Consider the following redox reaction:
$$
\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2
$$
The standard reduction potentials are given as below $\left(\mathrm{E}_{\text {red }}^{\circ}\right)$
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}
\end{aligned}
$$
If the equilibrium constant of the above reaction is given as $K_{e q}=10^x$, then the value of $x=$ (Nearest integer)
