Kimia Reaksi Anorganik

17/12/2011
Kimia Reaksi Anorganik
2nd Mid-Semester Topics
Yuniar Ponco Prananto
•
•
•
•
Frontier Orbitals (2 sks)
HSAB Application (5 sks)
Heterogeneous Acid Base Reaction (2 sks)
Reduction and Oxidation (12 sks):
- Extraction of the Elements
- Reduction Potential
- Redox Stability in Water
- Diagrammatic Presentation of Potential Data
Frontier Orbitals
Huheey, J.E., Keiter, E.A., and Keiter, R.L., 1993, Inorganic
Chemistry, Principles of Structure and Reactivity, 4th
ed., Harper Collins College Publisher, New York
Miessler, D. L. and Tarr, D. A., 2004, Inorganic Chemistry,
3rd ed., Prentice Hall International, USA
Atkins, P., Overton, T., Rourke, J., Shriver, D. F., Weller, M.,
and Amstrong, F., 2009, Shriver and Atkins’ Inorganic
Chemistry, 5th ed., Oxford University Press, UK
Gary Wulfsberg, 2000, Inorganic Chemistry, University
Science Book, California, USA
[email protected]
…..summary
• Frontier orbitals, which are HOMO (highest occupied
molecular orbital) and LUMO (lowest unoccupied molecular
orbital), has been focused nearly since the beginning of
HSAB theory.
• Koopman’s theorem, the energy of HOMO → the
ionization energy, while the energy of LUMO → the
electron affinity for a closed-shell spesies, in which both
orbitals are involved in the electronegativity and HSAB
relationships.
• Hard species → have a large HOMO – LUMO gap
• Soft species → have a small HOMO – LUMO gap.
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Frontier orbitals is the HOMO (filled or partly filled) and
the LUMO (completely or partly vacant) of a
MOLECULAR ENTITY.
HOMO
is the highest-energy molecular orbital of an atom or
molecule containing an electron. Most likely the first
orbital, from which an atom will lose an electron.
LUMO
is the orbital that can act as the electron acceptor,
since it is the innermost (lowest energy) orbital that
has room to accept an electron.
HSAB Principle
…..summary
Hard vs Soft
Pure Appl. Chem., Vol. 71, No. 10, pp. 1919-1981, 1999
• A structure-reactivity concept formulated in terms of
acid-base interaction.
• According to this principle soft acids react faster and
form stronger bonds with soft bases, whereas hard acids
react faster and form stronger bonds with hard bases,
everything else being assumed aproximately equal.
• The soft-soft preference is characteristic mostly of the
reactions controlled by the orbital interaction, and the
hard-hard preference relates to reactions in which
electrostatic factors prevail.
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Hardness is associated
with:
1. Small atomic/ionic radius
2. High oxidation state
3. low polarizability,
4. high electronegativity
and
5. energy low-lying HOMO
(bases) or energy highlying LUMO (acids)
Softness is related to:
1. large atomic/ionic radius
2. low or zero oxidation
state
3. high polarizability,
4. low electronegativity,
and
5. energy high-lying
HOMO (bases) or
energy low lying LUMO
(acids)
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1. Equilibrium direction (reactants or products)
ZnI2 + HgCl2
b/l A – SB
SA – b/l B
ZnCl2 + HgI2
b/l A – b/l B
SA – SB
• The I- is a soft base and Hg2+ is a soft acid,
although Zn2+ is a b/line acid and Cl- is a
b/line base (relative to Hg2+ and I-), both
Zn2+ and Cl- are the harder species, hence
the SA-SB is HgI2 and HA-HB is ZnCl2
→ products are favored
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CuI2 + Cu2O
b/l A – SB
SA – HB
2Cul + CuO
SA – SB b/l A – HB
• The I- is a soft base and O2- is a hard base,
since Cu+ is softer acid than Cu+2 therefore
the SA-SB is CuI and the HA-HB is CuO
→ products are favored
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Exercise 1
CdCI2 + CaSO4
b/l A – b/l B
HA – HB
CdSO4 + CaCl2
b/l A – HB
HA – b/l B
Predict the favorable species (reactants or products) in
the following equilibrium!
