Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & E_{\frac{{Z{n^{2 + }}}}{{Zn}}}^ \circ = - 0.76\,V \cr & E_{\frac{{F{e^{2 + }}}}{{Fe}}}^ \circ = - 0.44\,V \cr & {\text{Cell reaction,}} \cr & F{e^{2 + }} + Zn \to Z{n^{2 + }} + Fe \cr & E_{{\text{cell}}}^ \circ = E_{\left( {{\text{cathode}}} \right)}^ \circ - E_{\left( {{\text{anode}}} \right)}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\, = - 0.44 - \left( { - 0.76} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = + \,0.32\,V \cr} $$

$$\eqalign{ & BaC{l_2} \to B{a^{2 + }} + 2C{l^ - } \cr & \wedge _{BaC{l_2}}^ \circ = \wedge _{B{a^{2 + }}}^ \circ + 2 \wedge _{C{l^ - }}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 127 + \left( {2 \times 76} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 279\,oh{m^{ - 1}}\,c{m^2}\,e{q^{ - 1}} \cr & {\text{Equivalent conductivity}} \cr & = \frac{{279}}{2} = 139.5\,oh{m^{ - 1}}\,c{m^2}\,e{q^{ - 1}} \cr} $$

More the reduction potential, more is the deposition of metals at cathode. Cation having $${E^0}$$  value less than $$ - 0.83\,V$$  ( reduction potential of $${H_2}O$$  ) will not deposit from aqueous solution. Hence correct order of A deposition of the metal at the cathode is $$Ag > Hg > Cu$$

Corrosion of iron can be minimized by forming an impermeable barrier at its surface.

$$\eqalign{ & { \wedge _m} = \frac{{\kappa \times 1000}}{M} = \frac{{0.0110 \times 1000}}{{0.05}} \cr & = 220\,S\,c{m^2}\,mo{l^{ - 1}} \cr} $$

Electrochemical cell,
$$A\left| {{A^ + }\left( {xM} \right)} \right|\left| {{B^ + }\left( {yM} \right)} \right|B$$
The $$EMF$$  of cell is $$+ 0.20$$ $$V.$$  So, cell reaction is possible. The half-cell reactions are given as follows
$$\eqalign{ & {\text{(i) At negative pole}} \cr & A \to {A^ + } + {e^ - }\,\,\,{\text{(oxidation)}} \cr & {\text{(ii) At positive pole}} \cr & {B^ + } + {e^ - } \to B\,\,\,{\text{(reduction)}} \cr & {\text{Hence, cell reaction is}} \cr & A + {B^ + } \to {A^ + } + B,\,\,E_{cell}^ \circ = + 0.20\,V \cr} $$

Since, $$22400$$ $$mL$$  volume is occupied by $$1$$ $$mole$$  of $${O_2}$$  at $$STP.$$
$$\eqalign{ & {\text{Thus,}}\,5600\,mL\,{O_2}\,\,{\text{means}} \cr & = \frac{{5600}}{{22400}}mol\,{O_2} \cr & = \frac{1}{4}mol\,{O_2} \cr & \therefore \,\,{\text{Weight of}}\,{O_2} = \frac{1}{4} \times 32 = 8\,g \cr & {\text{According to problem,}} \cr & {\text{Equivalents of }}Ag = {\text{Equivalents of}}\,{O_2} \cr & = \frac{{{\text{Weight of }}Ag}}{{{\text{Equivalent weight of }}Ag}} \cr & = \frac{{{W_{{O_2}}}}}{{{\text{Equivalent weight of }}{O_2}}} \cr & \frac{{{W_{Ag}}}}{{\frac{{{M_{Ag}}}}{1}}} = \frac{{{W_{{O_2}}}}}{{\frac{{{M_{{O_2}}}}}{4}}} \cr & \therefore \,\,\frac{{{W_{Ag}}}}{{108}} \times 1 = \frac{8}{{32}} \times 4 \cr & \,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,2{H_2}O \to {O_2} + 4{H^ + } + 4{e^ - }} \right] \cr & \Rightarrow {W_{Ag}} = 108\,g \cr} $$

$$\eqalign{ & A{g^ + }\mathop { + {e^ - } \to }\limits_{1\,mol \equiv 1\,F} Ag \cr & C{u^{2 + }}\mathop { + 2{e^ - } \to }\limits_{2\,mol \equiv 2\,F} Cu \cr & F{e^{3 + }}\mathop { + 3{e^ - } \to }\limits_{3\,mol \equiv 3\,F} Fe \cr} $$
Grams of $$Ag$$  liberated $$ = \frac{{108}}{1} \times 5 = 540\,g$$
Grams of $$Cu$$  liberated $$ = \frac{{63.5}}{2} \times 5 = 158.75\,g$$
Grams of $$Fe$$  liberated $$ = \frac{{56}}{2} \times 5 = 93.3\,g$$

$$\eqalign{ & {\text{Cell constant}} \cr & = \kappa \times R \cr & = 0.212\,oh{m^{ - 1}}\,c{m^{ - 1}} \times 55\,ohm \cr & = 11.66\,c{m^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Here}}\,\,n = 4,\,and\left[ {{H^ + }} \right] = {10^{ - pH}} = {10^{ - 3}} \cr & {\text{Applying Nernst equation}} \cr & E = {E^o} - \frac{{0.059}}{n}\log \frac{{{{\left[ {F{e^{2 + }}} \right]}^2}}}{{{{\left[ {{H^ + }} \right]}^4}p{o_2}}} \cr & = 1.67 - \frac{{0.059}}{4}\log \frac{{{{\left( {{{10}^{ - 3}}} \right)}^2}}}{{{{\left( {{{10}^{ - 3}}} \right)}^4} \times 0.1}} \cr & = 1.67 - \frac{{0.03}}{2}\log {10^7} \cr & = 1.67 - 0.105 \cr & = 1.565\,V \cr} $$