If E° of calomel electrode Cl−Hg2Cl2Pt is 0.27V its potential when KCl=0.1M would be

If $E °$ of calomel electrode $({Cl}^{-} |{Hg}_{2} {Cl}_{2}| Pt)$ is $0.27 V$ its potential when $KCl = 0.1 M$ would be

A
0.25 V
B
0.26V
C
0.276 V
D
0.286 V

Solution:

The reaction is $\frac{1}{2} {Hg}_{2} {Cl}_{2} (s) + e^{-} \to Hg (l) + {Cl}^{-} (aq)$

Hence, $E = E^{\circ} - \frac{R T}{E} \ln ([{Cl}^{-}] / c^{\circ}) = 0.27 V - (0.059 V) \log (0.1) = 0.2759 V$

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