ligands. Its formation is
described by a formation constant( Kf)
 A hydrated metal ion is a complex ion with water molecules as
ligands, other ligands can displace the water in a stepwise process.
In most cases, the Kf value of each step is large, so the fully
substituted complex ion forms almost completely in the presence of
excess ligand
 Adding a solution containing a ligand increases the solubility of an
ionic precipitate if the cation forms a complex ion with the ligand
19.1 equilibria of acid base buffer systems
buffer- something that lessens the impact of an external force.
Acid-base buffer- a solution that lessens the impact of pH from the addiction
of acid or base.
Example: a shelter when it rains
The components of a buffer are the conjugate acid-base pair of a weak acid
(or base).
How a buffer works
Common-ion effect- it‘s how buffers work
Example : CH3COOH(aq) +H2O  H3O(aq) + CH3COO(aq)
From Le Chatelier principle equations shifts left if CH3COO is added
CH3COO is common ion in this case, it occurs when a given ion is added to an
equilibrium mixture that already contains that ion, and shifts the position away
from forming more of it.
Common ion suppresses the dissociation of CH3COOH, making the solution less
acidic.
%Dissociation = dissoc/init *100 [CH3COOH]
a Buffer consist of high concentrations of acidic (HA) and basic (A-) components
when H3O or OH are added small amounts of one buffer component to convert
into the other.
As long as the amount is small the added ions have little effect on the pH
because they are consumed by one or the other buffer components.
Ka= [CH3COO][H3O]  [H3O] = Ka *[CHCOOH]
[CH3COOH]
[ CHCOO]
[H3O] of solution depends directly on the buffer-component ratio
[CH3COOH]
[CH3COO]
The conversion of one component into the other produces a small change in the
buffer- component concentration ratio and consequently a small change in [H3O]
and in pH.
Henderson equation
Henderson-Hasselbalch equation
pH= pKa + log[base]/[acid]
can find [H3O] very quick
allows us to prepare a buffer of desired pH
Buffer capacity- is a measure of the ability to resist pH change and depends on
both the absolute and relative component concentration
The more concentrated the components of a buffer, the greater the buffer
capacity.
The pH of a buffer is distinct from its buffer capacity.
For the given addition of acid or bases, the buffer-component concentration ratio
changes less when the concentrations are similar than when they are different.
*a buffer has the highest capacity when the component concentrations are
equal
 [A]/[HA] =1
A buffer whose pH is equal to or near the pKa of its acid component has the
highest buffer capacity
buffer range- the pH range over which the buffer acts effectively, and its related
to the relative component concentration.
Buffers have a usable range with +- 1 pH unit of pKa of acid component
Preparing a buffer
Step 1. Decide on the conjugate acid-base pair. Based off desired pH.
pH is close to pKa
step 2. Find the ratio [A-]/[HA] that gives the desired pH, using the
Henderson Hasselbalch equation
pH = pKa +log[A]/[HA]
step 3. Choose the buffer concentration and calculate the amount to mix .
find the amount of other component using the buffer- component
concentration ratio
step 4. Mix the amounts together and adjust the buffer pH to the desired value.
Add small amounts of strong acids or strong base while monitoring the solution
with a pH meter
19.2 acid-base titration curves
we track the pH of titration with an acid –base titration curve
a plot of pH vs. volume of titrant added
monitoring pH with an indicator
what is an acid base indicator?
A weak organic acid (Denoted as Hln) that has different color than its
conjugate base (IN)
HIn + H2O H3O + In
Ka= [H3O][In]/[HIn]
[H3O]/Ka = [HIn]/[In]
An indicator changes color in units of 2pH
Strong acids strong base titration curve
Features of the curve
1. pH starts out low, shows high [H3O], goes up gradually as base is added
2. sudden, steep pH rise , dues to moles of OH equaling mole H3O
3. after steep there is an increase but slowly until equilibrium
equivalence point- number of moles OH is equal to number of mole H3O
at this point the solution consist of the anion of a strong acid and the cation
of a strong base
ions do not react with water at this point so the solution is neutral.
Before the titration begins , we add an indicator to acid to signal when
equivalence point.
The end point occurs when the indicator changes color
An indicator with an end point close to the equilibrium point
**** The visible change in color of indicator(end point) indicates point where
moles of base equal mole of acid originally.******
calculating the pH of a system
1. original solution of strong HA. pH=-log[H3O]
2. before the equivalence point. Initial amount of H3O – change= which equals
amount of OH added. Concentration [H3O]
3. at the equilibrium point.
Weak acid- strong base titration
features of the curve
1.the initial pH is higher , because weak acid dissociates slightly
2. a gradual rise portion of curve called the buffer region, appears before the
steep rise to the equivalence point, the pH equal the pKa at half the original HPr
reacted
3. the pH at the equivalence point is greater than 7.00
4. beyond the equivalent point the pH increases slowly as excess OH is added
19.3 equilibria of slightly soluble ionic compounds
ion-product expression(Qsp) and the solubility-product constant (Ksp)
equilibrium exists between solid solute and aqueous ions.
PbSO4 (s) Pb2+ (aq)+SO42-(aq)
Qc= [Pb2+][SO42-]/[PbSO4]solid
Ion-product expression Qsp= Qc[PbSO4] =[Pb2+][SO42-]
When [PbSO4] reaches equilibrium then we get Ksp solubility equilibrium
constant
Depends on the temperature not the concentration
Also equals the subscript of each ion in the compounds formula
Cu(OH)2 Cu2+ + 2OHKsp= [Cu][OH]2
Exceptions include insoluble metal sulfides . Sulfur ion is so basic it is not soluble
in water
MnS Mn +S
S + H2O  HS + OH
MnS + H2O  Mn + HS + OH
Ksp=[Mn][HS][OH]
The value of Ksp indicates how far to the right the dissolution proceeds at
equilibrium (saturation)
Calculations with Ksp
Most are in grams of solute dissolved in 100 grams H2O =100ml of solution
Then we convert grams per 100ml solution into molar solubility:
The amount of mol of solute dissolved per liter of solution.
Using Ksp values to compare solubilities
If compounds have the same total number of ions. Then the higher the Ksp the
greater the solubility
Effects of a common ion on solubility
The presence of a common ion decreases the solubility of a slightly soluble ionic
compound….. using le Chateliers principle
Effects of pH on solubility
If the compound contains the anion of a weak acid or CO3, addition of H3O(from
a strong acid) increases its solubility.
