Question

Calculate the electrode potential at ${25}^{\circ }\text{C}$ of $C{r}^{+3},C{r}_2{O}_7^{2-}$ electrode at $pOH=11$ in a solution of $0.01\text{M}$ both in $C{r}^{3+}$ and $C{r}_2{O}_7^{2-}$.
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
$E^{\circ }=1.33\text{V}$

• A. $0.847$
• B. $0.892$
• C. $0.936$
• D. $0.789$

Answer 1

The correct option is C $0.936$

Given
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O;E^{\circ }=1.33\text{V}$
$E=E^{\circ }-\frac{0.059}{n}\log \frac{{\left[C{r}^{3+}\right]}^2}{\left[C{r}_2{O}_7^{2-}\right]{\left[H^{+}\right]}^{14}}$
$\therefore pH=14-pOH$
$pH=14-11=3$
$\therefore pH=-\log \left[H^{+}\right]$ so, $\left[H^{+}\right]={10}^{-3}$
On substituting all values in formula we get.
$E=1.33-\frac{0.059}{6}\log \frac{[0.01{]}^2}{\left[0.01\right][{10}^{-3}{]}^{14}}$
$E=1.33-\frac{0.059}{6}\log ({10}^{-2}\times {10}^{42})$
$E=1.33-\frac{0.059}{6}\times 40$
$E=1.33-0.0098\times 40$
$E=0.936\text{V}$