Column $I$	Column $II$
$(A)$. Kohlrausch law can calculate	$(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$
$(B)$. Molar conductance $\Lambda_m$	$(Q)$. $\frac{1}{R} \times \frac{l}{A}$
$(C)$. Specific conductance $\kappa$	$(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$
$(D)$. Degree of ionization of weak electrolyte	$(S)$. $\frac{\kappa \times 1000}{M}$

Which of the following options shows the correct matches?

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).

For the fuel cell reaction:
$2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$; $\Delta_fH^o_{298}(H_2O_{(l)}) = -285.5 \ kJ/mol$
What is $\Delta S^o_{298}$ for the given fuel cell reaction?
Given: $O_{2(g)} + 4H^+_{(aq)} + 4e^- \to 2H_2O_{(l)}$; $E^o = 1.23 \ V$

Which of the following statements is not correct?

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: $F = 96500 \text{ C mol}^{-1}$

The electrochemical cell shown below is a concentration cell.
$M \mid M^{2+} (\text{saturated solution of a sparingly soluble salt, } MX_2) \mid M^{2+} (0.001 \ mol \ dm^{-3}) \mid M$
The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \ K$ is $0.059 \ V$.
$1.$ The solubility product $(K_{sp}; \ mol^3 \ dm^{-9})$ of $MX_2$ at $298 \ K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 \ V$):
$(A) \ 1 \times 10^{-15} \quad (B) \ 4 \times 10^{-15}$
$(C) \ 1 \times 10^{-12} \quad (D) \ 4 \times 10^{-12}$
$2.$ The value of $\Delta G \ (kJ \ mol^{-1})$ for the given cell is (take $1 \ F = 96500 \ C \ mol^{-1}$):
$(A) \ -5.7 \quad (B) \ 5.7 \quad (C) \ 11.4 \quad (D) \ -11.4$
Give the answer for question $1$ and $2$.