Question

In acidic medium $C{r}_2{O}_7^{-2}$ is an oxiding agent
$C{r}_2{O}_7^{-2}+14H^{+}+6e^{-}\leftarrow 2C{r}^{+3}+7H_{2}O$
$[E_{C{r}_2{O}_7^{-2}/C{r}^{+3}}^{{}^{\circ }}=1.33V]$
The electrode potential of half cell at $pH=1$ keeping concentration of other species unity is:

• A. 2.1574 V
• B. 0.5026 V
• C. 1.1921 V
• D. 1.4679 V

Answer 1

The correct option is D 1.4679 V

For given cell, $E=E^{\circ }-\frac{0.059}{6}\log \frac{{\left[C{r}^{3+}\right]}^2}{\left[H^{+}{]}^{14}\right[C{r}_2{O}_7^{2-}]}$

$\because \left[C{r}^{3+}\right]=\left[C{r}_2{O}_7^{2-}\right]=1,\left[H^{+}\right]={10}^{-1}$

$\therefore E=1.33-\frac{0.059}{6}\log \frac{1}{{10}^{-14}}=1.4679V$