Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & m = \frac{{E.wt \times Q}}{{96500}}; \cr & \therefore \,\,E.wt = \frac{{w \times 96500}}{Q} \cr & = \frac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}} \cr & = 60.3 \cr & {\text{Oxidation state}} = \frac{{At\,wt.}}{{Eq.\,wt.}} \cr & = \frac{{177}}{{60.3}} \cr & = 3 \cr} $$

$$\eqalign{ & {H^ + } + {e^ - } \to \frac{1}{2}{H_2};E = {E^o} - \frac{{0.059}}{1}\log \frac{{P_{{H_2}}^{\frac{1}{2}}}}{{\left[ {{H^ + }} \right]}} \cr & {\text{Now}}\,{\text{if}}\,{P_{{H_2}}} = 2\,{\text{atm}}\,{\text{and}}\,\left[ {{H^ + }} \right] = 1M \cr & {\text{then}}\,E = 0 - \frac{{0.059}}{1}\log \frac{{{2^{\frac{1}{2}}}}}{1} = - 2 \cr} $$

$${H_2}O$$  is more readily reduced at cathode than $$N{a^ + }.$$  It is also more readily oxidized at anode than $$SO_4^{2 - }.$$  Hence, the electrode reactions are
$$\eqalign{ & 2{H_2}O + 2{e^ - } \to {H_2} \uparrow + 2O{H^ - }\left[ {{\text{at cathode}}} \right] \cr & {H_2}O \to \frac{1}{2}{O_2} \uparrow + 2{H^ + } + 2{e^ - }\left[ {{\text{at anode}}} \right] \cr} $$

$$\left( {\text{A}} \right)\log \,K = \frac{{nFE_{{\text{cell}}}^ \circ }}{{2.303\,RT}}$$
$$\left( {\text{B}} \right){E_{{\text{cell}}}} = E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{2}\log \frac{{\left[ {{a_1}} \right]}}{{\left[ {{a_2}} \right]}}$$
$$\left( {\text{D}} \right)$$ Expression at $$298\,K,$$  is $$E = {E^ \circ } + \frac{{0.0591}}{n}\log \left[ {{M^{n + }}} \right]$$

$$AgN{O_3}\left( {aq} \right) + KCl\left( {aq} \right) \to AgCl\left( s \right) + KN{O_3}\left( {aq} \right)$$
Conductivity of the solution is almost compensated due to formation of $$KN{O_3}\left( {aq} \right).$$ However, after at end point, conductivity increases more rapidly due to audition of excess $$AgN{O_3}$$  solution.

$$\eqalign{ & {\text{For}}\,\,Z{n^{2 + }} \to Zn \cr & {E_{\frac{{Z{n^{2 + }}}}{{Zn}}}} = E_{\frac{{Z{n^{2 + }}}}{{Zn}}}^ \circ - \frac{{2.303RT}}{{nF}}\log \frac{{\left[ {Zn} \right]}}{{\left[ {Z{n^{2 + }}} \right]}} \cr & = - 0.76 - \frac{{0.06}}{2}\log \frac{1}{{\left[ {0.1} \right]}} \cr & = - 0.76 - 0.03 \cr & {E_{\frac{{Z{n^{2 + }}}}{{Zn}}}} = - 0.79\,V \cr} $$

$$\eqalign{ & {\text{From second law of Faraday}} \cr & \,\,\,\,\,\,\frac{{{m_{Al}}}}{{{m_H}}} = \frac{{{E_{Al}}}}{{{E_H}}} \cr & \,\,\,\,\,\,\frac{{4.5}}{{{m_H}}} = \frac{{\frac{{27}}{3}}}{1} \cr & {\text{or}}\,\,\,{m_H} = 0.5\,g \cr & \because \,\,{\text{Volume of}}\,\,2\,g\,{H_2}\,{\text{at}}\,STP = 22.4\,L \cr & \therefore \,\,{\text{Volume of}}\,\,0.5\,g\,{H_2}\,{\text{at}}\,STP \cr & \,\,\,\,\,\, = \frac{{22.4 \times 0.5}}{2}L \cr & \,\,\,\,\,\, = 5.6\,L \cr} $$

Faraday's law is used in calculating the amount of metal deposited on cathode.

Substance which have higher reduction potential are stronger oxidising agent.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - },$$       $${E^ \circ } = - 0.35\,V$$
$$\eqalign{ & F{e^{2 + }} \to F{e^{3 + }} + {e^ - },\,{E^ \circ } = - 0.77\,V \cr & \because \,\,E_{oxi}^ \circ = - E_{red}^ \circ \cr} $$
$$\therefore \,{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }},$$        $${E^ \circ } = 0.35\,V$$
     $$F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,{E^ \circ } = 0.77\,V$$
Hence, $$F{e^{3 + }}$$  has maximum tendency to reduced, so it is the strongest oxidising agent.

$$\eqalign{ & {\text{Using Faraday's second law of electrolysis,}} \cr & \frac{{{\text{Weight of }}Cu{\text{ deposited}}}}{{{\text{Weight of }}Ag{\text{ deposited}}}} = \frac{{Equ.\,wt.\,{\text{of}}\,Cu}}{{Equ.\,wt.\,{\text{of}}\,Ag}} \cr & \Rightarrow \frac{{{w_{Cu}}}}{{0.18}} = \frac{{63.5}}{2} \times \frac{1}{{108}} \cr & \Rightarrow {w_{Cu}} = \frac{{63.5 \times 18}}{{2 \times 108 \times 100}} \cr & = 0.0529\,g. \cr} $$