For the following electrochemical cell at 298K,

$Pt\left(s\right)\left|H_2(g,1bar)\right|\left|H^{+}(aq,1M)\right|\left|M^{4+}\left(aq\right),M^{2+}\left(aq\right)\right|\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}t\left(s\right)$

$E_{cell}=0.092V$ when $\frac{\left[M^{2+}\left(aq\right)\right]}{\left[M^{4+}\left(aq\right)\right]}={10}^x$

Given:  $E_{M^{4+}/M^{2+}}^0=0.151V;2.303\frac{RT}{F}=0.059V$  The value of  $x$  is

$$\begin{gathered} Pt,H_2\left(1bar\right)\left|H+\left(1M\right)\right|\left|M^{+4}\right|\left|M^{+2}\right|Pt \\ {H}_2\to 2H^{+}+2e^{-},E^0=0 \\ {M}^{+4}+2e^{-}\to M^{+2},E^0=0.151 \\ {H}_2+{M^{+4}}_{\left(aqu\right)}\to 2H^{+}+{M^{+2}}_{\left(aqu\right)};E^0=0.151 \end{gathered}$$

$$\begin{gathered} E=E^0-\frac{2.303RT}{F}\log \left(\frac{{\left[H^{+}\right]}^2\left[M^{+2}\right]}{M^{+4}}\right) \\ \Rightarrow 0.092=0.151-\frac{2.303RT}{2F}\log \left(\frac{\left[{M^{+2}}_{\left(aqu\right)}\right]}{\left({M^{+4}}_{\left(aqu\right)}\right)}\right)\Rightarrow 0.059=\frac{2.303RT}{2F}\log \left(\frac{\left[M^{+2}\right]}{\left(M^{+4}\right)}\right) \\ \Rightarrow 0.059=\frac{0.059}{2F}\log \left(\frac{\left[M^{+2}\right]}{\left(M^{+4}\right)}\right)\Rightarrow \left(\frac{\left[M^{+2}\right]}{\left(M^{+4}\right)}\right) \end{gathered}$$

$={10}^2={10}^x\Rightarrow x=2$