Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$2F{e^{3 + }} + 3{I^ - } \rightleftharpoons 2F{e^{2 + }} + I_3^ - $$
For the above change at equilibrium, $$E = 0.$$
Using the relation, $$E = {E^ \circ } - \frac{{0.059}}{2}\log \,{K_c}$$
$${\text{or}}$$   $$0 = E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^ \circ - E_{\frac{{I_3^ - }}{{{I^ - }}}}^ \circ - \frac{{0.059}}{2}\log \,{K_c}$$
$${\text{or}}$$   $$0 = 0.77 - 0.54 - \frac{{0.059}}{2}\log \,{K_c}$$
$${\text{or}}$$   $$0 = 0.23 - \frac{{0.059}}{2}\log \,{K_c}$$
$${\text{or}}\,\,\frac{{0.059}}{2}\log \,{K_c} = 0.23$$      $${\text{or}}\,\,\log \,{K_c} = \frac{{0.23 \times 2}}{{0.059}}$$
$${\text{or}}\,\,\log \,{K_c} = 7.796\,\,$$    $${\text{or}}\,\,{K_c} = 6.25 \times {10^7}$$

$$\eqalign{ & MnO_4^ - = 96500 \times 5\,F \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 482500\,C \cr & C{r_2}O_7^{2 - } = 96500 \times 6\,F \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 579000\,C \cr & F{e^{3 + }} = 96500 \times 1\,F \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 96500\,C \cr & C{l_2} = 96500 \times 2\,F \cr & \,\,\,\,\,\,\,\,\,\, = 193000\,C \cr} $$

No explanation is given for this question. Let's discuss the answer together.

$$\eqalign{ & {\text{At}}\,{\text{anode}}\,:{H_2}\left( g \right) \rightleftharpoons 2{H^ + }\left( {aq} \right) + 2{e^ - } \cr & {\text{At}}\,{\text{cathode}}\,{\text{:}}\,{M^{4 + }}\left( {aq} \right) + 2{e^ - } \rightleftharpoons {M^2}\left( {aq} \right) \cr & {\text{Net cell reaction}}\,:\,{H_2}\left( g \right) + {M^{4 + }}\left( {aq} \right) \rightleftharpoons 2{H^ + }\left( {aq} \right) + {M^{2 + }}\left( {aq} \right) \cr & {\text{Now,}}\,{E_{cell}} = \left( {E_{\frac{{{M^{4 + }}}}{{{M^{2 + }}}}}^{^o} - E_{\frac{{{H^ + }}}{{{H_2}}}}^{^o}} \right) - \frac{{0.059}}{n}.\log \frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{M^{2 + }}} \right]}}{{{P_{{H_2}}}.\left[ {{M^{4 + }}} \right]}} \cr & {\text{or}},\,0.092 = \left( {0.151 - 0} \right) - \frac{{0.059}}{2}.\log \frac{{{{\text{l}}^2} \times \left[ {{M^{2 + }}} \right]}}{{1 \times \left[ {{M^{4 + }}} \right]}} \cr & \therefore \,\,\frac{{\left[ {{M^{2 + }}} \right]}}{{\left[ {{M^{4 + }}} \right]}} = {10^2} \Rightarrow x = 2 \cr} $$

A cation having maximum ( positive ) value of standard reduction potential is the strongest oxidising agent. Hence, $$M{g^{2 + }}$$  is the strongest oxidising agent

$$ \wedge _{eq}^\infty \left( {N{H_4}OH} \right) = \wedge _{eq}^\infty \left( {N{H_4}Cl} \right)$$       $$ + \wedge _{eq}^\infty \left( {NaOH} \right) - \wedge _{eq}^\infty \left( {NaCl} \right)$$

$$\eqalign{ & = 129.8 + 217.8 - 109.3 \cr & = 238.3\,oh{m^{ - 1}}c{m^2}e{q^{ - 1}} \cr & \alpha = \frac{{{ \wedge _{eq}}}}{{ \wedge _{eq}^\infty }} = \frac{{9.30}}{{238.3}} = 0.04 \cr} $$

Since concentration of ions is the same hence $${E_{cell}} = E_{cell\,\, \cdot }^ \circ $$

The given values show that $$Cr$$ has maximum oxidation potental, therefore its oxidation will be easiest. (Change the sign to get the oxidation values)

$$C{H_3}OH\left( l \right) + \frac{3}{2}{O_2}\left( g \right) \to $$      $$C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$$
$$\Delta {G_r} = \Delta {G_f}\left( {C{O_2},g} \right) + 2\Delta {G_f}\left( {{H_2}O,\ell } \right)$$        $$ - \Delta {G_f}\left( {C{H_3}OH,\ell } \right) - \frac{3}{2}\Delta {G_f}\left( {{O_2},g} \right)$$
$$\eqalign{ & = - 394.4 + 2\left( { - 237.2} \right) - \left( { - 166.2} \right) - 0 \cr & = - 394.4 - 474.4 + 166.2 \cr & = - 702.6\,k\,J \cr & \% \,{\text{efficiency}} = \frac{{702.6}}{{726}} \times 100 = 97\% \cr} $$

$${ \wedge _{eq}} = \frac{{\kappa \times 1000}}{N}$$