CaCO3(s) Ca(aq) +CO3(aq)
Add some strong acid , reacts with CO3 to form weak HCO3
CO3(aq) + H3O(aq) HCO (aq) + H2O(l)
More CaCO3 dissolves

CaCO3 Ca + CO3-H3O->HCO3-H3O-> H2CO3 CO2+H2O+Ca
Predicting the formation of a precipitate: Qsp vs. Ksp
Qsp=Ksp when the system is saturated
Qsp>Ksp precipitate forms until solution is saturated( need more water)
Qsp<Ksp Solution is unsaturated and no precipitate forms(can add more solid)
ACID RAIN
Substances involved
1. sulfurous acid(H2SO3)- coal burns to form sulfurous acid
a. H2O2+H2SO3 H2SO4 +H2O
2. sulfuric acid(H2SO4)- forms through atmosphere oxidation of SO2
3. nitric acid- from car engines
a. CO2+ 2H2O H3O +HCO3
Lakes bounded by limestone-rich soils form buffer system that prevent harmful
acidification by acid rain
19.4 equilibria involving complex
complex ion consist of central metal ion covalently bonded o two or more anions
or molecules called ligands
OH, Cl, CN,
all complex ions are Lewis adducts the metal ion acts as a Lewis acid
formation of complex ion
M(H2O)4(aq) + 4NH3(aq)  M(NH3)4(aq) + 4H2O(l)
Kc=[M(NH3)4][H2O]4/[M(H2O)4][NH3]4
formation constant (kf)= Kc/[H2O]=[M(NH4)]/[M(H2O)4][NH]4
Kf=Kf1*Kf2*Kf3…..
Complex ion and the solubility of precipitates
Ligand increases the solubility of a slightly soluble ionic compound if it
forms a complex ion with the cation
Chapter 20
Thermodynamics: entropy, free energy and the
direction of chemical reactions
20.1 the second law of thermodynamics: predicting spontaneous change
 A change in spontaneous under specified conditions if it occurs in a
 Neither the first law of thermodynamics nor the sign ΔH predicts the
direction of spontaneous change
 All spontaneous processes involve an increase in the dispersion of energy
 Entropy is a state function that measures the extant of energy dispersal
into the number of microstates possible for a system, which is related to
the freedom of motion of its particles
 The second law of thermodynamics states that, in a spontaneous process,
the entropy of the universe( system plus surroundings) increases
 Absolute entropy values can be found because perfect crystals have zero
entropy at 0K(third law)
 Standard molar entropy S°(j/mol*K) is affected by temperature, phase
changes, dissolution and atomic size or molecular complexity
20.2 calculating the change in entropy of a reaction
 The standard entropy of reaction, ΔSrxn, is calculated from S values.
 When the amount (mol) of gas (Δngas) increases in a reaction, usually
ΔSrxn>0
 The value of ΔSsurr is related directly to ΔHsys and inversely to T at which
the change occurs
 In a spontaneous change, the entropy of the system can decrease only if
the entropy of the surroundings increases even more
 For a system at equilibrium, ΔSuniv=0
20.3 entropy, free energy and work
 The sign of the free energy change ΔG = ΔH – TΔS, is directly related to
reaction spontaneity: a negative ΔG corresponds to a positive ΔS univ
 We use the standard free energy of formation ΔGf to calculate ΔGrxn at 25C
 The maximum work a system can do is never obtained from a real
(irreversible) process because some free energy is always converted to
heat
 The magnitude of T influences the spontaneity of temperature-dependent
reactions( same signs of ΔH &ΔS) by affecting the size of TΔS. For such
reactions, the T at which the reaction becomes spontaneous can be
estimated by setting ΔG=0
 A nonspontaneous reaction (ΔG >0) can be coupled to a more
spontaneous one (ΔG<<0) to make it occur. For example, in organisms,
the hydrolysis of ATP drives many reactions with a positive ΔG
20.4 free energy, equilibrium, and reaction direction
 Two ways of predicting reaction direction are from the value of ΔG and
from the relation of Q o K. these variables represent different aspects of
the same phenomenon and are related to each other by ΔG = RT ln Q/K.
When Q=K the system can release no more free energy
 Beginning with Q at the standard state, the free energy change is ΔG o, and
it is related to the equilibrium constant ΔGo = -RT ln K. For nonstandard
conditions, ΔG has two components:
ΔGo & RT ln K
 And non-equilibrium mixture of reactants and products move
spontaneously (ΔG<0) toward the equilibrium mixture. A product-favored
reaction goes predominantly towards products and has K>1 and ΔGo <0
20.1 the second law of thermodynamics: predicting spontaneous change
spontaneous change- occurs by itself under specified conditions, without a
continuous input of energy from outside the system,; can be physical, chemical or
change in position
example: burning something- just need a spark then the system takes over
pushing a book off a desk, a force to start but then system takes over
A non-spontaneous change occurs when the surrounding supply the system with
constant energy.
Example: picking a book up , constant supply of energy
Putting a fire out, constant supply of energy
*********A system can only be spontaneous in one direction, it cannot be
spontaneous in both**********
spontaneous is not instantaneous , it implies the system acting by itself
example: Ripening, rust
A chemical reaction proceeding towards equilibrium is an example of spontaneous
change
The directions the equation moves it’s the direction it is spontaneous
Limitations of first law of thermodynamics
First law (law of conservation of energy)- states that internal energy (E), sum of
kinetic and potential energy, changes when heat or work are gained or lost
ΔE=q+w
Eunivers=Esystem+Esurrounding
(q+w)system= - (q+w)surround
because total energy of the universe is conserved=0
the first law does not help tell what direction the system is going (no spontaneous)
the sign of ΔH cannot predict spontaneous change
freedom of particle motion and dispersal of particle energy
disperse- spread over more quantized energy levels
less freedom of particle motion more freedom of particle motion
localized energy of motion  dispersed energy of motion
phase change:
solid liquid  gas
dissolving of salt:
crystalline solid + liquid  ions in solution
chemical change:
crystalline solids  gases +ions in solution
************* in thermodynamic terms a change in freedom of motion of particles in
a system, that is, in the dispersal of their energy of motion, is a key factor
determining the direction of a spontaneous process**********
entropy and the number of microstates
what is a microstate?
