Copper reduces $NO_{3}^{-}$ into $NO$ and $NO_{2}$ depending upon the concentration of $HNO_{3}$ in solution. Assuming fixed $[Cu^{2+}]$ and $P_{NO} = P_{NO_{2}} = 1 \ bar$,the $HNO_{3}$ concentration at which the thermodynamic tendency for reduction of $NO_{3}^{-}$ into $NO$ and $NO_{2}$ by copper is the same is $10^{x} \ M$. The value of $2x$ is ...... .
$[Given: E_{Cu^{2+} / Cu}^{\circ} = 0.34 \ V, E_{NO_{3}^{-} / NO}^{\circ} = 0.96 \ V, E_{NO_{3}^{-} / NO_{2}}^{\circ} = 0.79 \ V$ and at $298 \ K, \frac{RT}{F}(2.303) = 0.059]$

For a saturated solution of $Ag_{2}CrO_{4}$ at infinite dilution,$\lambda_{m}^{\infty}(Ag^{+}) = 127 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{m}^{\infty}(CrO_{4}^{2-}) = 246 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$. If the specific conductance of the solution is $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$,calculate the solubility product $(K_{sp})$ of $Ag_{2}CrO_{4}$.

Given:
$\lambda _{ClCH_2COONa} = 224 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{NaCl} = 38.2 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{HCl} = 203 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
What is the value of $\lambda _{ClCH_2COOH}$ in $\Omega ^{-1} \ cm^2 \ gmeq^{-1}$?

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 $2$.

Consider the following electrochemical cell at standard condition: $Au_{(s)} | QH_2, Q | NH_4X(0.01 \ M) || Ag^{+}(1 \ M) | Ag_{(s)}$. Given $E_{\text{cell}} = +0.4 \ V$. The couple $QH_2 / Q$ represents the quinhydrone electrode,and the half-cell reaction is given as: $Q + 2e^- + 2H^+ \rightarrow QH_2$ with $E^o_{Q/QH_2} = +0.7 \ V$. Given: $E^o_{Ag^+/Ag} = +0.8 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$. The $pK_b$ value of the ammonium halide salt $(NH_4X)$ used here is $.........$ (nearest integer).

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 $, kJ$)?