1. Hg(CN)2 + Ca(ClO4) 2
• Both Cd+2 and Cl- are b/line acid and b/line
base, while both Ca+2 and O-2 are hard acid
and hard base, therefore those species in
the left are favorable, due to its stronger
interactions, than those species in the right
→ reactants are favored
2. 3FeO + Fe2S3
3. PbCO3 + 2NaBr
4. MgCl2 + AgI
5. MnS + NiF2
6. Ti(NO3)2 + ZnCl2
7. Nb2S5 + 5HgO
Ca(CN) 2 + Hg(ClO4) 2
3FeS + Fe2O3
Na2CO3 + PbBr2
MgI2 + AgCl
NiS + MnF2
Zn(NO3) 2 + TiCl2
Nb2O5 + 5HgS
2. Solubility of Halides and Chalcogenides
[Ag(H2O)n]+ + [Cl(H2O)m]SA – HB
b/l B – HA
AgCl(s) + (m+n)H2O
SA – b/l B
HA – HB
Hence:
We can predict that “the chlorides, bromides, and iodides of the
soft acids are insoluble in water” → most of the predictions are
generally true for the +1 and +2 ions of the soft-acid metals, i.e:
CuCl, AgCl, AuCl, TlCl, Hg2Cl2, OsCl2, IrCl2, PdCl2, PtCl2, and
PbCl2, except CuCl2, AuCl3, HgCl2, PtCl4.
Similar predictions of the solubility of pseudohalide salts, i.e:
soft base → cyanide (CN), thiocyanate (SCN),
b/l base → azide (-N=N+=N-)
→ “the cyanides, thiocyanates, and azides of the soft acids
• Special case: the sulfides, selenides and tellurides of
the soft and borderline acids are insoluble in water due
to the combination of softness and some strength in
bases (S2-, Se2- and Te2- ions are stronger bases than
Cl-, Br- and I- ions)
• Moreover…
– for a given soft acid, the solubility of the iodide <
bromide < chloride,
– similarly, for a given soft acid, the solubility of
telluride < selenide < sulfide
are insoluble in water”
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Exercise 2
• Due to the absence of soft base involvement in the
precipitation equilibrium, the HSAB principle is
irrelevant to the water solubility’s prediction of the salts
from hard bases F- (weak base), OH- and O2-(strong
bases)
→ “the fluorides, oxides, and hydroxides of acidic
cations are insoluble in water”
• Because there is no soft acid is involved in the
solubility equilibrium, the solubility of the sulfides,
selenides and tellurides of the hard acids cannot be
predicted by the HSAB principle.
• Identify all insoluble compounds and
explain it according to HSAB principle:
a. PtAs2
e. TiO2
b. AgI
f. CdTe
c. CaF2
g. AgF
d. KI
h. CoSO4
3. The Qualitative Analysis Scheme for Metal Ions
• Group I
→ precipitated with 0.3 M HCl
• Group II
→ precipitated with H2S from acidic solution
• Group III
→ precipitated with (NH4)2S from basic solution
• Group IV
→ precipitated with (NH4)2CO3
• Group V
→ remains in solution
?
.: the member of each group does not belong to
certain / similar period or group in the periodic
table of elements
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Group I and II
• The chloride ion is a borderline base that will combine
strongly to soft acid, and if a large excess of Cl- is
present, only Ag+, Hg2+, Tl+ and Pb2+ will be
precipitated while others like Cu, Pd, Os, Ir, Pt and Au
will form soluble chloro anions [MCln]x-.
• The sulfide ion is a very soft and a stronger base than
Cl-. Therefore when it reacts with Cu, Pd, Os, Ir, Pt
and Au, it will produce more insoluble sulfides.
Note: the scheme should be done properly and in order!!
Group III
• The sulfide ion is present in much higher concentration
and also competing with another strong base plus
much harder hydroxide ion.
