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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 [ Zn2+] = 1.0 M and [Pb2+] = 2.0⋅10−4 M. Pb2+ (aq) + Zn (s) → Zn2+ (aq) + Pb (s)

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Answer: 0.52V

Explanation:

Ecell = Ecell(standard) - [(0.0592 logQ)/n]

Q = product of the quotient

n = no of electrons transferred = 2

Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]

Ecell = 0.63 - 0.0194

Ecell = 0.5205V

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