$$\eqalign{ & { \wedge _m} = \frac{{\kappa \times 1000}}{M} \cr & \,\,\,\,\,\,\,\,\, = \frac{{1.52 \times {{10}^{ - 2}} \times 1000}}{{0.15}} \cr & \,\,\,\,\,\,\,\,\, = 101\,{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} \cr} $$

$$\eqalign{ & C{u^{2 + }} + {e^ - } \to C{u^ + }; \cr & E_1^ \circ = 0.15\,V;\,\,\Delta G_1^ \circ = - {n_1}E_1^ \circ F \cr & C{u^{2 + }} + 2e \to Cu; \cr & E_2^ \circ = 0.34\,V;\,\Delta G_2^ \circ = - {n_2}E_2^ \circ F \cr & {\text{On subracting eq}}{\text{.(i) from eq}}{\text{. (ii) we get}} \cr & C{u^ + } + {e^ - } \to Cu;\,\Delta {G^ \circ } = \Delta G_2^ \circ - \Delta G_1^ \circ \cr & - n{E^ \circ }F = - \left( {{n_2}{E^ \circ }F - {n_1}E_1^ \circ F} \right) \cr & {E^ \circ } = \frac{{{n_2}E_2^ \circ F - {n_1}E_1^ \circ F}}{{nF}} \cr & = \frac{{2 \times 0.34 - 0.15}}{1} \cr & = 0.53\,V \cr} $$

$$\eqalign{ & 2{H^ + } + 2{e^ - } \to {H_2} \cr & {E_H}\left( {Eq.\,wt} \right) = \frac{2}{2} = 1 \cr & 1\,g = \frac{{22400}}{2} \cr & \,\,\,\,\,\,\,\, = 11200\,mL\,\left( {STP} \right) \cr & {\text{Total charge passed}} \cr & = \frac{{96500 \times 112}}{{11200}} \cr & = 965 \cr & Q = It = 965 \cr & I = \frac{{965}}{{965}} \cr & = 1\,amp. \cr} $$

Since, $$HCl$$  is strong acid and dissociates completely. Hence, it conducts electricity best in its aqueous solution.

Kohlrausch Law, $${ \wedge _{eq}}\left( {NaF} \right) + { \wedge _{eq}}\left( {HCl} \right) - { \wedge _{eq}}\left( {NaCl} \right) = { \wedge _{eq}}\left( {HF} \right)$$

$$ \wedge _m^\infty \left( {NaBr} \right) = $$    $$ \wedge _m^\infty \left( {NaCl} \right) + \wedge _m^\infty KBr - \wedge _m^\infty \left( {KCl} \right)$$
$$\lambda _m^\infty \left( {N{a^ + }} \right) + \lambda _m^\infty \left( {B{r^ - }} \right) $$     $$ = 126.5 + 151.5 - 150$$
$$\lambda _m^\infty \left( {N{a^ + }} \right) = 50\,S\,c{m^2}e{q^{ - 1}}.$$

Sodium chloride in water dissociates as
$$\eqalign{ & NaCl \rightleftharpoons N{a^ + } + C{l^ - } \cr & {H_2}O \rightleftharpoons {H^ + } + O{H^ - } \cr} $$
When electric current is passed through this solution using platinum electrodes, $$N{a^ + }$$  and $${H^ + }$$ move towards cathode whereas $$C{l^ - }$$ and $$O{H^ - }$$  $$ions$$  move towards anode.
At cathode
$$\eqalign{ & {H^ + } + {e^ - } \to H \cr & H + H \to {H_2} \cr} $$
At anode
$$\eqalign{ & C{l^ - } \to Cl + {e^ - } \cr & Cl + Cl \to C{l_2} \cr} $$
If mercury is used as cathode, $${H^ + }$$ $$ions$$  are not discharged at mercury cathode because mercury has a high hydrogen over voltage. $$N{a^ + }$$  $$ions$$  are discharged at cathode in preference of $${H^ + }$$ $$ions$$  yielding sodium, which dissolves in mercury to form sodium amalgam.

$$\eqalign{ & E_X^ \circ = - 1.2\,V, \cr & E_Y^ \circ = + 0.5\,V, \cr & E_Z^ \circ = - 3.0\,V, \cr} $$
$$\therefore \,\,Z > X > Y\,$$   [ $$\because $$  higher the reduction potential, lesser the reducing power. ]

The metal with higher negative standard reduction potential, have higher tendency to get reduced.
$$\eqalign{ & Z{n^{2 + }} + 2{e^ - } \to Zn;\,{E^ \circ } = - 0.76\,V \cr & F{e^{2 + }} + 2{e^ - } \to Fe;\,{E^ \circ } = - 0.44\,V \cr} $$
Here, in galvanised iron, $$Zn$$  has higher negative reduction potential means $$Zn$$  takes electrons given by iron and itself gets reduced.
Thus, $$Zn$$  works as anode and protects iron from rusting by making iron as cathode.

$$\frac{4}{3}Al + {O_2} \to \frac{2}{3}A{l_2}{O_3},$$     $$\Delta G = - 827\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{ & 12{e^ - } + 6\,{O^{2 - }} \to 3\,{O_2} \cr & 4\,A{l^{3 + }} \to Al + 12{e^ - } \cr & {\text{or}}\,\,\,\frac{4}{3}A{l^{3 + }} \to \frac{4}{3}Al + 4{e^ - } \cr & 4{e^ - } + 2{O^{ - 2}} \to {O_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta G = - nE{F^ \circ }\,\,\,\left( {n = 4} \right) \cr & - 827 \times {10^3}J = - 4 \times {E^ \circ } \times 96500 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E = \frac{{827 \times {{10}^3}}}{{4 \times 96500}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = 2.14\,V \cr} $$