Calculate the equilibrium constant for each of the reactions at 25°C. Standard Electrode Potentials at 25 °C Reduction Half-Reaction E° (V) Fe+(aq) + 3 e → Fe(s) -0.036 Sn²+ (aq) + 2 e → Sn(s) -0.14 2+ Cu" (ag) + 2 е — Cu(s) 0.16 O2 (9) + 2 H2O(1) + 4 e → 4 OH (aq) 0.40 Cl2(9) + 2 е — 2 CI 1.36 I2(s) + 2 e → 2I- 0.54

Chemistry: Principles and Practice
3rd Edition
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Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.98QE
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О2(9) + 2 НэО(1) + 2 Cu(s) — 4 ОН (ад) + 2 Cu+ (ag)
Express your answer using two significant figures.
ΑΣΦ
?
K =
Transcribed Image Text:О2(9) + 2 НэО(1) + 2 Cu(s) — 4 ОН (ад) + 2 Cu+ (ag) Express your answer using two significant figures. ΑΣΦ ? K =
Calculate the equilibrium constant for each of the
reactions at 25°C.
Standard Electrode Potentials at 25 °C
Reduction Half-Reaction
E° (V)
Fe+(aq) + 3 e → Fe(s)
-0.036
Sn²+ (aq) + 2 e → Sn(s)
-0.14
2+
Cu" (ag) + 2 е — Cu(s)
0.16
O2 (g) + 2 H2O(1) +4 e¯ →4 OH (aq)
0.40
Cl2(9) + 2 е —2 CI
1.36
I2(s) + 2 e → 2 1
0.54
Transcribed Image Text:Calculate the equilibrium constant for each of the reactions at 25°C. Standard Electrode Potentials at 25 °C Reduction Half-Reaction E° (V) Fe+(aq) + 3 e → Fe(s) -0.036 Sn²+ (aq) + 2 e → Sn(s) -0.14 2+ Cu" (ag) + 2 е — Cu(s) 0.16 O2 (g) + 2 H2O(1) +4 e¯ →4 OH (aq) 0.40 Cl2(9) + 2 е —2 CI 1.36 I2(s) + 2 e → 2 1 0.54
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