Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & E_{MnO_4^ - /M{n^{2 + }}}^o = 1.51V \cr & E_{C{l_2}/C{l^ - }}^o = 1.36V \cr & E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o = 1.33V \cr & E_{C{r^{3 + }}/Cr}^o = - 0.74 \cr} $$
More negative the $${E^ \circ }$$ value of the species, more stronger is the reducing agent. Since $$C{r^{3 + }}$$  is having least reducing potential, so $$Cr$$  is the best reducing agent.

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

$$pH = - \log \left[ {{H^ + }} \right] = 14 \Rightarrow \left[ {{H^ + }} \right]$$       $$ = {10^{ - 14}}$$
$$\eqalign{ & {\text{Hence,}}\,\,\left[ {O{H^ - }} \right] = 1\,M \cr & \therefore \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {10^{ - 14}} \cr & Cu{\left( {OH} \right)_2} \to C{u^{2 + }} + 2O{H^ - } \cr & {K_{sp}} = \left[ {C{u^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2} \cr & {\text{For the reaction,}} \cr & C{u^{2 + }} + 2{e^ - } \to Cu,n = 2 \cr & {E_{{\text{cell}}}} = E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{n}\log \frac{1}{{\left[ {C{u^{2 + }}} \right]}} \cr & {E_{{\text{cell}}}} = 0.34 - \frac{{0.0591}}{2}\log \frac{1}{{1 \times {{10}^{ - 19}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - 0.22\,V \cr} $$

$$\eqalign{ & \kappa = \frac{{{\text{Cell constant}}}}{R} = \frac{{1.15}}{{250}} \cr & { \wedge _{eq}} = \frac{{1.15}}{{250}} \times \frac{{1000}}{1} = 4.6 \cr} $$

$${B_2}{H_6} + 3{O_2} \to {B_2}{O_3} + 3{H_2}O$$
$$27.66 g$$  of $${B_2}{H_6}$$  (1 mole) requires 3 moles of oxygen $$\left( {{O_2}} \right)$$  (3 moles) is obtained by electrolysis of 6 moles of $${H_2}O$$
On electrolysis : $$6{H_2}O \to 6{H_2} + 3{O_2}$$
Number of Faradays = 12 = Amount of charge
$$\eqalign{ & 12 \times 96500 = i \times t \cr & 12 \times 96500 = 100 \times t \cr & t = \frac{{12 \times 96500}}{{100}}{\text{second}} = \frac{{12 \times 96500}}{{100 \times 3600}}{\text{hour}} \cr & = 3.2\,{\text{hours}} \cr} $$

$$Cu\left( s \right) + 2\,A{g^{2 + }}\left( {aq} \right) \to $$     $$C{u^{2 + }}\left( {aq} \right) + 2\,Ag\left( s \right)$$
$$\eqalign{ & {E^ \circ } = 0.46\,V\,{\text{at}}\,\,298\,K \cr & RT\,\,{\text{ln}}\,\,K = nF{E^ \circ } \cr & {\text{ln}}\,\,K = \frac{{nF{E^ \circ }}}{{RT}} \cr & {\text{ln}}\,\,K = \frac{{2 \times 0.46}}{{0.0591}} \cr & K = 3.68 \times {10^{15}} \simeq 4 \times {10^{15}} \cr} $$

Fuel cell is a device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy.
Electrolytic cell converts electrical energy into chemical energy.
Dynamo is an electrical generator that produces direct current with the use of a commutator.
$$Ni{\text{ - }}Cd$$   cell is a type of rechargeable battery which consists of a cadmium anode and a metal grid containing $$Ni{O_2}$$  acting as a cathode.

$$\eqalign{ & \kappa = \frac{1}{R} \times \frac{l}{A} = \frac{1}{{40}} \times \frac{2}{5} \cr & { \wedge _{eq}} = \kappa \times \frac{{100}}{N} = \frac{1}{{40}} \times \frac{2}{5} \times \frac{{1000}}{{0.5}} \cr & \,\,\,\,\,\,\,\,\, = 20\,oh{m^{ - 1}}c{m^2}e{q^{ - 1}} \cr} $$

$$Zn + 2A{g^ + } \to Z{n^{2 + }} + 2Ag$$       can be represented as $$Z{n_{\left( s \right)}}\left| {Zn_{\left( {aq} \right)}^{2 + }} \right|\left| {Ag_{\left( {aq} \right)}^ + } \right|A{g_{\left( s \right)}}$$

$$\eqalign{ & {\text{Given for 0}}{\text{.2 M solution}} \cr & R = 50\,\Omega \cr & \kappa = 1.4\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}S\,c{m^{ - 1}} \cr & {\text{Now}},\,R = \rho \frac{\ell }{a} = \frac{1}{\kappa } \times \frac{\ell }{a} \cr & \Rightarrow \frac{\ell }{a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}} \cr & {\text{For 0}}{\text{.5 M solution}} \cr & R = 280\,\Omega \,\,;\,\,\kappa = ? \cr & \frac{\ell }{a} = 50 \times 1.4 \times {10^{ - 2}} \cr & \Rightarrow R = \rho \frac{\ell }{a} = \frac{1}{\kappa } \times \frac{\ell }{a} \cr & \Rightarrow \kappa = \frac{1}{{280}} \times 50 \times 1.4 \times {10^{ - 2}} \cr & = \frac{1}{{280}} \times 70 \times {10^{ - 2}} \cr & = 2.5 \times {10^{ - 3}}S\,c{m^{ - 1}} \cr & {\text{Now}},\,\,{\Lambda _m} = \frac{{\kappa \times 1000}}{M} \cr & = \frac{{2.5 \times {{10}^{ - 3}} \times 1000}}{{0.5}} \cr & = 5\,S\,c{m^2}mo{l^{ - 1}} \cr & = 5 \times {10^{ - 4}}S\,{m^2}mo{l^{ - 1}} \cr} $$