$E^{\circ}$ for $H X_{3} AsO X_{4} + 2 H X^{+} + 2 e X^{-} H X_{3} AsO X_{3} + H X_{2} O$ is $0.56 V$ and $E^{\circ} (I X_{2} / I X^{-}) = 0.54 V$ at 298 K. All dissolved species are at $1 M$ except $H X^{+}$ , and $2.303 RT / F = 0.0591 V$ . Which statements are correct?

Answer: A, B, C

E (As) = 0.56 + \frac{0.0591}{2} lo g [H X^{+}]^{2} = 0.56 - 0.0591 pH

, while

E (I X_{2} / I X^{-}) = 0.54

at all pH since no

H X^{+}

is involved (D is wrong). At pH 0,

0.56 > 0.54

so As(V) oxidises iodide (A). At pH 7,

E (As) = 0.56 - 0.414 = 0.146 V < 0.54

, so the direction reverses and

I X_{2}

oxidises As(III) (B). Equality:

0.56 - 0.0591 pH = 0.54

gives

pH = 0.02/0.0591 = 0.34

(C).