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 ago
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