Reason : Transparent bead $(NaBO_2 + B_2O_3)$ forms coloured bead with coloured cation.
$N{{a}_{2}}{{B}_{4}}{{O}_{7}}\,\xrightarrow{740\,{{\,}^{o}}C}\underbrace{\underset{\begin{smallmatrix}
\,\,\,\text{sodium } \\
\text{metaborate}
\end{smallmatrix}}{\mathop{2NaB{{O}_{2}}}}\,\,+\,\,\underset{\begin{smallmatrix}
\text{ boric} \\
anhydride
\end{smallmatrix}}{\mathop{{{B}_{2}}{{O}_{3}}}}\,}_{\text{glassy bead}}$
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$\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
The standard reduction potentials are given as below $\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)$
$\mathrm{E}_{\mathrm{MmO}_4^{-} / \mathrm{Mm}^{2+}}^{\circ}=+1.51 \mathrm{~V}$
$\mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}$
If the equilibrium constant of the above reaction is given as $K_{\text {eq }}=10^x$, then the value of $x=$____ (nearest integer)