Ox: 3Pb( s )3Pb2+ (aq) +6e E 0 = 0.13v Red: 2MnO4^-(aq)+8H+(aq)+6e 2MnO2( s )+4H

Question

Ox: 3Pb(s)3Pb2+(aq)+6e                                                  E 0 = 0.13v
Red: 2MnO4^-(aq)+8H+(aq)+6e 2MnO2(s)+4H2O(l)               E0 = 1.68v

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3Pb(s) + 2MnO4^- (aq)+ 8H^+(aq) ------------> 3Pb^2+ (aq) + 2MnO2(s) + 4H2O(l)    E0cell = 1.81v

      n = 6

Ecell      =   E0cell - 0.0592/n logQ

               = 1.81 - 0.0592/6 log[Pb^2+]^3/[MnO4^-]^2 [H^+]^8

            = 1.81 - 0.00986log(0.21)^3/(1.8)^2*(1.5)^8

             = 1.81- 0.00986log0.0001115

              = 1.81-0.00986*-3.9527

              = 1.849v >>>>answer

Explanation / Answer

QUESTION 5

An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)Pb2+(aq, 0.21 M )+2e
Red: MnO4(aq, 1.80 M )+4H+(aq, 1.5 M )+3e MnO2(s)+2H2O(l)

Compute the cell potential at 25 C.?

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