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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
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…..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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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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