$$\eqalign{ & Moles{\text{ of }}NaOH{\text{ formed}} = \frac{{600 \times 1}}{{100}} = 0.6 \cr & {\text{or }}geq{\text{ of }}NaOH = 0.6{\text{ }}geq{\text{ of }}Cu{\text{ deposited}} \cr & {\text{ = }}\frac{{31.8}}{{\frac{{63.6}}{2}}} = 1.0 \cr & \therefore \,\,{\text{Current efficiency}} = \frac{{0.6 \times 100}}{1} = 60\% \cr} $$

$$\eqalign{ & {E_{cell}} = E_{cell}^ \circ - \frac{{0.0592}}{2}{\text{log}}\frac{{\left[ {Hg_2^{2 + }\left( {aq} \right)} \right]}}{{{{\left[ {A{g^ + }\left( {aq} \right)} \right]}^2}}} \cr & = \left( {0.80 - 0.785} \right) - \frac{{0.0592}}{2}{\text{log}}\frac{{{{10}^{ - 1}}}}{{{{\left( {{{10}^{ - 3}}} \right)}^2}}} \cr & = - 0.133\,V \cr} $$
hence backward reaction is spontaneous.

$$\eqalign{ & A{g^ + } + {e^ - } \to Ag \cr & {E_{{\text{cell}}}} = E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{1}\log \frac{1}{{\left[ {A{g^ + }} \right]}} \cr & {E_{{\text{cell}}}} = 0.80 - \frac{{0.0591}}{1}\log \frac{1}{{0.1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.80 - 0.0591 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.741\,V \cr} $$

TIPS/FORMULAE:
$$\eqalign{ & {\text{Use Nernst's equation; }} \cr & {\text{Cell}}\,{\text{reaction}}\,:\,Zn + F{e^{2 + }} \to Z{n^{2 + }} + Fe \cr & {\text{Using Nernst equation }} \cr & {E_{cell}} = {E^o}_{cell} - \frac{{0.0591}}{n}\log \left[ {\frac{{Z{n^{2 + }}}}{{F{e^{2 + }}}}} \right] \cr & E = {E^o} - \frac{{0.0591}}{2}\log \frac{{{{10}^{ - 2}}}}{{{{10}^{ - 3}}}} \cr & {E^o} = 0.2905 + \frac{{0.0591}}{2} = 0.32 \cr} $$
$${\text{or}}\,\,0.32 = \frac{{0.0591}}{2}\log {K_{eq.\,}}\,\,{\text{or}}$$       $${K_{eq}} = {10^{\frac{{0.32}}{{0.0295}}}}$$

$$A{g^ + } + \mathop {{e^ - }}\limits_{1\,F} \to \mathop {Ag}\limits_{108\,g} $$
Amount of $$Ag$$  deposited by $$9650\,C$$
$$\eqalign{ & = \frac{{108}}{{96500}} \times 9650 \cr & = 10.8\,g \cr} $$

$$\eqalign{ & E = {E^ \circ } - \frac{{0.0591}}{n}\log \,Q \cr & {E^ \circ } = 0\,\,{\text{for all concentration cells}} \cr & \,\,\,\,\,\,\,\, = 0 - \frac{{0.0591}}{1}\log \left( {\frac{{0.01}}{{0.05}}} \right) \cr & \,\,\,\,\,\,\,\, = 0.0413\,V \cr} $$

$$\eqalign{ & {\text{A}}{{\text{g}}^ + } + \mathop {{e^ - }}\limits_{1F} \to \mathop {Ag}\limits_{108\,g} \cr & 1\,F = 1\,mole\,{\text{of electrons}} = 96500\,C \cr & 0.01F = 1.08\,g\,Ag;Ag\,{\text{left}} = 1.08 - 1.08 = 0\,g \cr} $$

$$\eqalign{ & {\text{By Faraday's}}\,{I^{st}}\,{\text{law of electrolysis,}} \cr & \frac{W}{E} = \frac{Q}{{96500}}{\text{(where }}Q{\text{ = it = charge of ion)}} \cr & {\text{We know that no}}{\text{. of gram equivalent}} \cr & = \frac{W}{E} = \frac{{{\text{it}}}}{{96500}} = \frac{{1 \times 965}}{{96500}} = \frac{1}{{100}} \cr & \left( {{\text{where}}\,\,i = 1\,A,t = 16 \times 60 + 5 = 965\,\sec .} \right) \cr & {\text{Since, we know that}} \cr & {\text{Normality}} \cr & = \frac{{{\text{No}}{\text{. of gram equivalent}}}}{{{\text{Volume (in litre)}}}} \cr & = \frac{{\frac{1}{{100}}}}{1} \cr & = 0.01\,N \cr} $$

$${E_{red}} = E_{red}^ \circ + \frac{{0.591}}{n}{\text{log}}\left[ {{M^{n + }}} \right]$$
Lower the concentration of $${{M^{n + }},}$$  lower is the reduction potential.
Hence order of reduction potential is : $$Q > R > S > P$$

When an aqueous solution of $$NaCl$$  is electrolysed, hydrogen is liberated at cathode. Specific conductivity and molar conductivity are different terms. Silver nitrate solution cannot be stored in a copper container as silver will get precipitated because of high reactivity of $$Cu$$  than $$Ag.$$ 
The addition of liquid bromine to iodide solution turns it violet.
$$\mathop {B{r_{2\left( l \right)}} + 2I_{\left( {aq} \right)}^ - }\limits_{{\text{Reddish - brown}}} \to 2Br_{\left( {aq} \right)}^ - + \mathop {{I_{2\left( l \right)}}}\limits_{{\text{Violet}}} $$