Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{In reactivity series,}} \cr & Mg > Al > Zn > Fe > Cu \cr} $$
\[\xrightarrow{\text{Reactivity decreases}}\]
Hence, $$Mg$$  can displace $$Al, Al$$   can displace $$Zn$$  and so on.

For a spontaneous reaction $$\Delta G$$  must be $$-ve$$  Since $$\Delta G = - nF{E^ \circ }$$
Hence for $$\Delta G$$  to be $$-ve$$  $$\Delta {E^ \circ }$$  has to be positive. Which is possible when $$X = Zn,Y = Ni$$
$$Zn + N{i^{ + + }} \to Z{n^{ + + }} + Ni$$
$$E_{\frac{{Zn}}{{Z{n^{ + 2}}}}}^ \circ + E_{\frac{{N{i^{2 + }}}}{{Ni}}}^ \circ = 0.76 + \left( { - 0.23} \right)$$       $$ = + 0.53\,{\text{(positive)}}$$

$$\eqalign{ & \log \,{K_c} = \frac{{nE_{{\text{cell}}}^ \circ }}{{0.0591}} \cr & {\text{For the given reaction,}}\,n = 1 \cr & \log \,{K_c} = \frac{{1 \times 0.36}}{{0.0591}} = 6.09 \cr & {K_c} = {\text{antilog}}\,{\text{6}}{\text{.09}} = 1.2 \times {10^6} \cr} $$

Key Idea The relation between molar conductivity $$\left( {{\lambda _m}} \right)$$  and electrolytic conductivity $$\left( \kappa \right)$$  is given as
$$\eqalign{ & {\lambda _m} = \frac{{\kappa \times 1000}}{M} \cr & {\text{where, M is molarity of solution}}{\text{.}} \cr & {\text{Given, concentration of solution,}} \cr & M = 0.5\,mol/d{m^3} \cr & {\text{Electrolytic conductivity,}} \cr & \kappa = 5.76 \times {10^{ - 3}}S\,c{m^{ - 1}} \cr & {\text{Temperature,}}\,T = 298\,K \cr & \therefore {\text{Molar conductivity,}} \cr & {\lambda _m} = \frac{{\kappa \times 1000}}{M} \cr & \,\,\,\,\,\,\,\, = \frac{{5.76 \times {{10}^{ - 3}} \times 1000}}{{0.5}} \cr & \,\,\,\,\,\,\,\, = 11.52\,S\,c{m^2}/mol \cr} $$

Reactions take place at electrodes are :
At cathode : $$AuCl_4^ - + 3{e^ - } \to Au + 4C{l^ - }$$
At anode : $$C{l^ - } \to \frac{1}{2}C{l_2} + {e^ - }$$
For the deposition of $$197\,g\left( {1\,mole} \right)$$   of $$Au = 3F$$   of charge is required thus. for the deposition of $$78\,g$$  of $$Au,$$  charge required $$ = \frac{3}{{197}} \times 78 = 1.19\,F$$     $$ = 1.19 \times 96500$$    coulombs
$${\text{From}}\,\,Q = I \times t$$
$$t = \frac{Q}{I} = \frac{{1.19 \times 96500}}{6}$$     $$ = 19139.16\,\sec $$
By $$1\,F$$  charge, $$35.5\,g$$  of $$C{l_2}$$  gas is formed. thus from $$1.19\,F,$$  mass of chlorine gas formed $$ = 35.5 \times 1.19 = 42.24\,g$$

No explanation is given for this question. Let's discuss the answer together.

$$Cu \to C{u^{ + + }} + 2{e^ - }$$
i.e, to deposit $$1\,mole$$  of $$Cu$$  at cathode from $$C{u^{2 + }}SO_4^{2 - }$$   solution $$ = 2\,moles$$   of electrons are required
i.e, To deposit
$$6.35\,g = \frac{{6.35}}{{63.5}} \times 2 = \frac{2}{{10}} = \frac{1}{5}\,moles$$
Thus total no. of electrons required $$ = \frac{{{N_A}}}{5}$$

Apply Nernst equation to the reaction
$$\eqalign{ & Pb + S{n^{2 + }} \to P{b^2} + Sn \cr & {E_{cell}} = {E^ \circ } + \frac{{0.059}}{2}\log \frac{{\left[ {S{n^{2 + }}} \right]}}{{\left[ {P{b^{2 + }}} \right]}} \cr} $$
$${\text{or}}\,\,\,\log \frac{{\left[ {S{n^{2 + }}} \right]}}{{\left[ {P{b^{2 + }}} \right]}} = \frac{{0.01 \times 2}}{{0.059}} = 0.3$$       $$\left( {\because \,\,{E_{cell}} = 0} \right)$$
$${\text{or}}\,\,\,\frac{{\left[ {S{n^{2 + }}} \right]}}{{\left[ {P{b^{2 + }}} \right]}} = {\text{antilog}}\,\left( {0.3} \right)$$

$$\eqalign{ & W = Z \times I \times t \cr & \,\,\,\,\,\,\,\, = 4 \times {10^{ - 4}} \times 12 \times \frac{{75}}{{100}} \times 3 \times 3600 \cr & \,\,\,\,\,\,\,\, = 38.8\,g \cr} $$

$$\eqalign{ & CaC{l_2} \to C{a^{2 + }} + 2C{l^ - } \cr & C{a^{2 + }} + \mathop {2{e^ - }}\limits_{2\,moles} \to \mathop {Ca}\limits_{1\,mol\left( {40\,g} \right)} \cr & 40\,g\,\,{\text{of}}\,\,Ca \equiv 2\,F \cr & 100\,g\,\,{\text{of}}\,\,{\text{Ca}} \equiv 5\,F \cr} $$