E deg(Cl2/Cl-) = 1.36 V. The reduction potential of a chlorine electrode at 298 K with p_Cl2 = 0.1 bar and [Cl-] = 0.2 M is (take 2.303RT/F = 0.0591 V, log 2.5 = 0.398)

JEE Main levelSingle correct optionDifficulty 2 of 3ec083

$E^{\circ} (Cl X_{2} / Cl X^{-}) = 1.36 V$ . The reduction potential of a chlorine electrode at 298 K with $p_{Cl X_{2}} = 0.1 bar$ and $[Cl X^{-}] = 0.2 M$ is (take $2.303 RT / F = 0.0591 V$ , $lo g 2.5 = 0.398$ )

Half reaction

Cl X_{2} (g) + 2 e X^{-} 2 Cl X^{-}

n = 2

E = E^{\circ} + \frac{0.0591}{2} lo g \frac{p _{Cl X_{2}}}{[ Cl X ^{-} ] ^{2}}

\frac{0.1}{( 0.2 ) ^{2}} = 2.5

lo g 2.5 = 0.398

E = 1.36 + 0.02955 \times 0.398 = 1.36 + 0.0118 = 1.372 V

. Forgetting to square

[Cl X^{-}]

gives 1.351 V.