All the quantized states of the whole system molecules
At any instance the energy is dispersed about a microstate
A second later the energy is dispersed about a different microstate
Number of microstate for a system of 1 mole is 10^10^23
Microstates with relevance to thermal energy is the number of ways it can disperse
its thermal energy amongst various modes of motion of all its molecules
Boltzmann equation relating microstate (W) with entropy(s) of a system
S=k ln(W)
K is Boltzmann constant= R/Na=1.38*10-23 J/K
 A system with fewer microstates (smaller W) has a lower Entropy(lower S)
 A system with more microstates (larger W) has a larger Entropy (larger S)
Lower entropy (fewer microstates) higher entropy (more microstates)
Change in entropy
ΔSsystem = ΔSfinal - ΔSinitial
Quantities meaning of entropy change
1 mole neon (initial:1L and 298K) 1 mole neon(final:2 L and 298K)
two approaches
1. approach based on number of microstates
a. use S= k ln(W)
b. W in this case is Wfinal/Winitial = 2NA NA is number of particles
c. ΔSsys =Sf – Si = k lnWf – k lnWi = k ln (Wf - Wi)
2. approach based on change in heat
a. ΔSsys = qrev/T
b. T = temperature
q = heat absorbed
c. Rev refers to reversible process,
Entropy and the second law of thermodynamics
The second law of thermodynamics- considering both the change in system and
its surroundings, all real processes occur spontaneously in the direction that
increases the entropy of the universe( system plus surroundings)
ΔSuniverse = ΔSsystem + ΔSsurrounding > 0
Standard molar entropies and the third law
-We can measure change in enthalpy but not absolute enthalpy
-We can measure the absolute entropy of a substance using the third law
Third law of thermodynamics- A perfect crystal has zero entropy at a temperature
of absolute 0
Ssys = 0 @ 0K
Standard molar entropy (S°) is at 1 molar
Predicting relative S° values of a system
1. temperature change (S° increases as temperature increases)
T(K):
273
295
298
S°:
31.0
32.9
33.2
2. physical state and phase change
S° increases as we go
Na
S°(s or l)
51.4(s)
S° (q)
153.6(g)
Solid liquid  gas
H2O
C(graphite)
69.9(l)
5.7(s)
188.7(g)
158
3. dissolving a solid or liquid
S° increases from ions to solution
NaCl
AlCl3
S°(s or l)
72.1 (s)
167(s)
S° (aq)
115.1
-148
4. dissolving a gas
gasliquid/solid
CH3OH
127(s)
132
S° goes down
5. atomic size or molecular complexity
bigger the atom more microstates more entropy
Li
Na
K
Atomic radius 152
186
227
Molar mass 6.941
22.99
39.1
S° (s)
29.1
51.4
64.7
Rb
248
85.47
69.5
Cs
265
132.9
85.2
***** trends hold true for substances in the same state***
****states dominate over trends******
chapter 20.2 calculating the change in entropy of a reaction
standard entropy of reaction ΔSrxn - the entropy change that occurs when all
reactants and products are in their standard states.
When the number of moles of gas increases, ΔSrxn is usually positive
When the number of moles of gas decreases, ΔSrxn is usually negative
* we cannot predict the sign of entropy change unless the reaction involves a
change in the number of moles of gas***
ΔSrxn = ΣmSproducts – ΣnSreactants
Chang in entropy in surrounding
-For a spontaneous process, decreases in the entropy of the system can occur only
increases in the entropy of the surroundings outweigh them. ( second law)
- the surrounding’s role is to either ass het to the system or remove heat from it.
1. exothermic change
heat loss by system ending in entropy of surrounding increasing.
Qsys <0
Qsurr>0
ΔSsurr >0
2. endothermic change
heat gained by system ending with entropy of surrounding decreasing
Qsys >0
Qsurr <0
ΔSsurr <0
Analogy
if you are a light weight in drinking and you take a shot to start. Then you take
another shot now you have drunk 100% more.
If you are a heavy weight and are about 8 deep and you take another shot, it will not
affect you as much because you already have a lot in your system
The change in entropy of the surroundings is directly related to an opposite change
in the heat of the system and inversely related to an opposite change in heat of the
system and inversely related to the temperature at which the heat is transferred.
ΔSsurr – ΔΗsys
T
The entropy change of forward reaction is equal in magnitude but opposite in sign
to the entropy change of the reverse reaction. When a system is in equilibrium,
neither the forward nor the reverse reaction is spontaneous, so no net reaction in
either direction
Summary of spontaneous exothermic and endothermic reaction
1. for an exothermic reaction ΔHsys < 0. He is released increasing freedom of
movement thus ΔSsurr >0
a. if the reaching system yields whose entropy is greater than reactants
ΔSsys >0 then total entropy is positive ΔSsys +ΔSsurr
b. if the entropy of system decreases ΔSsys<0 the entropy of surrounding
must increase even more ΔSsurr>> 0 making total ΔS positive
2. for an endothermic reaction ΔHsys >0 heat lost by the surroundings decreases
movement resulting in decreases of entropy of surroundings. ΔSsurr <0. If ΔSsys>>0
and can outweigh ΔSsurr then it is positive
20.3 entropy, free energy, and work
the Gibbs free energy aka free energy (G) is a function that combines the system’s
enthalpy and entropy
free energy change ΔG is a measure of the spontaneity of a process and of the useful
energy available from it.
ΔGsys = ΔHsys – TΔSsys
The sign of ΔG tells if a reaction is spontaneous
 ΔSuniv for a spontaneous process
 ΔSuniv <0 for non-spontaneous process
 ΔSuniv = 0 for a process at equilibrium
Standard free energy change (ΔG) : occurs when all components of the system are in
their standard states. Adapting the Gibbs equation we have
ΔGo =ΔHosys – TΔSosys
Standard free energy of formation ΔGof occurs when 1 mole of compound is made
from its elements
ΔGof vales of reactant and product to calculate ΔGorxn no matter how reaction occurs
ΔGorxn = ΣmΔGof(products) –ΣnΔGof(rectants)
* ΔGof of an element in its standard state are zero
* an equation coefficient (m or n above) multiplies ΔGof by that number
* reversing a reaction changes the sign of ΔGof
a spontaneous process (ΔG <0 ) at constant T, P, and ΔG is the maximum useful work
obtainable from the system as a process takes place
ΔG= wmax
A nonspontaneous process (ΔG>0) at constant T, P, and ΔG is the minimum work
done to the system to make it process take place
ΔG =wmin
The maximum work is done by a spontaneous process only if it is carried out
reversibly.
we can never obtain the max work
No real process uses all the available free energy to do work because some is always
changed to heat
Table 20.1 reactions spontaneity and the signs of ΔH, ΔS, and ΔG
ΔH
ΔS
-TΔS
ΔG
description
+
spontaneous at all T
+
+
+
non spontaneous at all T
+
+
+ or- spontaneous at higher T;
nonspontaneous at lower T
+
+ or- spontaneous at lower T;
nonspontaneous at higher T
explanation
most exothermic reactions are spontaneous
the temperature of a reaction influences the magnitude of the TΔS term, so, the
overall spontaneity depends on the temperature
-temperature-independent cases (ΔH & ΔS have opposite signs) the reactions occurs
spontaneously at all temperatures or none
1. reaction is spontaneous at all temperatures: ΔH<0, ΔS>0. So -TΔS is negative; thus
ΔG is always negative
2. reactions are nonspontaneous at all temperatures: ΔH>0; ΔS<0 so –TΔS is
positive; thus ΔG is always positive
-temperature-dependent cases (ΔH&ΔS have the same signs)
3. reactions are spontaneous at high temperature and non-spontaneous at lower
temperature. ΔH >0 &ΔS >0. –TΔS I negative and ΔG is either positive or negative
depending on the temperature
4. reactions are spontaneous at lower temperature and nonspontaneous at higher
temperature. ΔH<0 &ΔS<0, o –TΔS is going to be positive. Making ΔG either positive
or negative depending on the temperature.
