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If S.R.P of Nickel and chlorine electrodes are -0.25V and 1.36V respectively. The emf of the cell $N i / N i^{+ 2} (0.01 M) / / C l^{-} (0.01 M) / C l_{2}, p t$ is

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1.62V

1.79V

1.85V

1.23V

answer is D.

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Detailed Solution

$E = E_{0} - \frac{0.059}{n} \log (o r) = 1.61 - 0.03 {\log 10^{- 2} \times 10^{- 4}} 1.61 + 0.18 = 1.79 v$

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