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	$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
$(i)$ $2Cr_{(s)} + 3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Cd_{(s)}$
$(ii)$ $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)}$
Calculate the $\Delta_r G^\Theta$ and equilibrium constant of the reactions.

At $298 \ K$,the equilibrium constant is $2 \times 10^{15}$ for the reaction:
$Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$
The equilibrium constant for the reaction $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$ is $x \times 10^{-8}$. The value of $x$ is (Nearest Integer).

Consider the following two half-cell reactions:
$CO_2 + 6H^+ + 6e^- \rightarrow CH_3OH + H_2O$ $(E^{\ominus} = 0.02 \text{ V})$
$\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O$ $(E^{\ominus} = 1.23 \text{ V})$
$A$ fuel cell was set up such that the cell operates under standard conditions. The fuel cell works with $80\%$ efficiency. If the work derived from the cell using $1 \text{ mol}$ of $CH_3OH$ is used to compress an ideal gas isothermally against a constant pressure of $1 \text{ kPa}$, then the change in the volume of the gas, $\Delta V =$ . . . . . . $\text{m}^3$. (nearest integer) Given: $F = 96500 \text{ C mol}^{-1}$

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: $Hg_2^{2+} \to Hg^{2+} + Hg_{(l)}$

Consider the change in oxidation state of Bromine corresponding to different $emf$ values as shown in the diagram below:
$BrO_4^{-}$ $\xrightarrow{1.82 \ V} BrO_3^{-}$ $\xrightarrow{1.5 \ V} HBrO$ $\xrightarrow{1.0652 \ V} Br_2$ $\xrightarrow{1.595 \ V} Br^{-}$
Then the species undergoing disproportionation is: