Question
In: Chemistry
What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+] = 0.46 M, [H+] = 0.047 M and PH2 = 1.0 atm ?
What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+] = 0.46 M, [H+] = 0.047 M and PH2 = 1.0 atm ?
____V
Solutions
Expert Solution
Pb(s) --------------> Pb^2+ (aq) + 2e^- E0 = 0.13v
2H^+ (aq) + 2e^- ------> H2(g) E0 = 0.00v
Pb(s) + 2H^+ (aq) ---------> Pb^2+ (aq)+H2(g) E0cell = 0.13v
n = 2
Ecell = E0cell- 0.0592/nlogQ
= 0.13-0.0592/2 log[Pb^2+]PH2/[H^+]^2
= 0.13 -0.0296log0.46*1/(0.047)^2
= 0.13 -0.0296*2.3185 = 0.0614v
queen_honey_blossom answered 2 years agoRelated Solutions
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