All the energy released from the reaction $X \rightarrow Y, \Delta_{r}G^0 = -193 \ kJ \ mol^{-1}$ is used for oxidizing $M^{+}$ as $M^{+} \rightarrow M^{3+} + 2e^-, E^0 = -0.25 \ V$. Under standard conditions,the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is $\left[F = 96500 \ C \ mol^{-1}\right]$.

Match the column $I$ with column $II$ and mark the appropriate choice.

Column $I$	Column $II$
$A$. Kohlrausch law	$i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$
$B$. Molar Conductivity	$ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$
$C$. Degree of Dissociation	$iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$
$D$. Dissociation Constant	$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

Consider the following redox reaction :
$MnO_4^{-} + H^{+} + H_2C_2O_4 \rightleftharpoons Mn^{2+} + H_2O + CO_2$
The standard reduction potentials are given as below $(E_{red}^{\circ})$ :
$E_{MnO_4^{-} / Mn^{2+}}^{\circ} = +1.51 \ V$
$E_{CO_2 / H_2C_2O_4}^{\circ} = -0.49 \ V$
If the equilibrium constant of the above reaction is given as $K_{eq} = 10^x$,then the value of $x = $ . . . . . . (nearest integer).

Given $E_{Fe^{3+}/Fe^{2+}}^{\circ} = +0.76 \ V$ and $E_{I_{2}/I^{-}}^{\circ} = +0.55 \ V$. The equilibrium constant for the reaction taking place in a galvanic cell consisting of the above two electrodes is (Given $\frac{2.303 \ RT}{F} = 0.06 \ V$)

Find the solubility product $(K_{sp})$ of a saturated solution of $Ag_2CrO_4$ in water at $298 \ K$,if the $emf$ of the cell $Ag | Ag^{+} (\text{satd. } Ag_2CrO_4 \text{ solution}) || Ag^{+} (0.1 \ M) | Ag$ is $0.591 \ V$ at $298 \ K$.

Given ${E^o}_{Hg^{2+}|Hg_2^{2+}} = 0.9 \ V$ and ${E^o}_{Hg_2^{2+}|Hg} = 0.8 \ V$. Calculate the value of ${\Delta _r}{G^o}$ at $25 \ ^oC$ in $kJ/mol$ for the reaction: $Hg_2^{2+} \to Hg^{2+} + Hg_{(l)}$