Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$R = 100\,\Omega ,\,\kappa = \frac{1}{R}\left( {\frac{1}{a}} \right),$$      $$\frac{1}{a}\left( {{\text{cell}}\,{\text{constant}}} \right) = 1.29 \times 100{m^{ - 1}}$$
$${\text{Given,}}\,R = 520\,\Omega ,\,C = 0.2M,$$       $$\mu {\text{(molar conductivity) = ?}}$$
$$\mu {\text{ = }}\kappa \times {\text{V}}$$     ( $$\kappa $$  can be calculated as $$\kappa = \frac{1}{R}\left( {\frac{1}{a}} \right)$$    now cell constant is known. )
$$\eqalign{ & {\text{Hence,}} \cr & \mu = \frac{1}{{520}} \times 129 \times \frac{{1000}}{{0.2}} \times {10^{ - 6}}{m^3} \cr & \,\,\,\,\, = 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}} \cr} $$

Water is reduced at the cathode and oxidized at the anode instead of $$N{a^ + }$$ and $$SO_4^{2 - }$$
$$\eqalign{ & {\text{Cathode}}\,:\,2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - } \cr & {\text{Anode}}:\,{H_2}O \to 2{H^ + } + \frac{1}{2}{O_2} + 2{e^ - } \cr} $$

From the given representation of the cell, $${E_{cell}}$$  found as follows.
$${E_{cell}} = E_{\frac{{F{e^{2 + }}}}{{Fe}}}^ \circ - E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ - \frac{{0.059}}{6}\log $$       $$\frac{{{{\left[ {C{r^{3 + }}} \right]}^2}}}{{{{\left[ {F{e^{2 + }}} \right]}^3}}}{\text{[Nernst - Equ}}{\text{.]}}$$
$$\eqalign{ & = - 0.42 - \left( { - 0.72} \right) - \frac{{0.059}}{6}\log \frac{{{{\left( {0.1} \right)}^2}}}{{{{\left( {0.01} \right)}^3}}} \cr & = - 0.42 + 0.72 - \frac{{0.059}}{6}\log \frac{{0.1 \times 0.1}}{{0.01 \times 0.01 \times 0.01}} \cr & = 0.3 - \frac{{0.059}}{6}\log \frac{{{{10}^{ - 2}}}}{{{{10}^{ - 6}}}} \cr & = 0.3 - \frac{{0.059}}{6} \times 4 \cr & = 0.30 - 0.0393 \cr & = 0.26\,V \cr & {\text{Hence option (D) is correct answer}}{\text{.}} \cr} $$

Salt bridge keeps the solutions in two half-cells electrically neutral. It prevents transference or diffusion of the ions from one half-cell to the other.

$$\eqalign{ & {\text{Cell reaction}} \cr & {\text{cathode :}}{H_2}O\left( l \right) + \frac{1}{2}{O_2}\left( g \right) + 2{e^ - } \to 2O{H^ - }\left( {aq} \right) \cr & \underline {{\text{anode :}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_2}\left( g \right) \to 2{H^ + }\left( {aq} \right) + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & {H_2}O\left( l \right) + \frac{1}{2}{O_2}\left( g \right) + {H_2}\left( g \right) \to 2{H^ + }\left( {aq} \right) + 2O{H^ - }\left( {aq} \right) \cr & {\text{Also we have}} \cr & {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right) \cr & \Delta G_f^ \circ = - 237.2\,kJ/mol \cr & {H_2}O\left( l \right) \to {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right); \cr & \Delta {G^ \circ } = 80\,kJ/mol \cr & {\text{Hence for cell reaction}} \cr & \Delta {G^ \circ } = - 237.2 + \left( {2 \times 80} \right) \cr & = - 77.20\,kJ/mol \cr & \therefore \,\,{E^ \circ } = - \frac{{\Delta {G^ \circ }}}{{nF}} = - \frac{{77200}}{{2 \times 96500}} = - 0.40\,V \cr} $$

$$\eqalign{ & MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O \cr & E = 1.51 - \frac{{0.059}}{5}\log \frac{{\left[ {M{n^{2 + }}} \right]}}{{\left[ {MnO_4^ - } \right]{{\left[ {{H^ + }} \right]}^8}}} \cr} $$
Taking $$M{n^{2 + }}$$  and $$MnO_4^ - $$  in standard state i.e. $$1\,M,$$
$$\eqalign{ & E = 1.51 - \frac{{0.059}}{5} \times 8\log \frac{1}{{\left[ {{H^ + }} \right]}} \cr & = 1.51 - \frac{{0.059}}{5} \times 8 \times 3 \cr & = 1.2268\,V \cr} $$
Hence at this $$pH,MnO_4^ - $$   will oxidise only $$B{r^ - }$$  and $${I^ - }$$  as $$SRP$$  of $$\frac{{C{l_2}}}{{C{l^ - }}}$$  is $$1.36\,V$$  which is greater than that for $$\frac{{MnO_4^ - }}{{M{n^{2 + }}}}.$$

$$\eqalign{ & Fe\left( s \right) \to F{e^{2 + }} + 2{e^ - };\,\,\,\,\,\,\,\,{E^o} = 0.44\,V \cr & \frac{{2{H^ + } + 2{e^ - } + \frac{1}{2}{O_2} \to {H_2}O\left( l \right);\,\,\,\,{E^o} = + 1.23\,V}}{{Fe\left( s \right) + 2{H^ + } + \frac{1}{2}{O_2} \to F{e^{2 + }} + {H_2}O;}} \cr & E_{cell}^o = 0.44 + 1.23 = 1.67V \cr & \therefore \,\,\Delta {G^o} = - nFE_{cell}^o = - 2 \times 96500 \times 1.67 = - 322kJ \cr} $$

$$\eqalign{ & \Delta {G^ \circ } = - nF{E^ \circ } \cr & {\text{From given data,}} \cr & \left( {\text{i}} \right)\,Cu\left( s \right) \to C{u^{2 + }}\left( {aq} \right) + 2{e^ - }, \cr & \Delta G_1^ \circ = - 2\left( { - 0.34} \right) \times F \cr & \left( {{\text{ii}}} \right)\,C{u^{2 + }}\left( {aq} \right) + {e^ - } \to C{u^ + }\left( {aq} \right), \cr & \Delta G_2^ \circ = - 1\left( {0.15} \right) \times F \cr & {\text{On addition,}} \cr & Cu\left( s \right) \to C{u^ + }\left( {aq} \right) + {e^ - }, \cr & \Delta G_3^ \circ = - 1 \times {E^ \circ } \times F \cr & {\text{and}}\,\,\Delta G_3^ \circ = \Delta G_1^ \circ + \Delta G_2^ \circ \cr & - {n_3}F{E^ \circ } = - {n_1}FE_1^ \circ - {n_2}F{E_2} \cr & - {E^ \circ } = - 2\left( { - 0.34} \right) - 1\left( {0.15} \right) \cr & = \left( { - 2 \times - 0.34} \right) + \left( { - 1 \times 0.15} \right) \cr & - {E^ \circ } = + 0.68 - 0.15 = 0.53 \cr & {\text{or}}\,\,\,{E^ \circ } = - 0.53\,V \cr & C{u^ + }\left( {aq} \right) \rightleftharpoons C{u^{2 + }}\left( {aq} \right) + {e^ - }; \cr & \Delta {G_1} = - 1 \times \left( { - 0.15} \right) \times F \cr & C{u^ + }\left( {aq} \right) + {e^ - } \rightleftharpoons Cu\left( s \right); \cr & \Delta {G_2} = - 1 \times \left( { - 0.53} \right) \times F \cr & {\text{On adding above equation we get,}} \cr & 2C{u^ + } \rightleftharpoons C{u^{2 + }} + Cu;\,\Delta G \cr & \Delta G = \Delta {G_1} + \Delta {G_2} \cr & - nF{E^ \circ } = 0.15F + \left( { - 0.53F} \right) \cr & - F{E^ \circ } = - 0.38F \cr & {E^ \circ } = 0.38\,V \cr} $$
Thus, for the result reaction $${E^ \circ }$$ value is $$0.38\,V{\text{.}}$$

At the anode, the following oxidation reactions are possible :
$$\left( {\text{D}} \right)Cl_{\left( {aq} \right)}^ - \to \frac{1}{2}C{l_{2\left( g \right)}} + {e^ - };$$      $$E_{{\text{cell}}}^ \circ = 1.36\,V$$
$$\left( {\text{B}} \right)2{H_2}{O_{\left( l \right)}} \to {O_{2\left( g \right)}} + 4H_{\left( {aq} \right)}^ + + 4{e^ - };$$       $$E_{{\text{cell}}}^ \circ = 1.23\,V$$
The reaction at anode with lower value of $${E^ \circ }$$  is preferred and therefore, water should get oxidised in preference to $$Cl_{\left( {aq} \right)}^ - $$
However, on account of overpotential of oxygen, reaction (D) is preferred.

$$\eqalign{ & \Lambda _{HCl}^\infty = 426.2\,\,\,...\left( {\text{i}} \right) \cr & \Lambda _{AcONa}^\infty = 91.0\,\,\,...\left( {{\text{ii}}} \right) \cr & \Lambda _{NaCl}^\infty = 126.5\,\,\,...\left( {{\text{iii}}} \right) \cr & \Lambda _{AcOH}^\infty = \left( {\text{i}} \right) + \left( {{\text{ii}}} \right) - \left( {{\text{iii}}} \right) \cr & = \left[ {426.2 + 91.0 - 126.5} \right] \cr & = 390.7 \cr} $$