$\text{ For }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}⟶2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O} {\mathrm{E}}^0+1.33\mathrm{V}.\text{ At }\left[{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right]=4.5$ millimole $\left[{\mathrm{Cr}}^{3+}\right]=15 \mathrm{millimole}, \mathrm{E} \mathrm{is} 1.067\mathrm{V}.\mathrm{The} \mathrm{pH} \mathrm{of}$ the solution is nearly equal to

The Nernst equation for reduction of dichromate is written as 

$E_{cell}=E_{cell}^o-\frac{0.0591}{n}\log \frac{{\left[C{r}^{3+}\right]}^2}{\left[C{r}_2{O}_7^{2-}\right]{\left[H^{+}\right]}^{14}}$

$$\begin{gathered} 1.067=1.33-\frac{0.0591}{6}\log \frac{{(15\times {10}^{-3})}^2}{(4.5\times {10}^{-3}){\left(H^{+}\right)}^{14}} \\ on solving \left[H^{+}\right]= {10}^{-2} \\ pH=2 \end{gathered}$$