Question

$C{r}_2{O}_7^{2-}(aq,1M)+14H^{+}\left(aq\right)+6e^{-}\to 2C{r}^{3+}(aq,1M)+7H_{2}O$. The standard electrode potential for the following reaction is + 1.33V. what is the potential at $pH=2.0$ ?

• A. + 1.82 V
• B. + 1.99 V
• C. + 1.61 V
• D. + 1.05 V

Answer 1

The correct option is D + 1.05 V

$pH=2.0\Rightarrow \left(H^{+}\right)={10}^{-2}M{Q}_C=\frac{[C{r}^{3+}{]}^2}{\left[C{r}_2{O}_7^{2+}\right][H^{+}{]}^{14}}{Q}_C=\frac{1^2}{1\times ({10}^{-2}{)}^{14}}={10}^{28}{E}_{C{r}_2{O}_7^{2-}/C{r}^{3+}}=E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^0+\frac{0.06}{n}log\frac{1}{Qc}=1.33+\frac{0.06}{6}log{10}^{-28}=1.05$