ΔH=TΔS & T=ΔH
ΔS
A non-spontaneous step is driven by a spontaneous step when coupling of reactions.
One step supplies enough free energy for the other to occur
Example
Cu2O(s) 2Cu(s) +1/2O2(g)
Oxidation of carbon
C(s) +1/2O2  CO(g)
ΔG375=140kJ
ΔG375=-143.8kJ
Coupling these steps to spontaneous ones is a life-sustaining strategy that is
common to all organisms. A key spontaneous biochemical reaction is the hydrolysis
of a high-energy molecule called :adenosine triphosphate (ATP) to adenosine
diphosphate(ADP)
ATP4- + H2O ADP3- +HPO42- +H+
ΔGo` =-30.5 kJ
20.4 free energy, equilibrium, and reaction directions
Recall from chapter 17
If Q<K (Q/K <1) the reaction as written proceeds to the right
If Q>K (Q/K >1) the reaction as written proceeds to the left
If Q=K (Q/K =1) the reaction is at equilibrium, no net change
ΔG and Q/K are related
 If Q/K<1, then ln Q/K <0; reaction proceeds to the right (ΔG <0)


If Q/K >1, then ln Q/K >0; reaction proceeds to the left (ΔG>0)
If Q/K =1, then ln Q/K =0; reaction is at equilibrium (ΔG =0)
ΔG = RT ln Q/K = RT ln Q – RT ln K
When all constants are at 1 mol or 1 atm then ΔG equals ΔGo
ΔGo = -RT ln K
Table 20.2 The relationship between ΔGo and K @ 298K
ΔGo (kJ)
K
significant
-36
200
9 *10
essentially no forward reactions;
100
3 *10-18
reverse reaction goes to completion
-9
50
2*10
10
2 *10-2
1
7 *10-1
0
1
forward reaction and reverse reaction proceed
-1
1.5
to the same extent
-10
50
-50
6 *108
-100
3 *1017
forward reaction goes to completion;
-200
1*1035
essentially no reverse reaction
ΔG = ΔGo + RT ln Q
Case
Q. K so ΔG<0
The free energy decreases as the reaction proceeds, until it reaches a minimum at
the equilibrium mixture: Q=K and ΔG=0
So for A B
ΔGoB is smaller than ΔGoA so ΔGo is negative so K>1 aka
product favored
Chapter 21
Electrochemistry: chemical change and electrical work
21.1 Redox reactions and electrochemical cells
 An oxidation-reduction (redox) reaction involves transfer of electrons from
reducing agent to an oxidizing agent
 The half-reaction methods of balancing and dividing the overall reaction into
half-reactions that are balanced separately and then recombined
 The two types of electrochemical cells are based on redox reactions. In a
voltaic cell, a spontaneous reaction generates electricity and does work to
drive a nonspontaneous reaction. In both types, two electrodes dip into
electrolyte solutions; oxidation occurs at the anode, and the reduction occurs
at the cathode.
21.2 voltaic cells: using spontaneous reactions to generate electrical energy
 A voltaic cell consist of oxidation (anode) and reduction (cathode) half-cells,
connected by a wire to conduct electrons and a salt bridge to maintain charge
neutrality as the cell operates
 Electrons move from anode (left) to cathode (right), while cations move from
the salt bridge into the cathode half-cell and anions from the salt bride into
the anode half-cell
 The cell notation shows the species and their phases in each half-cell, as well
as the direction of current flow
21.3 Cell potential: output of a voltaic cell
 The output of a cell is called cell potential (Ecell) and is measured in volts
(1V= 1J/C)
 When all substances are in their standard states, the output is the standard
cell potential (Eocell). Eocell >0 for a spontaneous reaction at standard-state
conditions
 By convention, a standard electrode potential (Eohalf-cell) refers to the
reduction half-reaction. Eocell equals Eohalf-cell of the cathodes minus Eohalf-cell of
anode
 Using a standard hydrogen electrode, other Eohalf-cell values can be measured
and used to rank oxidizing (or reducing) agent
 Spontaneous redox reactions combine stronger oxidizing and reducing
agents to form weaker ones
 A metal can reduce another species (H+, H2O or even ions of other metal) if
Eocell for the reaction is positive
21.4 free energy and electrical work
 A spontaneous process is indicated by a negative ΔG or a positive Ecell, which
are related:ΔG =-nFEcell. The ΔG of the cell reaction represents the maximum
about of electrical work the cell can do.
 The standard free energy change, ΔGo is related to Eocell and to K, we can use
Eocell to determine K
 At nonstandard conditions the first?equation shows the Ecell depends on Eocell
and a correction term based on Q. Ecell is high when Q is small (high
[reactant]), and it decreases as the cell operates. At equilibrium, ΔG and Ecell
are zero, which means that Q=K
 Concentration cells have identical half-reactions, but solutions of differing
concentration; they generate electrical energy as the concentrations become
equal. Ion-specific electrodes, such as pH electrode, measure the
concentration of one species
21.5 electrochemical processes in batteries
 Batteries contain several voltaic cells in series and are classified as
primary(eg. Alkaline, mercury, and silver), secondary ( eg. Lead-acid, nickelmetal hydride, and lithium-ion, or fuel cell
 Supplying electricity to a rechargeable battery reserves the redox reaction,
forming more reactant
 Fuel cells generate a current through the controlled oxidation of a fuel such
as H2
21.6 corrosion: the case of environmental electrochemistry
 Corrosion damages metal structures through a natural electrochemical
change
 Iron corrosion occurs in the presence of oxygen and moisture and is
increased by a high [H+], high [ion], or contact with a less active metal, such

as Cu. Fe is oxidized and O2 is reducing in one redox reaction, while rust
(hydrated from Fe2O3) is formed in another reaction that often takes place at
different location
Because Fe functions as both anode and cathode in the process. An iron or
steel object can be protected by physically covering its surface or joining it to
a more active metal (such as Zn, Mg, or Al) which act as the anode in place of
Fe
21.7 electrolytic cells: using electrical energy to drive nonspontaneous reactions
 An electrolytic cell uses electrical energy to drive a nonspontaneous
reactions
 Oxidation occurs at the anode and reduction at the cathode, but the direction
of electron flow and the charges of the electrodes are opposite those in
voltaic cells
 When two products can form at each electrode, the more easily oxidized
substance reacts at the anode and the more easily reduced at the cathode.
