QUESTION 5 An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.21 M...

QUESTION 5

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

Compute the cell potential at 25 ∘C.?

Expert Solution

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

-------------------------------------------------------------------------------------------

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

queen_honey_blossom answered 2 years ago

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