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Reduction potential of Iron electrode in contact with 0.01M ${FeSO}_{4}$ solution is $[E_{{Fe}^{+ 2} / Fe}^{o} = - 0 .44 V]$

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An Intiative by Sri Chaitanya

- 0.50 V

- 0.38 V

- 0.47 V

- 0.41 V

answer is D.

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Detailed Solution

$E_{{Fe}^{+ 2} / Fe} = E_{{Fe}^{2 +} / Fe}^{o} + \frac{0 .0591}{2} \log [{Fe}^{+ 2}]; E_{{Fe}^{+ 2} / Fe} = - 0.44 V + \frac{0 .0591}{2} \log (10^{- 2}); E_{{Fe}^{+ 2} / Fe} = - 0.50 V;$

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