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The standared electrode potential for the following reaction is +1.33 V. What is potential at pH=2.0?
${Cr}_{2} O_{7}^{2 -} (aq, 1 M) + 14 H^{+} (aq) + 6 e^{-} \to 2 {Cr}^{3 +} (aq, 1 M) + 7 H_{2} O (l)$

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+1.608 V

+1.990 V

+1.0542V

+1.820V

answer is D.

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Detailed Solution

$\begin{array}{l} E = E^{0} + (- 0 .0591 / n) \log Q \\ = 1 .33 - (0 .0591 / 6) \log (1)^{2} / (1) \times {(10^{- 2})}^{14} \\ = 1 .33 - 0 .2758 = 1 .0542 V \end{array}$

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