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 rusting of iron takes place as follows. Calculate $\Delta G^o$ for the net process in $kJ \ mol^{-1}$.
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$
$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$

At $25^\circ \text{C}$,given the following data: $Ag_{(s)} + I^-_{(aq)} \rightarrow AgI_{(s)} + e^-$,$E^o = 0.152 \ V$; $Ag_{(s)} \rightarrow Ag^+_{(aq)} + e^-$,$E^o = -0.800 \ V$. What is the value of $\log \ K_{sp}$ for $AgI$? (where $K_{sp} = \text{solubility product}$)

If the $\Delta G^o$ for the cell reaction $AgCl_{(s)} + \frac{1}{2} H_{2(g)} \rightarrow Ag_{(s)} + H^+ + Cl^-$ is $-21.52 \, kJ$,what will be the $\Delta G^o$ for the reaction $2AgCl_{(s)} + H_{2(g)} \rightarrow 2Ag_{(s)} + 2H^+ + 2Cl^-$ (in $, kJ$)?

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 $2$.

Match the following:

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+}|Cu$	$(II)$ $0.059$
$(C)$ $Zn|Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$