$\mathop {C{H_2}Cl.C{H_2}Cl}\limits_{{\text{ethylene dichloride}}} \xrightarrow{{{\text{aq}}{\text{. }}KOH}}\mathop {C{H_2}OH - C{H_2}OH}\limits_{{\text{glycol}}} $
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Given, $\lambda_{\left(\mathrm{Ag}^{+}\right)}^{\circ}=6 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \lambda_{\left(\mathrm{Br}^{-}\right)}^{\circ}=8 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \lambda_{\left(\mathrm{NO}_{\mathrm{j}}\right)}^{\circ}=7 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$.
$MnO_4^ - \, + \,x{e^ - }\,\xrightarrow[{(Alkaline\,\,medium)}]{}MnO_4^{2 - }$
$MnO_4^ - \, + \,y{e^ - }\,\xrightarrow{{(Acidic\,\,medium)}}M{n^{2 + }}$
$MnO_4^ - \, + \,z{e^ - }\,\xrightarrow{{(Neutral\,\,medium)}}Mn{O_2}$
$\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
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)
(At. No. $Ti = 22, Cr = 24, Co = 27, Zn = 30$)