 The reduction or oxidation of water takes place at nonstandard conditions
 Overvoltage causes the actual voltage to be unexpectedly high and can affect
electrode product that forms
 The industrial trial production of many elements, such as sodium, chlorine,
copper, and aluminum utilize electrolytic cells
 The amount of product that forms depends on the quantity of charge flowing
through the cell, which is related to the magnitude of the current and the
time it flows
21.1 redox reaction and electrochemical cells
electrochemistry is the study of the relationship between chemical change
and electrical work. It is typically investigated in electrochemical cells a
system that incorporates redox reactions
-electrochemical process always involves movement of electrons from one
chemical species to another through a redox reaction.
- oxidation is the loss of electrons
- reducing is the gain of electrons
-oxidizing agent is the species that reduces
- reducing agent is the species that oxidizes
After the reaction, the oxidized substances have a higher oxidation number
O.N., and the reducing substance has a lower one.



half-reaction method: divides the overall redox reaction into oxidation and
reduction half-reactions
reasons why to study electrochemistry
it separates the oxidation and reduction steps, which reflects their actual
physical separation in electrochemical cells
it is readily applied to redox reactions that take place in acidic or basic
solution, which is common in these cells
it does not require assigning O.N.s. ( in some cases when half-reactions are
not obvious, we assign O.N.s to determine which atoms undergo a change)
if the oxidizing form of a species is on the left side of the skeleton reaction, the
reducing form of that species must be on the right side, and vice versa
steps
step 1: divide skeleton structure into half-reaction, each contains oxidizing and
reduced forms of one species.
Step 2: balance the atoms and charges in each half-reaction
 atoms are balanced in order : atoms other than O and H, then O, and then H
 charge is balanced by adding electrons (e-). They are added to the left in the
reducing half-reaction because they are gains; they are added to the right in
the oxidation half-reactions because they are lost
step 3: if necessary, multiply one or both half-reactions by an integer to make the
number e- gained in the reduction equal the number lost in oxidation
step 4: add the balanced half-reactions, and include states of matter
step 5: check that the atoms and charges are balanced
balancing redox reaction in acidic solution
let’s balance the equation
Cr2O72- (aq) + I-(aq)  Cr3+ (aq) + I2(s) [acidic solution]
Step 1: divide the reaction into half-reactions
Cr2O72-  Cr3+
I-  I2
Step 2: balance atoms and charges in each half-reaction
Cr2O7/2Cr 3+ half reaction
a. balance atoms other than O and H.
Cr2O7 2Cr
b. balancing the O atoms by adding H2O molecules
Cr2O7—2Cr +7H2O
c. balance H atoms by adding H ions.
14H + Cr2O7 2Cr + 7 H2O
d. balance charges by adding electrons
6e- +14H+ + Cr2O72-  2Cr3+ + 7H2O
we can see the reaction is reducing because the electrons appear on the left
balancing the I-/I2 half reaction
a. balancing atoms other than O and H
2I- I2
b. balance O atoms with H2O : not needed
c. balance H atoms with H: not needed
d. balance charge with e2I-  I2 +2estep 3: multiply each half-reaction, if necessary, by an integer
3(2I-  I2 + 2e-)
6I-  3I2 + 6estep 4: adding the half-reaction together
6e- + 14 H+ + Cr2O72-  2Cr3+ +7H2O
6I-  3I2 + 6e6I(aq) + 14H+ + Cr2O7 2Cr3+ +7H2O + 3I2
step 5: check that atoms and charges balance
balancing redox reactions in basic solutions
both half reactions have been balanced as if they took place in acidic solution
we add one OH ion to both sides of the equation for every H+ ion present
Excess H2O molecules are cancelled
Half-reaction reveals a great deal and is essential to understanding electrochemistry
 any redox reaction can be treated as the sum of a reduction and an oxidation
half-reaction
 atoms and charge are conserved in each half-reaction
 electrons lost in one half-reaction are gained in the order

although the half-reactions are treated separately, electron los and electron
gain occur simultaneously
overview of electrochemical cells
1. voltaic cells( galvanic cell) – spontaneous reaction (ΔG <0) to generate
electricity
high level of energy to lower level of energy I converted to electric energy
System does work on surrounding
2. electrolytic cell – uses an electric energy to drive a nonspontaneous reaction
the surroundings do work on the system.
Two electrodes, which conduct the electricity between cell and surroundings, are
dipped into an electrolyte
An electrode is identified as either anode or cathode depending on the half-reaction
that takes place


the oxidation half-reaction occurs at the anode. Electrons are lost by the
substance being oxidized( reducing agent) and leave the cell at the anode
The reduction half-reaction occurs at the cathode. Electrons are gained by
the substance being reduced ( oxidizing agent) and enter the cell at the
cathode.
The relative charges of the electrodes are opposite in the two types of cells
Analogy:
1. the words anode and oxidation start with vowels; the words cathode and
reduction start with consonant
2. Alphabetically, the A in anode comes before the C in cathode, and the O in
oxidation comes before the R in reduction
3. Look at the first syllable and use your imagination
21.2 Voltaic cells: using spontaneous reactions to generate electrical energy
Cu2+ (aq)+ 2e- Cu (s)
[reduction]
2+
Zn(s) Zn (aq) + 2e
[oxidation]
2+
2+
Zn(s) + Cu (aq) Zn (aq) + Cu(s)
[overall reaction]
Construction and operation of a voltaic cell
In the situation above no electrical energy is generated because the oxidation agent
and the reduction agent are in the same beaker
If we split into half equations then electricity would be generates
The components of each half-reaction are placed in a separate container, or halfreaction, which consist of one electrode dipped into an electrolyte solution. The two
half cells are joined by a circuit which consist of a wire and a salt bridge
The oxidation half-cell(anode compartment) is shown on the left and the reduction
half-cell (cathode compartment) on the right. Here are the key points about the
Zn/Cu2+ voltaic cell:
1. the oxidation half-cell.
2. The reduction half-cell
3. Relative charges on the electrodes. The electrode charges are determines by
the source of electrons and the direction of electron flow through a circuit.
The electrons flow left to right through the wire to the cathode.
In any Voltaic cell the anode is negative and the cathode is positive
4. The purpose of a salt bridge.
To enable the cells to operate , the twohalf-cells are joined by a salt-bridge, which acts as a “liquid wire”
The circuit is completed as electrons move left to right through the
wire, while anions move right to left and cations move left to right through a
salt bridge
5. Active vs. inactive electrodes. The electrodes in Zn/Cu2+ cell are active
because the metal bars themselves are components of the half-reactions .
as the cell operates, the mass of the zinc electrode gradually decreases
and the[Zn2+ ] in the anode half-cell increases.
Inactive electrons are commonly rods of graphite or platinum: they conduct
electrons into or out of the half-cells but cannot take part in the half-reaction.
