For $Cr X_{2} O X_{7} X^{2 -} + 14 H X^{+} + 6 e X^{-} 2 Cr X^{3 +} + 7 H X_{2} O$ at 298 K, a plot of reduction potential against pH (other concentrations fixed) is a straight line of slope (take $2.303 RT / F = 0.0591 V$ )

Answer: A

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

. The

H X^{+}

term gives

\frac{14 \times 0.0591}{6} lo g [H X^{+}] = - 0.138 pH

. Slope

= - \frac{14}{6} \times 0.0591 = - 0.138 V

per pH unit, steeper than the 0.0946 V of permanganate.