Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$MnO_4^ - $$  will oxidise $$C{l^ - }$$ ion according to the following equation
$$2MnO_4^ - + 16{H^ + } + 10C{l^ - } \to 2M{n^{2 + }} + 8{H_2}O + 5C{l_2} \uparrow $$
The cell corresponding to this reaction is as follows :
$$\eqalign{ & Pt,C{l_2}\left( {{\text{l}}\,atm} \right)\left| {C{l^ - }} \right|\left| {MnO_4^ - ,M{n^{2 + }},{H^ + }} \right|Pt \cr & E_{cell}^ \circ = 1.51 - 1.40 = 0.11V \cr} $$
$$E_{cell}^ \circ \,$$ being $$+ve,$$  $$\Delta {G^ \circ }$$ will be $$-ve$$  and hence the above reaction is feasible. $$MnO_4^ - $$  will not only oxidise $$F{e^{2 + }}$$  ion but also $$C{l^ - }$$ ion simultaneously. So the quantitative estimation of $$aq$$  $$Fe{\left( {N{O_3}} \right)_2}$$   cannot be done by this.

$$E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ $$   has lowest reduction potential hence, $$C{r^{3 + }}$$  is the most stable oxidised species.

$$C{u^{2 + }} + {e^ - } \to C{u^ + },E_1^ \circ = 0.15\,V,$$       $$\Delta G_1^ \circ = - {n_1}E_1^ \circ F$$
$$C{u^ + } + {e^ - } \to Cu,E_2^ \circ = 0.50\,V,$$       $$\Delta G_2^ \circ = - {n_2}E_2^ \circ F$$
$$C{u^{2 + }} + 2{e^ - } \to Cu,{E^ \circ } = ?,$$      $$\Delta {G^ \circ } = - n{E^ \circ }F$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\Delta {G^ \circ } = \Delta G_1^ \circ + \Delta G_2^ \circ \cr & - n{E^ \circ }F = - {n_1}E_1^ \circ F - {n_2}E_2^ \circ F \cr} $$
$${\text{or}}\,\,\, - 2{E^ \circ }F = $$    $$ - 1F \times 0.15 + \left( { - 1\,F \times 0.50} \right)$$
$$\eqalign{ & {\text{or}}\,\,\, - 2{E^ \circ }F = - 0.15\,F - 0.50\,F \cr & {\text{or}}\,\,\, - 2F{E^ \circ } = - F\left( {0.15 + 0.50} \right) \cr & \therefore \,\,{E^ \circ } = \frac{{0.65}}{2} = 0.325\,V \cr} $$

$$\eqalign{ & 2{H_2}O \to {H_3}{O^ + } + O{H^ - } \cr & E_{{\text{cell}}}^ \circ = \frac{{0.0591}}{n}\log \,K \cr} $$
$$\log \,K = \frac{{E_{{\text{cell}}}^ \circ \times n}}{{0.0591}} = \frac{{ - 0.8277 \times 1}}{{0.0591}}$$       $$ = - 14 \Rightarrow K = {10^{ - 14}}$$

$$\eqalign{ & E_{{\text{cell}}}^ \circ = E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ - E_{\frac{{F{e^{2 + }}}}{{Fe}}}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.34\,V - \left( { - 0.44} \right)V \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.78\,V \cr} $$
$$\Delta {G^ \circ } = - nFE_{{\text{cell}}}^ \circ = - 2 \times 96500 \times 0.78$$        $$ = 150540\,J\,\,{\text{or}}\,\,150.5\,kJ$$

The ionic reactions are :
$$\eqalign{ & \frac{2}{3} \times \left( {2A{l^{3 + }}} \right) + 4{e^ - } \to \frac{4}{3}Al \cr & \frac{2}{3} \times \left( {3{O^{2 - }}} \right) \to {O_2} + 4{e^ - } \cr} $$
Thus, no. of electrons transferred, $$n = 4.$$
$$\eqalign{ & \Delta G = - nFE = - 4 \times 96500 \times E \cr & {\text{or}}\,\,966 \times {10^3} = - 4 \times 96500 \times E \cr & \Rightarrow E = - \frac{{966 \times {{10}^3}}}{{4 \times 96500}} = - 2.5\,V \cr} $$

Thermodynamic efficiency of a cell is given by the ratio of Gibbs function change to enthalpy change in the overall cell reaction.

$$\eqalign{ & E_{cell}^ \circ = E_{RP\left( {RHS} \right)}^ \circ - E_{RP\left( {LHS} \right)}^ \circ \cr & = 1.61 - 1.51 \Rightarrow 0.10\,V \cr & \Delta {G^ \circ } = - nF{E^ \circ } \Rightarrow - 5 \times 96500 \times 0.10\,J \cr & \Delta {G^ \circ } = - 48.25\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{The cell reaction is}} \cr & C{l_2}\left( g \right)\left( {{P_2}\,atm} \right) \to C{l_2}\left( {aq} \right)\left( {{P_1}\,atm} \right) \cr & {E_{cell}} = - \frac{{0.0592}}{2}\log \frac{{{P_1}}}{{{P_2}}} \cr & = \frac{{0.0592}}{2}\log \frac{{{P_2}}}{{{P_1}}} \cr & {E_{cell}}\,{\text{will be positive when}}\,{P_2} > {P_1}. \cr} $$

In concentration cell the spontaneous process is physical in nature involving transfer of matter from higher concentration to lower concentration in indirect manner.