• Only hard-acid metal ions and small group of borderline
acids from the +2 ions of the 1st row transition metals
(Zn, Ni, Co, Fe, and Mn) are in the solution and will be
precipitated as sulfides.
• The remains, which is strong - hard acid metal (also all
of the f-block elements), will react with the strong –
hard base OH- and gives precipitation of metal
hydroxides or metal oxides.
Group IV and V
• Only the non-acidic and weak acidic hard acid of the
1st two groups of the periodic table are present, and
then moderately-basic carbonate ion is added → only
the weak acidic cations will be precipitated.
• The rest will only be the non-acidic cations that prefer
to bind with the (non-basic) hard base water in the
form of hydrated-ion (to get the precipitation of these
metals, sometimes the test is continued by adding the
non-basic anions)
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Exercise 3
A new qualitative analysis scheme that separate
cations into five existing groups using novel
precipitation anions is prepared. Choose the best
substitute for:
1. HCl in Group I → HBr, HF, HClO, or HClO4
2. H2S in Group II → H2O, H2Te, H2SO3, or H2SO4
3. (NH4) 2CO3 in Group IV → NH4OH, NH4ClO4,
(NH4) 2SO4, (NH4) 4C
4. If you want to precipitate Group V, which
compound would do this → LiOH, LiClO4,
Li2SO4, Li4C.
4. The Geochemical Classification and
Differentiation of the Elements
Geochemical classification of the elements divided into
Primary and Secondary. The primary emphasizes
behavior under condition at the surface of the earth,
which are:
• Lithophiles
metals and nonmetals that tend to occur as cations in
oxides, silicates, sulfates and carbonates.
the anions posses oxygen as the donor atom hence the
metal ions that attached are hard acid metals,
the nonmetals are, or have been oxidized by
atmospheric oxygen to oxo-anions, hard bases.
Natroxalate
Na2C2O4
• Chalcophiles
Hard Acid Metal
Natrolite
Na2[Al2Si3O10]·2H2O
occur in nature as cations in sulfides (less commonly w/
other soft bases such as telluride, arsenic, etc). Mostly
from the borderline and some soft acids and many of
soft bases.
Magnesite
MgCO3
Gypsum
CaSO4
• Atmophiles
chemically unreactive nonmetals that occur in the
atmosphere in elemental form, i.e: N2 and noble gases
• Siderophiles
Fluorite CaF2
Sylvite KCl
soft acid metals that tend to occur native in elemental
form (subdivision of chalcopiles)
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Barite
BaSO4
Gaylussite
Na2Ca(CO3) 2·5(H2O)
(Meta)variscite AlPO4.2H2O
Corundum
Al2O3
Uvarovite
Ca3Cr2 (SiO4)3
Beryl Be3Al2Si6O18
Rutherfordine
(UO2)(CO3)
Rutile TiO2
Rhodochrosite
MnCO3
Molybdenite
MoS2
Erythrite
Co3(AsO4)2·8(H2O)
Hard – b/line Acid Metal
Hematite Fe2O3
Pyrite FeS2
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Siderite FeCO3
Strengite
FePO4.2H2O
Magnetite Fe3O4
Fayalite Fe2SiO4
Smithsonite
ZnCO3
Morenosite NiSO4·7H2O
Cassiterite
SnO3
Salesite Cu(IO3)(OH)
Greenockite CdCO3
Stibnite Sb2S3
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Soft Acid Metal
Cinnabar
HgS
Calaverite AuTe2
Elements
1. Sodium
Galena
PbS
2. Calcium
Iodargyrite AgI
Elements
8. Iron
9. Manganese
10. Zinc
11. Nickel
Marshite CuI
Mineral Names
