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in Chemistry by kratos

Calculate emf of the following cell at 298 K : Mg() | Mg2+ (0.1 M) || Cu2+ (0.01M) | Cu() [Given E°cell = +2.71 V, 1 Faraday = 96500 C mol–1]

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+1 vote
by kratos
 
Best answer

At anode : Mg → Mg2+ + 2e–

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