$E^{\circ} (Cr X_{2} O X_{7} X^{2 -}, H X^{+} / Cr X^{3 +}) = 1.33 V$ . The reduction potential of the dichromate couple at 298 K in a solution with $[Cr X_{2} O X_{7} X^{2 -}] = 0.01 M$ , $[Cr X^{3 +}] = 0.01 M$ and pH 2 is (take $2.303 RT / F = 0.0591 V$ )

Answer: A

E = 1.33 + \frac{0.0591}{6} lo g \frac{[ Cr X _{2} O X _{7} X ^{2 -} ] [ H X ^{+} ] ^{14}}{[ Cr X ^{3 +} ] ^{2}}

lo g Q = lo g (0.01) + 14 lo g (1 0^{- 2}) - 2 lo g (0.01) = - 2 - 28 + 4 = - 26

E = 1.33 + 0.00985 \times (- 26) = 1.33 - 0.256 = 1.074 V

. Forgetting to square

[Cr X^{3 +}]

gives

lo g Q = - 28

and 1.054 V.