Sartorite PbAs2S4
Hard / Soft – Acid / Base
Hard Acid Na + B/line Base Cl
Natron, Na2CO3.10H2O
…
Villiaumite, NaF
…
Caliche, NaIO3
…
Calcite, CaCO3
Hard Acid Ca + Hard Base O
Gypsum, CaSO4.2H2O
…
Apatite, Ca5(PO4)3OH
…
Fluorite, CaF2
…
3. Kalium
Sylvite, KCl
Hard Acid K + B/line Base Cl
4. Magnesium
Magnesite, MgCO3
Hard Acid Mg + Hard Base O
5. Alumunium
Bauxite, Al2O3
Hard Acid Al + Hard Base O
Cryolite, NaAlF4
…
Gibbsite, Al(OH)3
…
Gahnite, ZnAl2O4
…
6. Chromium
Chromite, FeCr2O4
…
7. Titanium
Rutile, TiO2
…
Ilmenite, FeTiO3
…
Hard / Soft – Acid / Base
Hematite, Fe2O3
Hard Acid Fe + Hard Base O
Magnetite, Fe3O4
…
Siderite, FeCO3
…
Pyrite, FeS2
…
Arsenopyrite, FeAsS
…
Pyrolusite, MnO2
Hard Acid Mn + Hard Base O
Psilomelane, BaMn5O11
…
Sphalerite, ZnS
B/line Acid Zn + Soft Base S
Smithsonite, ZnCO3
…
Pentlandite, (Ni,Fe)9S8
…
Heazlewoodite, Ni3S2
…
Melonite, NiTe2
…
Morenosite, NiSO4.7H2O
Hard Acid Ni + Hard Base O
Pararammelsbergite, NiAs2
…
Vaesite, NiS2
12. Tin
Cassiterite, SnO2
B/line Acid Sn + Hard Base O
13. Antimony
Stibnite, Sb2S3
B/line Acid Sb + Soft Base S
14. Molybdenum
Molybdenite, MoS2
B/line Acid Mo + Soft Base S
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Mineral Names
Halite, NaCl
Exercise 4
• Choose and explain the most likely
mineral sources of this elements based
on HSAB principle!
a. LaPO4, LaAs, LaI3, or LaCl3
b. ZrSiO4, ZrPbO4, or ZrCl4
c. PtAs2, PtN2, PtSiO4, or PtF2
d. CoAs2, CoCO3, CoSO4, or CoBr2
e. TeF4, Na2Te, PbTe, or TeI4
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5. Biochemistry, Biological Functions,
Toxicology, and Medicinal
• HSAB principle can organize the chemistry of the
metal ions in ecology, biochemistry and medicine
• The donor atom of biochemical ligands to which the
essential metal ions prefer to bind indicates the
general pattern of HSAB principle, as well as the
kind of biochemical sites at which toxic metal ions
are likely to bind.
• Mimicking the behavior of complicated biochemical
complexes with simpler coordination complexes
(bioinorganic chemistry) → catalysts, metal-activated
enzymes, template, etc.
Heterogeneous Acid-Base Reaction
Some of the most important reactions involving the
Lewis acidity of inorganic compounds occur at solid
surfaces.
For example, SURFACE ACIDS, which are solids with
a high surface area and Lewis acid sites, are used as
catalysts in the petrochemical industry for the
isomerization and alkylation of aromatic compounds.
The surfaces of many materials that are important in
the chemistry of soil and natural waters also have
Lewis acid sites.
Surface acidity
Alumina (Al2O3)
Aluminosilikat (Al2O3)
• Termasuk salah satu asam permukaan karena Al dgn biloks Al yang
tinggi (+3).
• Ketika alumunium oksida hidrat yang baru mengendap dipanaskan di
atas 150°C, proses dehidrasi terjadi dan permukaan mengalami
reaksi:
• Berbeda dgn alumina, aluminosilikat memperlihatkan sifat keasaman
Brownsted dimana pembentukkannya diperkirakan melalui
kondensasi unit Si(OH)4 dgn unit H2OAl(OH)3:
OH
OH
OH
Al
Al
Al
OH
O
Al
Al
Al
OH2
Si
Al
H
O
Si
Al
+ H2 O
+ H2O
• Reaksi tsb menghasilkan spesi Al3+ di permukaan, yg bertindak sbg
asam Lewis, begitu juga spesi O2-, yg bertindak sbg basa Lewis.
