Reduction potential of Chlorine electrode in contact with 0.001 M HCl solution is $(E_{Pt, \frac{1}{2} {Cl}_{2} / {Cl}^{-}} = + 1.36 V)$

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1.39 V

1.45 V

1.54 V

1.24 V

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {Cl}_{2} + e^{-} \to {Cl}^{-} \\ E_{RP} = E_{RP}^{0} - \frac{0.0591}{1} \log [{Cl}^{-}] \\ E_{RP} = + 1.36 V - \frac{0.0591}{1} \log (10^{- 3}) \\ E_{RP} = + 1.537 V \end{array}$

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