Document 6513582

HOW TO PREDICT WHETHER A REDOX REACTION WILL BE SPONTANEOUS: Looking at the table in the data booklet on page 8, ↑ INCREASING INCREASING ⏐ ⏐ TENDENCY TENDENCY TO ⏐ ⏐ TO REDUCE OXIDIZE ⏐ ⏐ = = ⏐ ⏐ ⏐ ⏐
INCREASING STRENGTH INCREASING STRENGTH ⏐ AS OXIDIZING AGENT AS REDUCING AGENT ↓ notes: 1. In general, metals are found on bottom half of table on the right (except for Cu, Ag, Hg, Au) 2. The halogens and oxyanions are on the top half of the table on the left. 3. Some metals have more than one common oxidation number, and therefore there will be more than one half reaction on the table. This means that can be found as oxidizing agents or reducing agents. (eg. Fe, Cu) 4. H2O2 at top left (strong oxidizing agent) and mid right (weaker reducing agent) 5. Use this table as you did the acid/base table on p. 6: rxns at the top have a tendency to go right and rxns at the bottom have a tendency to go left. 6. Each of the rxns can be written either forward or backward: Zn+2 + 2e  Zn OR Zn  Zn+2 + 2e 7. Use double arrows to show how the rxn could go either forwards or backwards eg. Zn+2 + 2e ⇄ Zn 8. Use single arrows when showing the half reaction of a redox reaction. TO DETERMINE IF A REACTION ACTUALLY HAPPENS: 1. 2. 3. a) b) Locate each reactant on the table. If both reactants are only found on the left or only on the right, then NO REACTION IS POSSIBLE. eg. Given the reactants Zn and Cu: Zn+2 + 2e  Zn +2
Cu + 2e  Cu Both Zn and Cu can only oxidize (go backwards from the reduction rxns written on the table) therefore there is no species present that can reduce, therefore NO REACTION IS POSSIBLE. LIKEWISE: eg. Given the reactants Zn+2 and Cu+2: Zn+2 + 2e  Zn Cu+2 + 2e  Cu +2
+2
Both Zn and Cu can only reduce (go forwards from the reduction rxns written on the table) therefore there is no species present that can oxidize, therefore NO REACTION IS POSSIBLE. If one reactant is on the left hand side of the table, and the other is on the right, consider two possible scenarios: The reactant to be reduced (left) is higher on the table than the reactant to be oxidized (right). A SPONTANEOUS REACTION OCCURS eg. using the four reactants used in #2, can you write what this spontaneous rxn would look like? RED: Cu+2 + 2e  Cu +2
OXID: Zn + 2e  Zn +2
+2
REDOX: Cu + Zn  Zn + Cu THE REACTION IS SPONTANEOUS AND GOES 100% TO COMPLETION. The reactant to be reduced (left) is lower on the table than the reactant to be oxidized (right). NO REACTION OCCURS Cu+2 + 2e  Cu +2
Zn + 2e  Zn Since this is the exact reverse of case b, it makes sense that it does not occur since we said that: REDOX: Cu+2 + Zn  Zn+2 + Cu proceeds 100% to the right. SUMMARY: a reaction will be spontaneous if and only if there is a reactant to be reduced (left side) that is ABOVE the reactant to be oxidized (right side) EXAMPLES: 1. Which of the following can be BOTH oxidized AND reduced? Na+ I-­‐ Cu+ For the next two questions, CLASSIFY AS SPONTANEOUS OR NO REACTION: (if spontaneous, write the overall balanced redox equation) 2. Ni+2 and Ag (s) 3. Zn+2 and Li (s) 4. Which is the stronger oxidizing agent? Zn+2 or Ca+2 5. Which is the stronger reducing agent? Cr+2 or Fe+2 6. Given the following four half reactions: D+2 + 2e ⇄ D(s) E+2 + 2e ⇄ E(s) F+2 + 2e ⇄ F(s) G+2 + 2e ⇄ G(s) It was found experimentally that: o
