3Ce4+ (aq) + 3 e- → 3Ce3+(aq) E° = +1.61 V ↑ E° → reduction → cathode
BiO+ (aq) + 2H+ (aq) + 3e- → Bi(s) + H2O(l) E° = 0.32 V ↓ E° → oxidation → anode
E°cell = 1.29 V
3Ce4+ (aq) + 3 e- → 3Ce3+(aq)
Bi(s) + H2O(l) → BiO+ (aq) + 2H+ (aq) + 3e-
3Ce4+ (aq) +Bi(s) + H2O(l) → 3Ce3+(aq) + BiO+(aq) + 2H+(aq)
n = 3
Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.
3Ce4+(aq) + Bi(s) + H2O(l) → 3Ce3+(aq) + BiO+(aq) + 2H+(aq)
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