Notation for a voltaic cell
Short hand notation for Zn(s) + Cu2+(aq)  Zn2+ (aq)+ Cu(s)
Zn(s) | Zn2+ (aq)|| Cu 2+ (aq)| Cu(s)
Key parts of the notation
--The components of the anode compartment( oxidation half-cell) are written to the
left of the components of the cathode compartment (reduction half-cell)
-- a single vertical line represents a phase boundary. For example Zn(s) | Zn 2+ (aq)
indicates that the solid Zn is a different phase from the aqueous Zn2+ . A comma
represents half-call components in the same phase.
Example
Graphite| I-(aq) | I2(s) || H+ (aq), MnO4- (aq) , Mn2+ (aq) | graphite
--half-cell components usually appear in the same order as in the half-reaction, and
electrodes appear at the far left and far right of the notation
-- a double vertical line indicated the separated half-cells and represents the phase
boundary on either side of the salt bridge ( the ion in the salt bridge is omitted
because they are not part of the reaction)
21.3 Cell potential: output of a voltaic cell
The purpose of a voltaic cell is to convert the free energy charge of a spontaneous
reaction into kinetic energy of the electrons moving through an external circuit. This
electrical energy is proportionate to the difference in electrical potential between
the two electrodes, called Cell potential (Ecell) also the voltage or electromotive
force(emf)
Ecell > 0 for a spontaneous process
Ecell <0 for a nonspontaneous process
Ecell = 0 when at equilibrium
Volt(v) electrical potential
Coulomb (C) electrical charge
When 2 electrodes differ by 1 volt of electrical potential, 1 joule of energy is released
1 V = 1 J/C
standard cell potentials
in order to compare the output of different cells we obtain standard cell potential
(Eocell) – the potential measured at a specified temperature with no current flow
and all components in their standard state
Zn (s) + Cu2+(aq; 1M)  Zn2+ (aq,1M) +Cu(s) Eocell = 1.1V
Standard electrode potential (Eohalf-cell)- potential associated with a given halfreaction when all the components are in their standard states
A standard electrode potential always refers to the half-reaction written as a
reduction
(Eozinc, anode compartment)
the standard cell potential is the difference between the standard electrode
potential of the cathode( reduction) half-cell and the standard electrode potential of
the anode ( oxidation ) half-cell
Eocell = Eocathode(reduction) – Eoanode(oxidation)
Half-cell potentials, such as Eozinc and Eocopper are not absolute values but they are
referenced to the change . the standard reference half-cell has its standard electrode
potential defined as zero (Eoreference = 0.00V)
Standard reference half-cell is a standard hydrogen electrode consist of
specially prepared platinum electrode immersed in a 1M aqueous solution of strong
acid, H+ (aq)
Reaction :
2H+(aq;1M) +2e-  H2(g; 1atm) Eoreference = 0.00V
Relative strength of oxidation and reduction agents
And their oxidation strengths by writing each half-reaction as a gain of
electrons(reduction with corresponding standard electrical potential
Cu2+ (aq) +2e-  Cu(s)
Eo = 0.34V
2H+ +2e-  H2
Eo = 0.00V
Zn2+(aq) + 2e-  Zn(S)
Eo = -0.76V
The more positive value, the more readily the reaction occurs
Note**** the list of half-reaction is in order of decreasing half-strength
Eo





all the values are related to the standard hydrogen (reference electrode:
by convention, the half-reaction are written as reductions, which means that
only reactants re oxidizing agents and only products are reducing agents
the more positive the Eohalf-cell the more readily the half-reaction occurs
half-reactions are shown with an equilibrium arrow because each can occur
as a reduction or an oxidation depending on Eohalf-cell of the other halfreaction
F2 is the strongest oxidizing agent but the weakest reducing agent, and Li+ = is
the strongest reducing agent but the weakest oxidizing agent
Writing spontaneous redox reaction
Every redox reaction is the sum of two half-reactions, so there is a reducing agent
and an oxidizing agent on each side.
Zn(s) +
Cu2+(aq)

Zn2+(aq)
+
Cu(s)
Stronger reducing stronger oxidizing weaker oxidizing
weaker reducing
agent
agent
agent
agent
*** the stronger reducing and oxidizing agent react spontaneously to for the weaker
ones****
it’s like acids and bases the stronger goes to the weaker
The stronger oxidizing agent has a half-reaction with a larger Eo value, and the
stronger reducing agent has a half-reaction with a smaller Eo value
For a spontaneous reaction to occur, the half-reaction higher in the list at the
cathode as written and the half-reaction lower in the list proceeds at the anode in
reverse
Ag+ (aq) +2e-  Ag(s)
Eosilver = 0.80V
Sn2+ (aq) + 2e-  Sn
Eotin = -0.14V
Steps
1. reverse one of the half reactions into an oxidation step such that the
difference of the electrode potential gives a positive Eocell
***************when we reverse the half-reaction we need not reverse
the sign of Eohalf-cell will be settled by Eocell= Eocathode – Eoanode)*******
2. Add the rearranged half-reactions to obtain a balanced overall equation.
Don’t forget to multiply by coefficients so that e- lost equals e- attained
Half-reactions written in the correct direction must make sure the electrons lost are
equal to the electrons gained .
Very important******When we double the coefficient of the silver half-reaction to
balance the number of electrons, we do not double the Eo value it remains 0.80V. It
is an intensive property like density
Relative reactivity of metals
1. metals that can displace H2 from acid
Fe(s)  Fe2+ (aq) +2e2H+ (aq) +2e-  H2 (g)
Fe(s) + 2H+(aq)Fe2+ (aq) + H2(g)
Eo = -0.44V [anode]
Eo = 0.00V [cathode]
Eocell= 0.00V- (-0.44V)=.44V
Explanation
If Eocell for the reduction of H+ is more positive for metal A than it is for metal
B, metal A is a stronger reducing agent than metal B and a more active metal.
2. metals that cannot displace the H2 from acid
if a metal is above the standard hydrogen (reference) half-reaction then it
cannot reduce H+ from acid.
