At $298 \mathrm{K}$, a $1$ litre solution containing $10 \mathrm{mmol}$ of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ and $100\mathrm{mmol}$ of ${\mathrm{Cr}}^{3+}$ shows a $\mathrm{pH}$ of $3.0$.

Given : ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}\to {\mathrm{Cr}}^{3+};{\mathrm{E}}^0=1.330 \mathrm{V}$ and $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V}$

The potential for the half cell reaction is $\mathrm{x}\times {10}^{-3} \mathrm{V}$ . The value of $\mathrm{x}$ is _____ .

The balanced half-reaction can be written as follows,

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

Using Nernst equation

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.059}{\mathrm{n}}\log \frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}$

For the above reaction, number of electrons involved n = 6.

$\mathrm{E}=1.33-\frac{0.059}{6}\log \frac{{\left(0.1\right)}^2}{\left({10}^{-2}\right){\left({10}^{-3}\right)}^{14}}$

$\mathrm{E}=1.33-\frac{0.059}{6}\times 42=0.917$

$\mathrm{E}=917\times {10}^{-3}$

$\mathrm{x}=917$.