The following cell is found to have EMF equal $\text{ to zero. }\mathrm{Pt},{\mathrm{H}}_2\left(x\mathrm{atm}\right)\left|0.01{\mathrm{MH}}^{+}\parallel 0.1{\mathrm{MH}}^{+}\right|{\mathrm{H}}_2$ (y atm), Pt The ratio x / y is 

$$\begin{gathered} \mathrm{the} \mathrm{overall} \mathrm{reaction} \mathrm{is} \\ {\mathrm{H}}_{2 (\mathrm{x} \mathrm{atm} )}+\hspace{0.17em}2\hspace{0.17em}{\mathrm{H}}_{(0.1\mathrm{M}) \mathrm{cathode}}^{+}\iff {\mathrm{H}}_{2 (\mathrm{y} \mathrm{atm})\hspace{0.17em}}+\hspace{0.17em}2{\mathrm{H}}_{(0.01\mathrm{M})\mathrm{anode}}^{+} \\ \mathrm{number} \mathrm{of} \mathrm{electron} \mathrm{involved} , \mathrm{n}= 2 \\ \mathrm{EMF}\hspace{0.17em}= \frac{0.0591}{2}\log \frac{{[ {\mathrm{H}}_{\mathrm{anode}}^{+}]}^2 \mathrm{X} \mathrm{P}{{ }_{\mathrm{H}2}}_{\mathrm{y}}}{{\left[{\mathrm{H}}_{\mathrm{cathode}}^{+}\right]}^{2 } \mathrm{X}\hspace{0.17em}{\mathrm{P}}_{\mathrm{H}2 { }_{\mathrm{x}}}} \\ \mathrm{At} \mathrm{equilibrium} \mathrm{EMF}\hspace{0.17em}= 0 \\ 0=\frac{0.0591}{2} \log \frac{{(0.01)}^2}{{(0.1)}^2} \mathrm{X}\hspace{0.17em}\frac{\mathrm{y}}{\mathrm{x}} \\ \mathrm{on} \mathrm{solving} \\ \frac{\mathrm{x}}{\mathrm{y}}= {10}^{-2}=0.01 \end{gathered}$$