• Kedua asam-basa Lewis ini akan dihasilkan bersamaan, namun situs
asam Lewis lebih utama untuk katalisis permukaan.
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OH
• Reaksi tsb menghasilkan asam Brownsted yang kuat pada situs
dimana proton dipertahankan untuk menyeimbangkan muatan
positif yang lebih kecil dari Al3+.
• Karakter dan kekuatan asam permukaan dapat dipelajari melalui
metoda sederhana seperti titrasi dgn larutan basa. Selain itu,
spektrofotometri inframerah dari molekul yang teradsorb dapat juga
digunakan untuk mempelajari situs permukaan yang reaktif.
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Silika (SiO2)
REAKSI OKSIDASI – REDUKSI
• Permukaan silika tidak seketika membentuk situs asam Lewis
karena gugus OH terikat kuat pada permukaan turunan SiO2, dan
keasamaan Brownsted lebih dominan.
• Keasaman Brownsted permukaan silika relatif sedang dibandingkan
asam asetat, berbeda dgn aluminosilikat yang menunjukkan
keasamaan Brownsted yang kuat.
• Reaksi permukaan oleh situs asam Brownsted gel silika digunakan
untuk membuat lapisan tipis beberapa gugus organik, yang salah
satunya dipakai untuk memodifikasi permukaan fase diam kolom
kromatografi
Si
OH
Si
O
SiR
+ H O
+ HOSiR
3
3
Si
OH
+ ClSiR 3
Si
O
SiR 3
2
+ HCl
Ekstraksi Unsur
• Reduksi bijih logam
reaksi redoks tidak selalu mencapai
kesetimbangan sehingga hukum termodinamika
dapat digunakan untuk memprediksi reaksi
mana yang lebih disukai, yaitu berdasarkan nilai
ΔG yang negatif (pada T - P tetap), dimana nilai
ΔG standar dirumuskan:
•
•
•
•
Ekstraksi Unsur
Potensial Reduksi
Kestabilan Redoks dalam Air
Diagram Data Potensial
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ΔG° = - RT ln K
Nilai ΔG yang negatif menandakan bahwa nilai K
> 1 yang juga berarti reaksi bergeser ke kanan
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• Energi bebas (ΔG) reaksi reduksi oksida logam vs
suhu → Diagram Ellingham
Energi bebas standar yang disajikan dalam diagram ini adalah
untuk pembentukan oksida dari logam:
2/ M
2
ΔG°(M)
x
(s or l) + O2(g) → /x MOx(s)
dan tiga jenis reaksi oksidasi karbon:
2C(s) + O2(g) → 2CO(g)
2CO(g) + O2(g) → 2CO2(g)
C(s) + O2(g) → 2CO2(g)
ΔG°(C, CO)
ΔG°(CO, CO2)
ΔG°(C, CO2)
Latihan:
Pada 1000°C, tuliskan reaksi reduksi dan nilai energi Gibbs dari
logam Ag, Zn, Al, dan Ca!
Contoh Soal
Jika ΔG° suatu logam bernilai positif pada suhu
tertentu , maka dekomposisi termal suatu oksida
logam dapat terjadi tanpa adanya reduktor.
