What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a...

Question

Question

What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+] = 0.28 M, [H+] = 0.034 M and PH2 = 1.0 atm ?

science chemistry

Answer 1

Answer #1

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/n logQ

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

= 0.13 - 0.0296log1*0.28/(0.034)^2

= 0.13 - 0.0296log242.2

= 0.13 -0.0296*2.3841

= 0.05943v >>>answer

Answer 2

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