MCQ
Above conversion can be achieved by
- A$Zn(Hg), HCl$
- ✓$NH_2 - NH_2 /KOH/ \Delta$
- C$LiAlH_4$
- D$H_2/Ni$
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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)