o
o
F+2 reacts with D(s), E(s), and G(s) no rxn occurs between D+2 and any of the metals G+2 only reacts with D(s) Arrange the half reactions in order of decreasing strength as oxidizing agents. IMPORTANT! READ ALL OF CHAPTER 11 AS IT CLEARLY GOES THROUGH REDOX AND BALANCING ANSWERS: 1. Na+ reduced only I-­‐ oxidized only Cu+ BOTH 2. no reaction: the species to be reduced (Ni+2) is below the species to be oxidized (Ag) on the table. 3. spontaneous: 2Li + Zn+2  2Li+ + Zn 4. Zn+2 is higher on the table 5. Cr+2 is lower on the table 6. decreasing strength as oxidizing agents = start with greatest strength first Put the ion with the greatest tendency to reduce first Due to: It was found experimentally that F+2 reacts with D(s), E(s), and G(s) Then the greatest tendency to reduce is: F+2 Due to: no rxn occurs between D+2 and any of the metals Then the least tendency to reduce is: D+2 Due to: G+2 only reacts with D(s) Then G+2 is only above D+2 F+2 + 2e ⇄ F(s) ⇄ E+2 + 2e G+2 + 2e ⇄ G(s) D+2 + 2e ⇄ D(s) E(s) You have just created your own table of reduction half reactions. IMPORTANT! READ ALL OF CHAPTER 11 AS IT CLEARLY GOES THROUGH REDOX AND BALANCING ELECTROCHEMICAL CELLS Recall: an electrolyte is a substance whose aqueous solution conducts a current. A redox reaction is a chemical change in which two species exchange electrons Electrochemical cell: -­‐ a redox reaction in which electricity is produced -­‐that is, a potential difference between each of the half cells exists An electrode is the conductor at which a half reaction occurs An anode is the electrode to which the anions go A cathode is the electrode to which the cations go -­‐oxidation occurs at the anode half cell (anions go to the anode; anions are the ions o and a are both vowels which are capable of losing electrons -­‐ oxidation is loss of electrons) -­‐reduction occurs at the cathode half cell (cations go to the cathode; cations are the ions r and c are both consonants which are capable of gaining electrons -­‐ reduction is gain of electrons) IMPORTANT! READ ALL OF CHAPTER 12 AS IT CLEARLY GOES THROUGH ELECTROCHEMICAL AND ELECTROLYTIC CELLS AND APPLICATIONS Eg. a wet cell battery: Half reaction at the anode (oxidation)
Zn o (s)  Zn +2 (aq)
single line
in half rxn
equations
+ 2e
According to booklet p. 8
Eo = +0.76 V
because equation is backwards
from p. 8
Half reaction at the cathode (reduction)
2Ag + (aq) + 2e
 2Ag o (s)
single line
in half rxn
equations
According to booklet p. 8
Eo = +0.80 V
do not multiply X 2!!!
Net Redox reaction for this cell:
Zn o (s) +
Eo cell =
2Ag + (aq)
+0.76 V

+
Zn +2 (aq) + 2Ag o (s)
+0.80V =
+1.56 V
Shorthand method for writing this cell:
Zn o (s)
Zn +2 (aq) (1M)
ANODE (oxidation)
0 to +2
Ag + (aq) (1M)
salt
bridge
Ag o (s)
CATHODE (reduction)
+1 to 0
Eo =+1.56 V
Why do electrons flow in a certain direction??
