Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{By Nernst equation,}} \cr & {E_{cell}} = E_{cell}^ \circ - \frac{{2.303\,RT}}{{nF}}{\text{lo}}{{\text{g}}_{10}}\,K \cr & {\text{At equilibrium, }}{E_{cell}} = 0 \cr & {\text{Given that,}} \cr & R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr & T = {25^ \circ }C + 273 = 298\,K \cr & F = 96500\,C\,\,{\text{and}}\,\,n = 2 \cr & \therefore \,\,E_{cell}^ \circ = \frac{{2.303 \times 8.314 \times 298}}{{2 \times 96500}}{\text{lo}}{{\text{g}}_{10}}\,K \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.0591}}{2}{\text{lo}}{{\text{g}}_{10}}\,K \cr & {\text{Given that}}\,E_{cell}^ \circ = 0.295\,V \cr & \therefore \,\,0.295 = \frac{{0.0591}}{2}{\text{lo}}{{\text{g}}_{10}}\,K \cr & \,\,{\text{lo}}{{\text{g}}_{10}}\,K = \frac{{0.295 \times 2}}{{0.0591}} = 10 \cr & {\text{antilog}}\,\,{\text{lo}}{{\text{g}}_{10}}\,K = {\text{antilog}}\,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K = 1 \times {10^{10}} \cr} $$

$$\eqalign{ & 2M{n^{2 + }} \to 2M{n^{3 + }} + 2{e^ - },\Delta G_1^ \circ \cr & M{n^{2 + }} + 2{e^ - } \to Mn,\Delta G_2^ \circ \cr & \overline {3M{n^{2 + }}\left( {aq} \right) \to Mn\left( s \right) + 2M{n^{3 + }}\left( {aq} \right)} \cr} $$
$$ - 2 \times F \times E_3^ \circ = - 2 \times F \times \left[ { - 1.51} \right]$$       $$ - 2 \times F \times \left( { - 1.185} \right)$$
$$\eqalign{ & E_3^ \circ = - 2.695 \cr & E_3^ \circ = + \frac{{0.0591}}{2}\log {K_{eq}};{K_{eq}} \simeq 6.3 \times {10^{ - 92}}. \cr} $$

$$\eqalign{ & {\text{given}}\,S \propto \frac{{area \times conc}}{\ell } = \frac{{\kappa {m^2}mol}}{{m \times {m^3}}} \cr & \therefore \,\,\kappa = S{m^2}mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \wedge _{A{l_2}{{\left( {S{O_4}} \right)}_3}}^ \circ = 2 \wedge _{A{l^{3 + }}}^ \circ + 3 \wedge _{SO_4^{2 - }}^ \circ \cr & \wedge _{A{l^{3 + }}}^ \circ = \frac{{ \wedge _{A{l_2}{{\left( {S{O_4}} \right)}_3}}^ \circ - 3 \wedge _{SO_4^{2 - }}^ \circ }}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{858 - \left( {3 \times 160} \right)}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 189\,S\,c{m^2}\,mo{l^{ - 1}} \cr} $$

Lower the reduction potential, higher is the reducing power.
$$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$$

$${\text{Given,}}\,E_{cell}^ \circ = - ve$$
The relation between $$\Delta {G^ \circ }$$  and $$E_{cell}^ \circ $$  is given as
$$\Delta {G^ \circ } = - nF\,E_{cell}^ \circ \,\,\,...{\text{(i)}}$$
If $$E_{cell}^ \circ $$  is negative, so $$\Delta {G^ \circ }$$  comes out to be positive. Again, relation between $$\Delta {G^ \circ }$$  and $${K_{eq}}$$  is given as
$$\Delta {G^ \circ } = - 2.303nRT\,{\text{log}}\,{K_{eq}}\,\,\,...{\text{(ii)}}$$
From Eq. (i) we get that $$\Delta {G^ \circ }$$  is positive. Now, if $$\Delta {G^ \circ }$$  is positive then $${K_{eq}}$$  comes out to be negative from eq (ii).
$${\text{i}}{\text{.e}}.\,\,\,\Delta {G^ \circ } > 1\,\,\,{\text{and}}\,\,\,{K_{eq}} < 1$$
Short trick As $$E_{cell}^ \circ $$  is negative so reaction is non-spontaneous or you can say reaction is moving in backward direction. For non-spontaneous reaction, $$\Delta {G^ \circ }$$  is always positive and $${K_{eq}}$$  is always less than 1.

It is a dry cell in which zinc can is anode, carbon or graphite rod is cathode and a paste of $$N{H_4}Cl + C + Mn{O_2}$$     acts as an electrolyte.

If an external potential of $$1.1\,V$$  is applied to the cell, the reaction stops and no current flows through the cell. Any further increase in external potential again starts the reaction but in opposite direction and the cell functions as an electrolytic cell.

The given values show that $$Cr$$  has highest negative value of $${E^ \circ }_{{M^{3 + }}/{M^{2 + }}}$$   i.e., $$Cr$$  is the strongest reducing agent among given metals therefore its oxidation will be easiest

$$Cu$$  electrode loses mass and $$Ag$$  electrode gains mass.