Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\Lambda ^ \circ }\,NaCl = {\lambda ^ \circ }N{a^ + } + \lambda C{l^ - }\,\,\,\,\,\,....\left( {\text{i}} \right) \cr & {\Lambda ^ \circ }\,KBr = {\lambda ^ \circ }{K^ + } + {\lambda ^ \circ }B{r^ - }\,\,\,\,\,\,\,\,\,....\left( {{\text{ii}}} \right) \cr & {\Lambda ^ \circ }\,KCl = {\lambda ^ \circ }{K^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,\,....\left( {{\text{iii}}} \right) \cr & {\text{operating (i) + (ii) }} - {\text{ (iii)}} \cr & {\Lambda ^ \circ }NaBr = {\lambda ^ \circ }N{a^ + } + {\lambda ^ \circ }B{r^ - } \cr & = 126 + 152 - 150 \cr & = 128\,\,S\,c{m^2}\,mo{l^{ - 1}} \cr} $$

Amount of charge passed
$$\eqalign{ & = 1.40 \times 200 \cr & = 280\,{\text{Coulombs}} \cr} $$
No. of moles of $$Zn$$  deposited by passing $$280\,C$$  of charge
$$\eqalign{ & = \frac{1}{{2 \times 96500}} \times 280 \cr & = 0.00145 \cr} $$
Molarity of zinc after deposition of zinc
$$\eqalign{ & = 0.180 - \frac{{0.00145 \times 1000}}{{500}} \cr & = 0.180 - 0.0029 \cr & = 0.177\,M \cr} $$

$$E_{H{g^{2 + }}\left| {Hg} \right.}^ \circ > E_{Hg_2^{2 + }\left| {Hg} \right.}^ \circ > E_{F{e^{2 + }}\left| {Fe} \right.}^ \circ > E_{P{b^2}\left| {Pb} \right.}^ \circ .$$

Higher the reduction potential value, greater is the tendency. to undergo reduction i.e., stronger will be the oxidising agent. Thus, $$MnO_4^ - $$  is the strongest oxidising agent.

$$\eqalign{ & {E_{cell}} = E_{cell}^ \circ + \frac{{0.059}}{n}\log \frac{{\left[ {C{u^{ + 2}}} \right]}}{{\left[ {Z{n^{ + 2}}} \right]}} \cr & = 1.10 + \frac{{0.059}}{2}\log \left[ {0.1} \right] \cr & = 1.10 - 0.0295 \cr & = 1.07V \cr} $$

In $${H_2} - {O_2}$$  fuel cell, the combustion of $${H_2}$$  occurs to create potential difference between the two electrodes

$$\Delta G = - nF{E^ \circ }$$
For concentration cell, from Nernst equation,
$$\eqalign{ & E = E_{cell}^ \circ - \frac{{RT}}{{nF}}\,{\text{ln}}\frac{{{C_1}}}{{{C_2}}} \cr & E_{cell}^ \circ = 0.00\,V \cr & E = \frac{{ - RT}}{{nF}}\,{\text{ln}}\frac{{{C_2}}}{{{C_1}}} \cr & E = \frac{{RT}}{{nF}}\,{\text{ln}}\frac{{{C_2}}}{{{C_1}}} \cr & {C_2} > {C_1} \cr} $$
$$E = \frac{{RT}}{{nF}}\,{\text{ln}}\frac{{{C_2}}}{{{C_1}}}$$     $$\left( {R,T,n\,\,{\text{and}}\,\,F\,{\text{are constant}}} \right)$$
$$\eqalign{ & {\text{therefore, }}{E^o}\,\,{\text{is based upon ln}}\frac{{{C_2}}}{{{C_1}}} \cr & \Delta G = - nF{E^ \circ } \cr & \,\,\,\,\,\,\,\,\,\, = - nF \times \frac{{RT}}{{nF}}{\text{ln}}\frac{{{C_2}}}{{{C_1}}} \cr & \,\,\,\,\,\,\,\,\,\, = - RT\,\,{\text{ln}}\frac{{{C_2}}}{{{C_1}}} \cr} $$
Hence, at constant temperature Gibbs free energy $$\Delta G$$  depends upon $${\text{ln}}\frac{{{C_2}}}{{{C_1}}}.$$

According to Debye Huckle onsager equation,
$${\lambda _C} = {\lambda _\infty } + \left( B \right)\sqrt C $$

$$\eqalign{ & {\text{Molarity = }}0.01{\text{ }}M{\text{ ; Resistance = }}40{\text{ }}ohm; \cr & {\text{Cell constant}}\,\,\frac{l}{A} = 0.4\,c{m^{ - 1}}. \cr & {\text{Specific conductivity}}\,\left( \kappa \right) \cr & = \frac{{{\text{cell constant}}}}{{{\text{resistance}}}} \cr & = \frac{{0.4}}{{40}} \cr & = 0.01\,oh{m^{ - 1}}\,c{m^{ - 1}} \cr & {\text{Molar conductance}}\,\left( {{ \wedge _m}} \right) = \frac{{1000\kappa }}{{{\text{Molarity}}}} \cr & = \frac{{1000 \times 0.01}}{{0.01}} \cr & = {10^3}oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}} \cr} $$

Iron coated with zinc does not get rusted even if cracks appear on the surface because $$Zn$$  will take part in redox reaction not $$Fe$$  as $$Zn$$  is more reactive than $$Fe.$$  If iron is coated with tin and cracks appear on the surface, $$Fe$$  will take part in redox reaction because $$Sn$$  is less reactive than $$Fe.$$