Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

Lower the reduction potential, more is the reducing power.
$$B{r^ - } < F{e^{2 + }} < Al$$

For hydrogen electrode, oxidation half reaction is
$$\eqalign{ & \mathop {{H_2}}\limits_{\left( {1\,\,atm} \right)} \to \mathop {2{H^ + }}\limits_{\left( {{\text{At}}\,pH\,10} \right)} + 2{e^ - } \cr & {\text{If}}\,\,pH = 10 \cr & {H^ + } = 1 \times {10^{ - pH}} = 1 \times {10^{ - 10}} \cr & {\text{From Nemst equation,}} \cr & {E_{cell}} = {E^ \circ }_{cell} - \frac{{0.0591}}{2}{\text{log}}\frac{{{{\left[ {{H^ + }} \right]}^2}}}{{{p_{{H_2}}}}} \cr & {\text{For hydrogen electrode,}}\,{E^ \circ }_{cell} = 0 \cr & {E_{cell}} = - \frac{{0.0591}}{2}{\text{log}}\frac{{{{\left( {{{10}^{ - 10}}} \right)}^2}}}{1} \cr & = + \frac{{0.0591 \times 2}}{2}{\text{log}}\frac{1}{{{{10}^{ - 10}}}} \cr & = 0.0591\,{\text{log}}\,{10^{10}} \cr & = 0.0591 \times 10 \cr & = 0.591\,V \cr} $$

$$\eqalign{ & \lambda _{HCOOH}^ \circ = \lambda _{{H^ + }}^ \circ + \lambda _{HCO{O^ - }}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 349.6 + 54.6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 404.2\,S\,c{m^2}\,mo{l^{ - 1}} \cr & \alpha = \frac{{{ \wedge _m}}}{{ \wedge _m^ \circ }} \cr & \,\,\,\, = \frac{{46.1}}{{404.2}} \cr & \,\,\,\, = 11.4\% \cr & {K_a} = \frac{{C{\alpha ^2}}}{{1 - \alpha }} \cr & \,\,\,\,\,\,\,\, = \frac{{0.025 \times {{\left( {0.114} \right)}^2}}}{{1 - 0.114}} \cr & \,\,\,\,\,\,\,\, = \frac{{0.025 \times 0.114 \times 0.114}}{{0.886}} \cr & \,\,\,\,\,\,\,\, = 3.67 \times {10^{ - 4}}\,mol\,{L^{ - 1}} \cr} $$

$$E = {E^ \circ } + \frac{{0.059}}{2}\log \left[ {M{g^{2 + }}} \right]$$
Hence, plot of $$E\,\,{\text{vs}}\,\,\log \left[ {M{g^{2 + }}} \right]$$   will be linear with positive slope and intercept $$ = {E^ \circ }$$

$${{\text{E}}_{cell}} = 0;$$   when cell is completely discharged.
$$\eqalign{ & {{\text{E}}_{cell}} = {{\text{E}}^o}_{cell} - \frac{{0.059}}{2}\log \left( {\frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}} \right) \cr & or\,0 = 1.1 - \frac{{0.059}}{2}\log \left( {\frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}} \right) \cr & \log \left( {\frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}} \right) \cr & = \frac{{2 \times 1.1}}{{0.059}} \cr & = 37.3 \cr & \therefore \,\,\left( {\frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}} \right) = {10^{37.3}} \cr} $$

In the electrolysis of dilute $${H_2}S{O_4},{O_2}$$    is liberated at anode and $${H_2}$$  at cathode.

The $$EMF$$  of the cell will not be zero because concentration of $${H^ + }$$ $$ions$$  in two electrolytic solutions is different. Mean $$HCl$$  is strong acid where, acetic acid is weak acid and gives different $$pH.$$

$${E_{red}} = E_{red}^ \circ + \frac{{0.059}}{n}\log \frac{{\left[ {{M^{\left( {x + n} \right)}}} \right]}}{{\left[ {{M^{x + }}} \right]}}$$

$$\eqalign{ & E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V,E_{\frac{{C{o^{2 + }}}}{{Co}}}^ \circ = - 0.28\,V \cr & {\text{The given cell reaction is}} \cr & Cr\left| {C{r^{3 + }}\left( {1.0\,M} \right)} \right|\left| {C{o^{2 + }}\left( {1.0M} \right)} \right|Co \cr & \therefore \,\,Cr\,\,{\text{is anode and }}Co{\text{ is cathode}} \cr & E_{cell}^ \circ = E_C^ \circ - E_A^ \circ \cr & = - 0.28 - \left( { - 0.74} \right) \cr & = - 0.28 + 0.74 \cr & = + 0.46\,V \cr} $$

The amount. of copper deposited at cathode by reduction of $$C{u^{2 + }}\,ions$$   is $$\frac{{3.2}}{{63}} = 0.05\,moles.$$
The same amount $$0.05\,mole$$   of $$C{u^{2 + }}$$  must pass into solution from anode by oxidation.