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The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is __________V when the concentration of Zn2+ = 1.0 M and the


concentration of Pb2+ = 2.0 x 10-4 M.



Pb2+(aq) + Zn(s) ---> Zn2+(aq) + Pb(s)

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Answer:

0.52 V

Explanation:

From Nernst equation;

Ecell =E°cell - 0.0592/n log Q

E°cell = +0.63 V

[Zn2+] =1.0 M

[Pb2+] = 2.0 x 10-4 M

n = 2

Ecell = 0.63 - 0.0592/2 log[1.0/2.0 x 10-4]

Ecell = 0.63 - 0.11

Ecell = 0.52 V

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