Namun kebanyakan oksida logam membutuhkan
reduktor, misalnya karbon:
C(s) + 2/x MOx(s) → 2/x M(l) + CO2(g)
ΔG° = ΔG° (C) - ΔG° (M)
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• Berapakah suhu minimum yang diperlukan untuk
mereduksi (a) ZnO dan (b) MgO menjadi logamnya
dengan reduktor karbon ? (lihat diagram Ellingham)
Jawab:
C(s) + ZnO(s) → Zn(l) + CO(g)
C(s) + MgO(s) → Mg(l) + CO(g)
C(s) + 2MgO(s) → 2Mg(l) + CO2(g)
T > 950°C
T > 1850°C
T > 2250°C
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• Diagram Ellingham juga dapat dipakai untuk
menentukan apakah suatu logam (M’) dapat dipakai
sebagai reduktor untuk oksida logam lain (M),
sebagai pengganti karbon:
M’ + MOx → M + M’Ox
Maka: ΔG° = ΔG° (M’) - ΔG° (M)
• Contoh:
Mg → Al2O3 pada suhu di bawah 1300°C
Mg → TiO2 pada suhu di bawah 1800°C
Mg → SiO2 pada suhu di bawah 2200°C
• Proses industri untuk mendapatkan logam melalui
reaksi reduksi memberikan banyak kemungkinan
dibandingkan analisa termodinamika yang ada, mulai
dari reduksi logam yang mudah, sedang, dan sulit.
Mudah ekstraksi hidrometalurgi tembaga
Sedang ekstraksi besi dalam tanur suhu tinggi
Sulit ekstraksi silikon dari pasir silika atau quarts
Note: see Atkin's for detail!!
Reduksi Elektrolitik
• Reduksi secara langsung Al2O3 dgn C dapat dilakukan pada
suhu di atas 2000°C (sesuai diagram Ellingham) sehingga
relatif mahal dan sulit pada 1886 digunakan proses Hall
- Heroult
• Proses ini merupakan suatu teknik untuk mengatur suatu
reduksi dgn merangkaikannya (melalui elektrode dan
sirkuit eksternal) ke sebuah reaksi dgn nilai ΔG yang lebih
negatif.
• Energi bebas yang diperoleh dari sumber luar dapat
diamati dari perbedaan potensial E yang dihasilkan
sepanjang elektrode tsb menggunakan hubungan
termodinamika
ΔG = - n.F.E
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• Energi bebas yang diperoleh dari sumber luar dapat
diamati dari perbedaan potensial E yang dihasilkan
sepanjang elektrode tsb. menggunakan rumus
ΔG = - n.F.E
dimana:
n = mol elektron yang ditransfer
F = konstanta Faraday = 96500 C/mol
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• Sementara itu, perubahan energi bebas Gibbs total dari
pasangan proses dalam dan luar sistem adalah
ΔG + ΔG(luar) = ΔG - n.F.E(luar)
• Apabila beda potensial dari luar berlebih maka:
E(luar) = ΔG : nF
ket:
proses reduksi dapat berlangsung spontan
selanjutnya keseluruhan proses berjalan
berkurangnya energi bebas
dan
dgn
CONTOH
Tentukan beda potensial minimal yang diperlukan untuk
mereduksi alumina pada 500°C!
2/ Al O → 4/ Al + O
3 2 3
3
2
Jawab:
Diagram Ellingham → ΔG = +960kJ (setiap mol O2)
Jumlah mol elektron = 4/3 x Al3+ = 4 mol
maka
E(luar) = 960 kJ : (4 mol x 96,5 kC/mol) = 2,49V
.:. Setidaknya beda potensial 2,5V harus digunakan untuk
mereduksi alumina pada 500°C.
Ekstraksi Unsur dgn Oksidasi
• Ekstraksi unsur dgn oksidasi yang paling penting
adalah golongan halogen.
• Energi bebas standar dari oksidasi Cl- dalam air
bernilai sangat positif, yang mengindikasikan bhw
diperlukan elektrolisis untuk reaksi berjalan ke kanan:
2Cl-(aq) +
2H2O(l) →
2OH-
(aq)
+ H2(g) + Cl2(g)
ΔG° = +422 kJ
• Perbedaan potensial minimum yang diperlukan
sekitar 2,2V (dimana n = 2 mol untuk reaksi di atas) --tunjukkan buktinya!
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• Namun, ketika beda potensial yang digunakan hanya 1,2V
(atau setara dgn n = 4), maka akan terjadi kompetisi dgn
reaksi berikut:
2H2O(l) → 2H2(aq) + O2(g) ΔG° = +414 kJ
Hal ini mengisyaratkan pentingnya faktor kinetika
• Kecepatan oksidasi air sangatlah rendah pada potensial
dimana reaksi tsb mulai spontan secara termodinamika reduksi memiliki overpotential yang tinggi.