Before connecting each half cell -­‐there is a small tendency for the metals to oxidize and produce positive ions -­‐any electrons given off remain behind on the metal -­‐Zn has a greater tendency to oxidize than Ag (see table p. 8) OR IN OTHER WORDS: -­‐certain metals hold their electrons very strongly -­‐in the above reaction the silver ions have a greater electron attracting ability (tendency to reduce -­‐ gain electrons) than the Zinc ions (Ag+ is higher on the table of reduction potentials on p. 8) -­‐certain metallic ions attract electrons more strongly than others. After the half cells are connected -­‐An excess of electrons has accumulated on the Zn electrode (because of last statement above) -­‐this causes electrons to flow from the Zn to the Ag electrode. -­‐as a result, the Zn oxidizes (loses electrons) and Zn becomes the ANODE. OR IN OTHER WORDS: -­‐In this example the electrons are being supplied to the Ag o (s) by the Zn o (s) -­‐The Ag + (aq) ions have a greater electron attracting ability than the Zn +2 (aq) ions. -­‐Ag + (aq) ions can take electrons from a piece of Zn o (s) but Zn +2 (aq) ions can not take electrons from a piece of Ag o (s) (Ag+ is higher on the the table of reduction potentials on p. 8 than Zn+2 is -­‐ relates to last day's notes -­‐ Any oxidizing agent on the left will react with a reducing agent on the right that is LOWER on the list.) ELECTRONS FLOW FROM THE ANODE TO THE CATHODE A Voltmeter can be attached to the wire to measure the direction of electron flow The salt bridge -­‐Water and certain ions are able to pass through the salt bridge -­‐the salt bridge allows the NO3-­‐ ions to move toward the Zn/Zn+2 half cell -­‐this beaker tends to become more positive as the Zn+2 (aq) ions are produced in solution. -­‐the K+ ions in the salt bridge move to the Ag/Ag+ half cell -­‐This solution becomes less positive or more negative as Ag+(aq) ions are reduced to Ago(s) Without the salt bridge: Zn+2 (aq) beaker becomes more and more positive Ag beaker becomes less and less positive Therefore the reaction stops almost immediately (<1s) unless the salt bridge is present to even out the charge The tendency of electrons to flow in an electrochemical cell is called the VOLTAGE or ELECTRICAL POTENTIAL to do work. The VOLTAGE is the work done per electron transferred. STANDARD HYDROGEN HALF CELL: OXIDATION: (anode) REDUCTION: (cathode) REDOX:  2H+ (aq) + 2e Eo= 0.0V (a zero point is arbitrarily defined for the H half cell) 2Ag + (aq) + 2e  2Ag o (s) Eo = +0.80 V H2 (g) H2 (g) + 2Ag + (aq)  2H+ (aq) + 2Ag o (s) Eo = +0.80 V STANDARD STATE CONDITIONS for an electrochemical cell: 25oC 101.3 kPa (all gases) all elements in standard states (normal phase for 25oC -­‐ see pink and blue periodic table which tells you when elements are s, l. g at room temp.) all solution involved in the half cell are 1 M solutions The STANDARD REDUCTION POTENTIAL Eo is measured in volts and implies that this is a standard state. DETERMINING THE Eo FOR A PARTICULAR REACTION AND DETERMINING WHETHER THE REACTION IS SPONTANEOUS TO THE RIGHT OR LEFT Consider the reaction: Cuo(s) + 2Ag+(aq) ⇄ Cu+2 (aq) + 2Ago (s) OXIDATION HALF RXN Cuo(s)  Cu+2 (aq) + 2e (anode) REDUCTION HALF RXN 2Ag+(aq) + 2e  2Ago (s) (cathode) REDOX: Cuo(s) + 2Ag+(aq) ⇄ Cu+2 (aq) + 2Ago (s) Eo = -­‐0.34V Eo = +0.80V Eo = +0.46V Since Eo rxn or Eo cell is positive, then the reaction is spontaneous to the right.
The more positive Eo cell is, the more the reaction goes to the right.
An Eo rxn or Eo cell of ) means the system is at equilibrium. (the rate of the forward reaction = the rate of
the reverse reaction).
A negative Eo rxn or Eo cell means the reaction proceeds at a faster rate to the left. (spontaneous to the left
- we would call the forward reaction NON SPONTANEOUS).
