Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$ \wedge _{m\,\left( {N{H_4}Cl} \right)}^ \circ + \wedge _{m\,\left( {NaOH} \right)}^ \circ - \wedge _{m\,\left( {NaCl} \right)}^ \circ $$
$$ \wedge _{m\,\left( {NH_4^ + } \right)}^ \circ + \wedge _{m\,\left( {C{l^ - }} \right)}^ \circ + \wedge _{m\,\left( {N{a^ + }} \right)}^ \circ + $$      $$ \wedge _{m\,\left( {O{H^ - }} \right)}^ \circ - \wedge _{m\,\left( {N{a^ + }} \right)}^ \circ - \wedge _{m\,\left( {C{l^ - }} \right)}^ \circ $$
$$ = \wedge _{m\,\left( {NH_4^ + } \right)}^ \circ + \wedge _{m\,\left( {O{H^ - }} \right)}^ \circ = \wedge _{m\,\left( {N{H_4}OH} \right)}^ \circ $$

NOTE : For spontaneous reaction $$\Delta G$$ should be negative. Equilibrium constant should be more than one
$$\left( {\Delta G = - 2.303\,RT\log \,{K_c},\,{\text{If}}\,{K_c} = 1\,{\text{then}}\,\Delta G = 0\,;\,{\text{If}}\,{K_c} < 1\,{\text{then}}\,\Delta G = + ve} \right).$$               Again $$\Delta G = - nFE_{cell}^ \circ .$$
$$E_{cell}^ \circ $$  must be $$+ve$$  to have $$\Delta G - ve.$$

Thinking process Calculate the value of $${E_{cell}}$$  i.e. $${E_1}$$  and $${E_2}$$  by substituting the respective given values in the Nernst equation,
$${E_{cell}} = {E^ \circ } - \frac{{0.059}}{n}{\text{log}}\frac{{\left[ {Z{n^{2 + }}} \right]}}{{\left[ {C{u^{2 + }}} \right]}}$$
Compare the calculated values of $${E_1}$$  and $${E_2}$$  and find the correct relation.
$$\eqalign{ & {\text{For the electrochemical cells,}} \cr & Zn\left| {ZnS{O_4}\,\left. {\left( {0.01M} \right)} \right\|} \right.\,CuS{O_4}\left( {1M} \right)\left| {Cu} \right. \cr & {\text{Cell reaction :}} \cr & Zn + C{u^{2 + }} \to Z{n^{2 + }} + Cu;n = 2 \cr} $$
$${E_1} = {E^ \circ } - \frac{{0.059}}{2}{\text{log}}\frac{{Z{n^{2 + }}}}{{C{u^{2 + }}}}$$       $$ = {E^ \circ } - \frac{{0.059}}{2}{\text{log}}\frac{{0.01}}{1}$$
$${E_1} = {E^ \circ } - \frac{{0.059}}{2}{\text{log}}\frac{1}{{100}}$$     $$ = \left( {{E^ \circ } + 0.059} \right)$$
$$\eqalign{ & {\text{For cell,}} \cr & Zn\left| {ZnS{O_4}\left. {\left( {1M} \right)} \right\|} \right.CuS{O_4}\left( {0.01M} \right)\left| {Cu} \right. \cr & {E_2} = {E^ \circ } - \frac{{0.059}}{2}{\text{log}}\frac{1}{{0.01}} \cr & {E_2} = {E^ \circ } - \frac{{0.059}}{2}{\text{log}}\,\,100 \cr & \therefore \,\,\, = \left( {{E^ \circ } - 0.059} \right) \Rightarrow {E_1} > {E_2} \cr} $$

$$\eqalign{ & 1\,mole\,\,{\text{of}}\,{e^ - } = 1F = 96500\,C \cr & 27\,g\,\,{\text{of}}\,Al\,\,{\text{is deposited by}}\,{\text{3}} \times 96500\,C \cr & 5120\,g\,\,{\text{of}}\,Al\,\,{\text{will be deposited by}} \cr & = \frac{{3 \times 96500 \times 5120}}{{27}} \cr & = 5.49 \times {10^7}C \cr} $$

During charging, the lead storage battery behaves like an electrolytic cell. So, at anode the reaction is $$PbS{O_4} + 2{H_2}O \to $$     $$Pb{O_2} + 4{H^ + } + SO_4^{2 - } + 2{e^ - }$$

NOTE: Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode.
Cell representation :
Anode/ Anodic electrolyte || Cathodic electrolyte / Cathode
Reaction at Anode : $${H_2} \to 2{H^ + } + 2{e^ - }$$
Reaction at Cathode : $$AgCl + {e^ - } \to Ag + C{l^ - }$$^-

$$\eqalign{ & Q = I \times t \cr & \,\,\,\,\,\, = 1.5 \times 3 \times 60 \times 60 \cr & \,\,\,\,\,\, = 16200\,C \cr} $$
Charge on 1 electron = 1.602 × 10^-19 $$C$$
16200 $$C$$  charge is on $$\frac{{1 \times 16200}}{{1.602 \times {{10}^{ - 19}}}}$$   electrons = 1.01 × 10²³ electrons

$$\eqalign{ & {\text{Current,}}\,I = 5\,A \cr & {\text{time, }}t = 40\min = 40 \times 60{\text{ = }}2400{\text{ }}s \cr & {\text{Amount of electricity passed}} \cr & Q = It \cr & Q = 5 \times 2400 \cr & Q = 12000\,C. \cr & Z{n^{2 + }} + 2{e^ - } \to Zn \cr & n = 2{e^ - } \cr & {\text{From Faraday first law}} \cr & W = Z\,I\,t \cr} $$
   $$Z = {\text{equivalent mass}}$$     $$\left[ {65.39 = {\text{mass of zinc}}} \right]$$
$$\eqalign{ & \,\,\,\,\,\,\, = \frac{{Mass}}{{{e^ - } \times F}} \cr & \,\,\,\,\,\,\, = \frac{{65.39}}{{2 \times 96500}}\,g\,\,{\text{of }}zinc \cr & {\text{therefore, }}12000{\text{ }}C{\text{ charge will deposite}} \cr & {\text{ = }}\frac{{65.39 \times 12000}}{{2 \times 96500}} \cr & = 4.065\,g\,\,{\text{of }}zinc. \cr} $$

The reduction potentials (as given) of the ions are in the order :
$$A{g^ + } > H{g_2}^{2 + } > C{u^2} > M{g^{2 + }}$$
$$M{g^{2 + }}\left( {aq.} \right)$$   will not be reduced as its reduction potential is much lower than that of water $$(-0.83 V).$$
Hence the sequence of deposition of the metals will be $$Ag, Hg, Cu.$$

$$\eqalign{ & Cd\left( s \right) + 2A{g^ + }\left( {aq} \right) \to C{d^{2 + }}\left( {aq} \right) + Ag\left( s \right), \cr & {E_{cell}} = E_{cell}^ \circ - \frac{{0.059}}{2}{\text{log}}\frac{{\left[ {C{d^{2 + }}} \right]}}{{{{\left[ {A{g^ + }} \right]}^2}}} \cr} $$