• Overpotential (ή) merupakan beda potensial sebagai
tambahan terhadap nilai kesetimbangan yang dibutuhkan
untuk mencapai kecepatan reaksi yang signifikan.
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Potensial Reduksi
• Golongan halogen lain:
F2 elektrolisis campuran HF/KF anhidrat dilakukan dan
meleleh pada 72°C
Br2 dan I2 oksidasi larutan halida dgn gas klor (Cl2)
• Reaksi redoks dapat berjalan apabila ΔG negatif,
selain itu energi bebas Gibbs berhubungan dan
dapat juga dituliskan dalam bentuk lain yaitu
beda potensial (E).
• Logam lain:
khususnya yang secara alami berada dalam bentuk
unsurnya seperti Emas (Au), yaitu dgn melarutkan dengan
CN- dan mereduksinya dgn logam reaktif seperti Zn:
2[Au(CN)2]-(aq) + Zn(s) → [Zn(CN)4]-2(aq) + 2Au(s)
• Kedua hal tsb (ΔG dan E) dapat digabungkan dan
menghasilkan cara lain yang bermanfaat dalam
membahas reaksi redoks.
Reaksi setengah redoks
• Spesies oksidasi dan reduksi pada reaksi setengah
membentuk pasangan redoks, dimana spesies yang
teroksidasi ditulis di depan spesies yang tereduksi yaitu
OKS/RED, misalnya:
…..1)
Zn(s) → Zn2+(aq) + 2e+
…..2)
2H (aq) + 2e → H2(g)
maka pasangan redoks ditulis dgn:
rx 1 ==> Zn2+ / Zn
rx 2 ==> H+ / H2
Potensial elektoda standar
• Potensial reduksi standar(E0) adalah voltase yang
berkaitan dengan reaksi reduksi pada elektroda jika
konsentrasi semua zat terlarut 1 M dan semua gas pada 1
atm.
• Energ Gibbs dari reduksi ion H+ ditetapkan bernilai nol dan
potensial reduksi zat lain dihitung berdasarkan acuan ini.
Zn2+(aq) + H2(g) → 2H+(aq) + Zn(s)
ΔG° = +147kJ/mol
sehingga diperoleh nilai energi Gibbs untuk reduksi Zn
Zn2+(aq) + 2e-→ Zn(s)
ΔG° = +147kJ/mol
Kesepakatan umum: ditulis dalam bentuk reaksi reduksinya!!
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15
17/12/2011
Serial
elektrokimia
ΔG = - n.F.E
Nilai potensial elektroda yang setara dgn nilai ΔG° dari reaksi
setengah ditulis dgn E° dan disebut potensial reduksi standar.
Karena ΔG° dari reaksi reduksi H+ bernilai nol maka potensial
reduksi standar H+ / H2 juga nol pada semua suhu.
Dgn rumus ΔG = - n.F.E maka diperoleh hasil
2H+(aq) + 2e- → H2(g)
E° = 0 V
2+
Zn (aq) + 2e → Zn(s)
E° = -0,76 V
Sehingga reaksi akan berjalan spontan bila:
2H+(aq) + Zn(s) → Zn2+(aq) + H2(g)
E° = +0,76 V
Reaksi spontan bila ΔG < 0 atau E > 0
Kespontanan Reaksi Redoks
ΔG = - n.F.Esel
ΔG0 = - n.F.E°sel
n = jumlah mol elektron dalam reaksi
J
F = 96.500
= 96.500 C/mol
V • mol
ΔG0 = -RT ln K = = -n.F.Esel
∆G0 = -RT ln K
E°sel =
(8,314 J/K•mol)(298 K)
RT
ln K =
ln K
nF
n (96.500 J/V•mol)
∆G = -nFEsel
E°sel =
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0,0257 V
ln K
n
atau
E°sel =
0,0592 V
log K
n
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