The Eocell would be negative making it nonspontaneous
3. metals that can displace H2 from water
these metal lie below the half-reaction of H2 so V < -0.42V
example:
2Na(s)  2Na+ (aq) + 2e- Eo = -2.71 V (anode)
2H2O (l) + 2e-  2H+ (g)+ 2OH-(aq)
Eo =-0.42V (cathode)
2Na(s) + 2 H2O(l)  2Na+(aq) + 2H- (g) + 2OH- (aq)
Eo= -0.42-(2.71)=2.29V
group 1a and larger alkine earth metals can reduce water
4. metals that can displace other metals from solution
a metal that is lower than another metal on the list
example
Zn(s)  Zn2+ (aq)+ 2eEo =-0.76V [anode; oxidize]
Fe2+ (aq)+ 2e-  Fe (s)
Eo = -0.44V [cathode]
Zn(s) + Fe2+ (aq) Zn2+(aq) + Fe(s)
Eo= -0.44 –(-0.76) = .32V
21.4 Free energy and electrical work
explanation
we examine the relationship between useful work, free energy, and equilibrium
constants in terms of electrochemistry and see the effects
standard cell potential and equilibrium constant
free energy change ΔG < 0 for a spontaneous reactions and electrochemical reaction
Ecell >0 for a spontaneous reaction
 the sign of ΔG and Ecell are opposite for spontaneous reaction*
ΔG α -Ecell
Ecell is the maximum voltage possible for the cell, the work is the maximum work
possible (wmax) * for work done on the surroundings, this quantity is negative *
Wmax = -Ecell X charge
The change in 1 mole of electrons is the faraday constant (F)
F = 9.65 X 104 J / V*mol eSubstitute for charge
ΔG = -nFEcell
When all are in standard state
ΔGo = -nFEocell
Substitute for ΔGo
Eocell = RT ln(K)
nF



substitute the know values of *.314 J/(mol rxnK) for constant R
substitute the known value of 9.65 104 J/ ( Vmol e-) for constant F
substitute standard temp of 298.15 K for T, but remember cell can be at diff
temp
 multiply by 2.303 to convert from ln to log
we get new equations
o
E cell = 0.0592V log (K)
or log(K) = nEocell at 298.15K
n
0.0592
effects of concentration on cell potential
relation between cell potential and concentration based on free energy and
concentration
by dividing both sides by –nF we get Nerest equation
Ecell = Eocell – RT ln(Q)
nF
* when Q< 1 and thus [Reactants] > [Products], lnQ <0 , so Ecell > Eocell
* when Q=1 and thus [Reactants] = [products] lnQ = 0, so Ecell = Eocell
* when Q >1 and thus [reactants]<[products] lnQ >0, so Ecell < Eocell
Ecell = Eocell – 0.0592V log(Q)
N
Note that Q contains only those species with concentration or pressure that can vary
Cs(s) + 2Ag+(aq)  Cd2+ (aq) +2 Ag(s) Q = [Cd2+] / [Ag+]2
Change in potential during operation
We go through stages
Stage 1. Ecell > Eocell when Q<1: so [reactants] > [products] , now more products are
going to form so Ecell decreases
Stage 2. Ecell = Eocell when Q=1 so [reactants] = [products] Q = 1
Stage 3. Ecell < Eocell when Q>1 so [reactants] < [products]
Stage 4. Ecell = 0 when Q= K term becomes large then equals Eocell
This occurs when system reaches equilibrium and no more free energy is released,
so the cell can do no more work, the battery is “dead”
Change in Ecell and concentration
Stage in cell
Q
operation
1. E>Eo
<1
2. E=Eo
=1
o
3. E < E
>1
E=0
=K
Relative [P] and
[R]
[P] <[R]
[P] = [R]
[P]> [R]
[P]>>[R]
0.0592V log(Q)
n
<0
=0
>0
=Eo
Concentration cells
When mixing concentrations of salt with dilute solutions of salt you get an
intermediate value . A concentration cell does the same thing but with electrical
energy
The two half-cells are separated but they become equal
In the concentration cell the half-reactions are the same but the concentrations are
different
Eocell is equal to 0 , but Ecell is not equal to 0 because it is depend on ratio of
concentrations
Cu(s)  Cu2+ (aq;.10M) +2e[anode]
2+
Cu (aq; 1M)+2e  Cu (s)
[cathode]
2+
2+
Cu (aq; 1M)  Cu (aq;0.1M)
Ecell =?
n=2
Ecell=Eocell -0.0592V log[Cu2+]dil = 0V – (0.0592V log0.1M
N
[Cu2+]con
2
1.0M
=.0296V
applications of concentration cells
we can find pH using Ecell
example
H2(g;1atm)  2H+ (aq; unknown) +2eanode
+
2H (aq;1M) +2e  H2(g;1atm)
cathode
+
+
2H (aq;1M)  2H (aq; unknown)
Ecell=?
Ecell is not 0 but E0cell is 0
Using Nernst equation
Ecell = Eocell -0.0592V log(H+)2unknown
N
[H+]2stand
Ecell = 0v- 0.0592V log (H+]2unknown = -[0.0592V 2 log H+unknown
2
1
2
Ecell = -0.0592 log[H+]unknown
-log[H] = pH so Ecell = 0.0592V pH
21.5 electrochemical processes in batteries
battery – a self-contained group of voltaic cells arranged in a series (plus minus
plus minus etc.) so the individual voltages are added together
3 types of batteries ( primary, secondary, fuel cells)
Primary (non- rechargeable) batteries
It is discarded when the components have reached equilibrium
Alkaline batteries
Has a zinc anode case that houses a MnO2 mixture and a paste of KOH and H2O
mercury and silver(button) batteries
found in watches, cameras, heart pacemakers, and hearing aids
have a zinc container as cathode and HgO as anode, the silver uses Ag2O can uses
steel can around cathode.
Secondary (rechargeable) batteries
Recharges by supplying electrical energy to reverse the cell reaction, in simpler
terms the cells flip when recharged
Example: car batteries
Lead acid battery
Overall reaction during discharge
PbO2(s) + Pb + 2H2SO4 (aq)  2PbSO4 (aq) + 2H2O (l)
Overall reaction of recharge
2PbSO4 + 2H2O  PbO2 + Pb + 2H2SO4
Ecell =2.1 V
Fuel cells
Sometimes called a flow battery, is not self-contained
Uses combustion to produce electricity
Does not burn because half-reactions are separated and electrons transfer through
an external circuit.
