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Question

Find the emf of the cell in which the following reaction takes place at 298 K

Ni(s) + 2Ag+ (0.001 M) $\to$ Ni2+ (0.001 M) + 2Ag(s)

(Given that E$_{cell}^o$ = 10.5 V, ${{2.303\,RT} \over F} = 0.059$ at 298 K)

10.4115 V

10.385 V

0.9615 V

10.05 V

Solution

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