If $63.5 \ g$ of $Cu$ is deposited on the electrode from a $CuSO_4$ solution,what is the number of electrons involved?

Based on the following information,arrange four metals $A, B, C$ and $D$ in order of decreasing ability to act as reducing agents:
$[I]$ Only $A, B$ and $C$ react with $1 \ M \ HCl$ to give $H_2 \ (g)$.
$[II]$ When $C$ is added to solutions of the other metal ions,metallic $B$ and $D$ are formed.
$[III]$ Metal $C$ does not reduce $A^{n+}$.

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

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 $2$.

The molar conductivity of acetic acid solution at infinite dilution is $390 \ S \ cm^2 \ mol^{-1}$. What is the molar conductivity of $0.01 \ M$ acetic acid solution (in $S \ cm^2 \ mol^{-1}$)? (Given: $K_{a}(CH_3COOH) = 1.8 \times 10^{-5}$,assume $1-\alpha \approx 1$)

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 $298 \ K, \frac{RT}{F}(2.303) = 0.059]$