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Practice question

MCQ

Question

A galvanic cell has E°cell = 0.62 V . If the cell potential becomes 0.68 V at 298 K when [Oxidized] = 0.01 M and [Reduced] = 0.1 M , how many electrons are transferred?

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Explanation

Ecell = E°cell - (0.059/n) log ([Ox]/[Red]) , 0.68 = 0.62 - (0.059/n) log (0.01/0.1) . 0.06 = (0.059/n) × 1 , n = (0.059/0.06) ≈ 1 .

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