The standard electrode potential for the foll | vedclass

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(D) The Nernst equation for the reaction is: $E = E^{\circ} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}}$Given $E^{\circ} =

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$+1.054 \ V$

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(D) The Nernst equation for the reaction is: $E = E^{\circ} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}}$Given $E^{\circ} = 1.33 \ V$,$n = 6$,$[Cr^{3+}] = 1 \ M$,$[Cr_2O_7^{2-}] = 1 \ M$,and $pH = 2.0$,so $[H^{+}] = 10^{-2} \ M$.Substituting the values: $E = 1.33 - \frac{0.0591}{6} \log \frac{1^2}{1 	imes (10^{-2})^{14}}$$E = 1.33 - \frac{0.0591}{6} \log (10^{28})$$E = 1.33 - \frac{0.0591}{6} 	imes 28$$E = 1.33 - 0.2758 = 1.0542 \ V$.

The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

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