Question

Enter your answer in the provided box. What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+] = 0.96 M, [H+] = 0.017 M and PH2 = 1.0 atm ?

Answer 1

Pb(s) /Pb^2+ (aq) //H^+ (aq) /H2 /Pt

Pb(s) -----------------> Pb^2+ (aq) + 2e^-      E0 = 0.13v

2H^+ (aq) + 2e^- -----------> H2(g)                 E0 = 0.00v

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Pb(s) + 2H^+ (aq) ------------> Pb^2+ (aq) + H2(g)      E0 cell = 0.13v

n = 2

Ecell     =   E0cell - 0.0592/n logQ

              = 0.13 - 0.0592/2 logPH2 *[Pb^2+]/[H^+]^2

             = 0.13 -0.0296log1*0.96/(0.017)^2

            = 0.13 -0.0296*3.52

             = 0.0258V >>>>answer