For ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-\mathrm{yields}}⟶2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O},E^0=1.33 \mathrm{V}$ at $\left[{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right]=4.5$ millimole, $\left[{\mathrm{Cr}}^{3+}\right]=1.5$ millimole and $\mathrm{E}=1.067 \mathrm{V}$, then calculate the $\mathrm{pH}$ of the solution.

According to the Nernst's equation:

$\mathrm{E} = {\mathrm{E}}^{°}-\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{lnQ}$

So, by given value of concentrations the above formula can be modified as

$\mathrm{E} = {\mathrm{E}}^{°}-\frac{0.059}{\mathrm{nF}}\log \frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{\left[{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}\right]{\left[{\mathrm{H}}^{+}\right]}^{14}}$

$1.067 = 1.33-\frac{0.06}{6\times 96500}\log \frac{{\left(15\times {10}^{-3}\right)}^2}{4.5\times {10}^{-3}{\left[{\mathrm{H}}^{+}\right]}^{14}}$

$\left[{\mathrm{H}}^{+}\right] = {10}^{-2} \mathrm{M}$

Hence, the concentration of ${\mathrm{H}}^{+}$ ion will be

$$\begin{gathered} \mathrm{pH} =-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left[{10}^{-2}\right]=-(-2)\times {\log }_{10}\left(10\right) \\ \mathrm{pH} = 2 \end{gathered}$$