Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & 2{H^ + } + 2{e^ - } \to {H_2} \cr & {E_{cell}} = 0.06\,\log \left[ {{H^ + }} \right] \cr & = 0.06 \times \log \sqrt {{K_a}C} \cr & = 0.06 \times \log \,{10^{ - 3}} \cr & = - 0.18\,V \cr} $$

Lower the value of reduction potential, stronger is the reducing agent i.e., undergoes oxidation most easily.
$$Cr \to C{r^{3 + }} + 3{e^ - }{\text{(oxidation)}}$$
Hence, $$Cr$$  is the strongest reducing agent.

$$\eqalign{ & {\text{Solubility}} \cr & = \frac{{\kappa \times 1000}}{{ \wedge _m^ \circ }} \cr & = \frac{{1.821 \times {{10}^{ - 5}} \times 1000}}{{130.26}} \cr & = 13.97 \times {10^{ - 5}}\,mol\,{L^{ - 1}} \cr} $$
  $$ = 13.97 \times {10^{ - 5}} \times 143.5$$       $$\left( {AgCl = 108 + 35.5 = 143.5} \right)$$
  $$ = 2.004 \times {10^{ - 2}}\,g\,{L^{ - 1}}$$

Half cell reaction : $$PbS{O_4} \to P{b^{4 + }} + 2{e^ - }$$
According to the reaction :
$$PbS{O_4} \to P{b^{4 + }} + 2{e^ - }$$
We require $$2F$$ for the electrolysis of 1 mol or $$303 g$$  of $$PbS{O_4}$$
∴ Amount of $$PbS{O_4}$$  electrolysed by $$0.05F$$
$$\eqalign{ & = \frac{{303}}{2} \times .05 \cr & = 7.575\,g \approx 7.6\,g \cr} $$

$$\eqalign{ & A{l_2}{O_3}\,\,{\text{ionises as,}} \cr & A{l_2}{O_3} \rightleftharpoons \mathop {A{l^{3 + }}}\limits_{{\text{Cathode}}} + \mathop {AlO_3^{3 - }}\limits_{{\text{Anode}}} \cr & {\text{At cathode}} \cr & A{l^{3 + }} + \mathop {3{e^ - }}\limits_{3\,F} \to \mathop {Al}\limits_{27\,g} \cr} $$
$$\because $$  Mass of aluminium deposited by $$3$$ $$F$$ of electricity $$=27$$ $$g$$
$$\therefore $$  Mass of aluminium deposited by
$$\eqalign{ & 4.0 \times {10^4} \times 6 \times 3600\,C\,\,{\text{of electricity}} \cr & {\text{ = }}\frac{{27 \times 4.0 \times {{10}^4} \times 6 \times 3600}}{{3\,F}}g \cr & = \frac{{27 \times 4.0 \times {{10}^4} \times 6 \times 3600}}{{3 \times 96500}}g \cr & = 8.1 \times {10^4}g \cr} $$

$$\eqalign{ & \Delta {G^ \circ } = - nFE_{cell}^ \circ \cr & = - 2 \times \left( {96000} \right) \times 2\,V \cr & = - 384000J/mol \cr & = - 384\,kJ/mol \cr} $$

$${E_{cell}} = $$  Reduction potential of cathode (right) - Reduction potential of anode (Left)
$$\,\,\,\,\,\,\,\,\,\,\,\,= {E_{{\text{right}}}} - {E_{{\text{left}}}}$$

On dilution number of ions decrease in unit volume hence specific conductance decreases. But separation between ions also increase hence equivalent conductance increases.

$${K_a} = C{\alpha ^2} = 0.01 \times {\left( {0.05} \right)^2}$$       $$ = 2.5 \times {10^{ - 5}}$$
$${K_a} = C{\alpha ^2}$$
$$2.5 \times {10^{ - 5}} = 0.05 \times {\alpha ^2} \Rightarrow \alpha $$       $$ = 0.0223$$
$$\alpha = \frac{{ \wedge _m^c}}{{ \wedge _m^\infty }}$$
$$\eqalign{ & \wedge _m^c = 0.0223 \times 4 \times {10^{ - 2}} \cr & \,\,\,\,\,\,\,\,\, = 8.92 \times {10^{ - 4}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}} \cr} $$

When 96500 coulomb of electricity is passed through the electroplating bath the amount of $$Ag$$  deposited = $$108g$$
∴ when 9650 coulomb of electricity is passed deposited $$Ag.$$
$$\eqalign{ & = \frac{{108}}{{96500}} \times 9650 \cr & = 10.8g \cr} $$