Most common is a proton exchange membrane (PEM) cell
Example:
2H2(g) 4H+ (aq) + 4 eanode
+
O2 + 4H (aq) +4e  2H2O (g)
cathode
2H2 (g)+ O2(g)  2H2O (g)
Ecell = 1.2V
lower rates than in other batteries
electrocataylist decreases the activation energy
seen in plane , the water and electricity
21.6 corrosion: a case of environmental electrochemistry
corrosion, the natural redox process that oxidizes metals to their oxides and sulfides
iron
25% of all steel is made to replace corroded steel
once irons loses electrons damage is done and pit starts to form
reaction
Fe(s) Fe2+ (aq) +2e-anode
O2(g) + 4H+(aq) + 2e-  2 H2O(l) cathode
2Fe(s) + O2(g) + 4H+(aq)  2Fe2+(aq) + 2H2O (l)
how to prevent corrosion
wash road salt off bodies
paint over iron so prevent oxidation
make iron act as a cathode because it only corrodes as an anode
21.7 electrolytic cells: using electrical energy to drive nonspontaneous reactions
voltaic cells are the exact opposite of electrolytic cells
electrolytic cells use electric energy from an external source to drive a nonspontaneous reaction
oxidation takes place at the anode and reduction takes place at the cathode, but the
direction of electron flow and the signs of electrodes are reversed
* in a voltaic cell, electrons are generated at the anode, so it is negative and electrons
are consumed at the cathode, so it’s positive
* in an electrolytic cell, electrons come from the external power source, which
supplies them to the cathode, so it is negative, and removes them from the anode, so
it’s positive
comparison of voltaic cells and electrolytic cells
Cell type
Voltaic
Voltaic
Electrolytic
Electrolytic
ΔG
<0
<0
>0
>0
Ecell
>0
>0
<0
<0
Name
Anode
Cathode
Anode
Cathode
Process
Oxidation
Reduction
Oxidation
Reduction
Sign
+
+
-
Electrolysis – the splitting of a substance by the input of electrical energy, often
used to decompose a compound into elements
Electrolysis of molten salts and the industrial production of sodium
Predicting the product at each electrode is simple if the salt is pure because the
cation will be reduced and the anion oxidized
Reaction
2Cl- (l) Cl2(g) +2eCa2+(l) +2e- Ca(s)
The dry solid i crushed and fused (melted) in an electrolytic apparatus called downs
cells
Electrolysis of water
Extremely pure water is difficult to electrolyze because very few ions are present to
conduct a current
Equation
2H2O(l)  O2(g) + 4H+(aq) +4eAnode
E=0.82V
2H2O(l)+2e-  H2(g) + 2OH-(aq) cathode
E=-0.42V
2H20(l)  2H2(g) +O2(g)
total equation
Ecell = -1.24V
* these electrode potentials are not written with a degree sign because they are not
standard electrode potentials*
Electrolysis of aqueous salt solutions; overvoltage and the chlori-alkali process
Two half-reactions are possible
* the reduction with the less negative (more positive) electrode potential occurs
* the oxidation with the less positive( more negative) electrode potential occurs
electrolyzing KI solution
case 1
K+ (aq) +e-K(s)
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
Eo = -2.93V
E=-0.42 V [reduction]
case 2
2I-(aq) I2(s) +2e-
Eo = 0.53V
2H2O(l)  O2 (g)+ 4H+ +4e-
E=0.82V
[reduction]
the increment above the expected required voltage is called overvoltage
overvoltage plays a key role in chlor-alkli process for industrial production of
chlorine
electrolytic oxidation of Cl- ion from concentrated aqueous NaCl solution
the asbestos diagram separates the anode and cathode compartments
Must go through the polymeric membrane
Membrane allows cations to move through and only from anode to cathode
compartment
Rules from aqueous solutions
1. cations of less active metals are reduced to the metal, including gold, silver,
copper, chromium, platinum, and cadmium
2. cations of more active metals are not reduced, including those of group 1A
and 2A and Al. Water is reduced to H2 and OH- instead
3. anions that are oxidized, because of overvoltage from O2 formation, include
the halides([Cl] must be high], except F
4. anions are not oxidized including F- and common oxoanions, such as SO42-,
CO32-,NO3-, PO43-, because the central nonmetal in these oxoanions is already
in its highest oxidation state. Water is oxidized into O2 and H+ instead
purification is accomplished by electrorefining, which involves the oxidation of Cu to
form Cu2+ ions in solution
1. copper and the more active impurities are oxidized to their cations, while the
less active ones are not.
2. Because Cu is much less active than Fe Ni impurities, Cu ions are reduced at
the cathode, but Fe and Ni ions remain to solution:
Cu2+(aq) +2e-  cu(s) Eo =0.34 V
Ni2+ (aq)+ 2e-  Ni(s) Eo= -0.25
Fe2+ (aq)+2e-  Fe (s) Eo=-0.44V
Copper obtained by electrorefining is over 99.99% pure
How to isolate aluminum
Bauxite, a mixed oxide-hydroxide that is the major ore of aluminum is used
Step 1. The mineral oxide, Al2O3 is separated from bauxite
Step 2. Oxide is converted to metal
Al2O3 electrolyzed at 2030 oC so we have to lower than by using molten cryolite
(Na3AlF6) to electrolyte around 1000oC
This process is called the hall-heroult process, and takes place in a graphite furnace
Example
2Al2O3 + 1AlF6 – 3Al2O2F42- (l)
cathode reduction
AlF62-(l) +3e-  Al(l) + 6F-(l) [cathode reduction]
Anode oxidation
Al2O2F42-(l) + 8F-(l) +C (graphite)  2AlF63- (l)+ CO2 (g)+ 4e- [anode; oxidation]
Anodes are consumed in half reactions fast and must be replaced quickly
Aluminum production accounts for more than 5% of all electricity used in America
The stoichiometry of electrolysis: the relationship between amounts and charge of
product
Faraday’s law of electrolysis: the amount of substance produced at each electrode is
directly proportional to the quantity of charge flowing through the cell
Questions asked
How much material will form a result of given quantity of charge?
How much charge us needed to produce a given amount of material?
Use faradays law
1. balance the half-reactions to find the number of moles of electrons
needed per mole of product
2. use the faraday constant (F = 9.65 *104 C/mol e-) to find the
corresponding charge
3. use the molar mass to find the charge needed for a given mass of
product
we measure current, the charge of flowing per unit time
the Si unit is ampere(A) = 1 coulomb flowing through a conductor in 1
second
1 Ampere = 1 coulomb/second
or
1A= 1C/s
we find the charge by measuring the current and the time during which
the current flows
Mass(g)
Of
substanc
e
oxidized
or
reduced

M(g/mol)
Amount(
mol) of
substanc
e
oxidized
or
reduced
Current
(A)

balanced
HalfReaction
Amount
(mol) of
electrons
transferr
ed

Faraday
constant
(C/mol e-)
Charge
(C)

time(s)
here is an example problem
How long does it take to produce 3grams of Cl(g) by electrolysis of
aqueous NaCl using a power supply with a current of 12A?
We know the mass of Cl2 produced, so we can find the amount of mol of
Cl2
The half-reaction tells us the loss of 2 moles of electrons produced 1
mole of chlorine gas
2Cl- (aq) Cl2(g) + 2ewe use this as a conversion factor in the faraday equation
Charge (C) = 3.0 g Cl2 * 1 mole Cl2 * 2 mole e- * 9.65*104C = 8.2*103C
70.90g Cl2
1 mol Cl2 1 mole ewe use the relationship between charge and current
Time (s) = Charge (C) =8.2 *103 C * 1s = 6.8 *102s(11min)
Current (A, or C/s)
12C
Grams Cl2  moles of Cl2  moles of e-  coulombs  seconds