Consider our original Ag/ Zn example. Given time, the voltage drops, approaching a value of 0. It is then at equilibrium and there no longer a net electron flow. Therefore there is no potential difference and therefore no useful work can be done. With time the [Ag+] decreases and the [Zn+2] increases, and according to LeChatelier's Principle: Zn o (s) + 2Ag + (aq) ⇄ Zn +2 (aq) + 2Ag o (s) Forward rxn decreases and reverse reaction increases until both are equal -­‐ i.e. at a state of equilbrium. DETERMINING THE Eo FOR A PARTICULAR REACTION AND DETERMINING WHETHER THE REACTION IS SPONTANEOUS TO THE RIGHT OR LEFT Consider the reaction: Cuo(s) + 2Ag+(aq) ⇄ Cu+2 (aq) + 2Ago (s) OXIDATION HALF RXN Eo = V (anode) REDUCTION HALF RXN Eo = V (cathode) REDOX: Eo = V If Eo rxn or Eo cell is positive, then the reaction is spontaneous to the right. The more positive Eo cell is, the more the reaction goes to the right. An Eo rxn or Eo cell of 0 means the system is at equilibrium. (the rate of the forward reaction = the rate of the reverse reaction). A negative Eo rxn or Eo cell means the reaction proceeds at a faster rate to the left. (spontaneous to the left-­‐ we would call the forward reaction NON SPONTANEOUS). Consider our original Ag/ Zn example. Given time, the voltage drops, approaching a value of 0. It is then at equilibrium and there no longer a net electron flow. Therefore there is no potential difference and therefore no useful work can be done. With time the [Ag+] decreases and the [Zn+2] increases, and according to LeChatelier's Principle: Zn o (s) + 2Ag + (aq) ⇄ Zn +2 (aq) + 2Ag o (s) Forward rxn decreases and reverse reaction increases until both are equal -­‐ i.e. at a state of equilbrium. SELECTING PREFERRED REACTIONS -HELPS US TELL WHICH ELECTROLYTE IS OXIDIZED AND REDUCED IN LAB 5.1
1)
List the possible half reactions with corresponding reduction potential values (Eo)
2)
When several different reduction half reactions can occur, the half reaction
having the highest tendency to accept electrons (highest reduction potential)
will occur (REDUCE) preferentially.
3)
When several different oxidation half reactions can occur, the half reaction
having the highest tendency to lose electrons (lowest reduction potential)
will occur (OXIDIZE) preferentially.
example: A beaker contains an iron nail wrapped with both a copper wire and a piece of magnesium ribbon, immersed in an aqueous solution containing CuSO4 and some dissolved Cl2(g). What is the overall preferred reaction? PROCEDURE: 1) List the species present, making sure that all ionic compounds are broken up into ions. Fe, Cu, Mg, Cu+2, SO4-­‐2, Cl2, H2O 2) Start at the UPPER LEFT (reduction side) of the table and look down the table until you find the first ion that you listed in #1 (it may not be the first ion that you listed, but it will be the first one that you find). THIS IS THE SPECIES THAT WILL HAVE THE GREATEST TENDENCY TO REDUCE since it is highest on the list of reduction potentials. Cl2 + 2e  2Cl-­‐ ; Eo = 1.36 V 3) 4) Start at the LOWER RIGHT (oxidation side) of the table and look UP the table until you find a match from #1. THIS IS THE SPECIES THAT WILL HAVE THE GREATEST TENDENCY TO OXIDIZE since it is stronger on the list of reducing agents. Mg+2 + 2e  Mg ; Eo = -­‐2.37 V ADDING THE PREFERRED REACTIONS GIVES THE OVERALL REACTION: Cl2 + Mg  Mg+2 + 2Cl-­‐ NOW TRY THE CHAPTER 12 SECTION END QUESTIONS! SELECTING PREFERRED REACTIONS -HELPS US TELL WHICH ELECTROLYTE IS OXIDIZED AND REDUCED IN LAB 5.1
1)
List the possible half reactions with corresponding reduction potential values (Eo)
2)
When several different reduction half reactions can occur, the half reaction
having the highest tendency to accept electrons (highest reduction potential)
will occur (REDUCE) preferentially.
