Problem: 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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FREE Expert Solution

Step 1:

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


Step 2

E°cell = E°cathode - E°anodeE°cell =1.61 V - 0.32 V

cell = 1.29 V


Step 3

3Ce4+ (aq) + 3 e-  3Ce3+(aq)  

Bi(s) + H2O(l) → BiO+ (aq) + 2H+ (aq) + 3e-

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3Ce4+ (aq) +Bi(s) + H2O(l)  3Ce3+(aq) + BiO+(aq) + 2H+(aq)


n = 3


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Problem Details

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)