Question

Design a voltaic cell with the following two reduction half-reactions: Ag+(aq) + e−  ⟶ Ag(s) Eo = 0.80 V Pb2+(aq) + 2 e− ⟶ Pb(s) Eo = −0.13 V Calculate Eocell and the equilibrium constant K for the voltaic cell at 298 K. Click here for a copy of Final Exam cover sheet.

Answer 1

oxidation half cell:

Pb(s) ---------------- Pb(2+)(aq) + 2e-, E0(oxidation) = 0.13 V

reduction half cell:

Ag+(aq) + e- ----------------- Ag(s), E0(reduction) = 0.80 V

Net balanced reaction:

Pb(s) + 2Ag+(aq) ---------------- Pb(2+)(aq) + 2Ag(s)

E0(cell) = E0(oxidation) + E0(reduction) = 0.80 + 0.13 = 0.93 V

Delta G = -nFE(cell) = -(2)(96500)(0.93) = -179.49 kJ

Delta G = -RT ln(K)

-179490 = -(8.314)(298) ln(K)

ln(K) = 72.445

K = e^(72.445) = 2.9 * 10^(31)

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