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Calculate the e.m.f. of the cell :
Cr | Cr3+ (0.1M) || Fe2+ (0.01 M) | Fe
Given : E© (Cr3+ | Cr) = -0.75 V, E© (Fe2+ | Fe) = -0.45 V
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Explanation:
Cr acts as anode and Fe act's as cathode
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