3)
When several different oxidation half reactions can occur, the half reaction
having the highest tendency to lose electrons (lowest reduction potential)
will occur (OXIDIZE) preferentially.
example: A beaker contains an iron nail wrapped with both a copper wire and a piece of magnesium ribbon, immersed in an aqueous solution containing CuSO4 and some dissolved Cl2(g). What is the overall preferred reaction? PROCEDURE: 1) List the species present, making sure that all ionic compounds are broken up into ions. 2) Start at the UPPER LEFT (reduction side) of the table and look down the table until you find the first ion that you listed in #1 (it may not be the first ion that you listed, but it will be the first one that you find). THIS IS THE SPECIES THAT WILL HAVE THE GREATEST TENDENCY TO REDUCE since it is highest on the list of reduction potentials. 3) 4) Start at the LOWER RIGHT (oxidation side) of the table and look UP the table until you find a match from #1. THIS IS THE SPECIES THAT WILL HAVE THE GREATEST TENDENCY TO OXIDIZE since it is stronger on the list of reducing agents. ADDING THE PREFERRED REACTIONS GIVES THE OVERALL REACTION: NOW TRY THE CHAPTER 12 SECTION END QUESTIONS! PRACTICAL APPLICATIONS OF ELECTROCHEMICAL CONCEPTS: THE BREATHALYSER Hebden p. 229 LEAD ACID STORAGE BATTERY Hebden p. 230 ZINC-­‐CARBON BATTERY Hebden p. 231 ALKALINE DRY CELL Hebden p. 232 FUEL CELLS Hebden p. 233 BATTERIES Text p. 528 FUEL CELLS Text p. 532 CORROSION: -­‐A spontaneous electrochemical process. -­‐There is a tendency for most metals to lose electrons and exist in nature as oxides and sulphides. -­‐Most of the iron produced in Canada is to replace or repair items which are destroyed by corrosion (i.e. rusted iron -­‐ rust is used to describe iron oxidation; other oxidized metals are described in terms of corrosion). -­‐Two reagents necessary for corrosion: H20 and O2 The reactions usually begin on an exposed surface of stressed metal (head of a nail). At the centre of the drop, there is an oxygen poor region -­‐ this is where the iron oxidizes: Fe (s)  Fe+2 + 2e The Fe+2 ions formed migrate from the anode to the cathode. This exposes fresh Fe(s) for continued oxidation. The electrons released at the anode migrate through the metal to the cathode site where water and oxygen are reduced: 1/2 O2 + H20 + 2e  2OH-­‐ In the presence of oxygen, the Fe+2 ions are further oxidized to Fe+3 ions. continued next page The Fe+3 ions react with the hydroxide ions to produce the flaky brown substance we call rust: Fe+3 + 3OH-­‐  Fe(OH)3 (s) precipitation OR: Fe+3 + 6OH-­‐  Fe2O3 (s) + 3H20 RUST Both Fe(OH)3 (s) and Fe2O3 (s) are considered components of rust. Or rust can be represented by: Fe2O3 • xH20 Different numbers of water molecules attached to the iron (III) oxide can change the colour of rust. PREVENTING CORROSION: 1) 2) 3) 4) Apply a protective layer (paint, grease, plastic or another metal) to the surface so that oxygen and water can not get to it. Use an alloy such as stainless steel (this can be quite costly) which contains a corrosion-­‐free metal. CATHODIC PROTECTION: the process of protecting a substance from unwanted oxidation by connecting it to a substance having a higher tendency to oxidize. (eg. If Cu and Fe come into contact and they are exposed to oxygen and water, this will encourage the corrosion of the iron). For example, if magnesium is in contact with iron, and an oxidizing agent is present, the magnesium has a higher tendency to oxidize than the iron, thus the magnesium will corrode but the iron will not. (Mg acts as the anode -­‐ OXIDATION; and Fe acts as the cathode -­‐ REDUCTION) becomes Mg+2 remains as Fe (s) GALVANIZED IRON consists of a zinc coating on iron objects. Rusting of the iron only occurs when all of the zinc is oxidized away. Change the conditions in the chemical surroundings so that the tendency of the surroundings to reduce is lowered. o remove oxygen from the solution o lower the [H+] by adding OH-­‐ ions o connect metal object to a source of direct electric current (the metal becomes the cathode in relation to the surrounding environment, and this inhibits the loss of electrons to the environment).