Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

The oxidation potential
$$ \propto \frac{1}{{{\text{Concentration of ions}}}}$$      and reduction potential $$ \propto $$  concentration of $$ions.$$  The cell voltage can be increased by decreasing the concentration of ions around anode or by increasing the concentration of ions around cathode

The most convenient method to protect the bottom of ship made of iron is white tin plating which prevents the build up of barnacles.

$$EMF$$  of galvanic cell $$ = 1.1\,volt$$
If $${E_{ext}} < EMF$$    then electron flows steadily from anode to cathode while If $${E_{ext}} > EMF$$    then electron flows from cathode to anode as polarity is changed.

Lower the reduction potential greater is the reducing power. Hence, increasing order of reducing power is $$Ag < Cr < Mg < K$$

$$\eqalign{ & {\text{Given}}\,\,\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}} = + 0.77\,V \cr & {\text{and}}\,\,\frac{{{I_2}}}{{2{I^ - }}} = 0.536\,V \cr & 2\left( {{e^ - } + F{e^{3 + }} \to F{e^{2 + }}} \right){E^ \circ } = 0.77\,V \cr & 2{I^ - } \to {I_2} + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.536\,V \cr & 2F{e^{3 + }} + 2{I^ - } \to 2F{e^{2 + }} + {I_2} \cr & {E^ \circ } = E_{ox}^ \circ + E_{red}^ \circ \cr & = 0.77 - 0.536 = 0.164\,V \cr & {\text{So, reaction will take place}}{\text{.}} \cr} $$

$$1\,F$$  deposits $$108\,g$$  of $$Ag\left( {A{g^ + } + {e^ - } \to Ag} \right)$$
$$54\,g$$  of $$Ag$$  will be deposited by $$\frac{1}{{108}} \times 54 = \frac{1}{2}F$$
$$3\,F$$  deposit $$27\,g$$  of $$Al$$     $$\left( {A{l^{3 + }} + 3{e^ - } \to Al} \right)$$
$$\frac{1}{2}F$$  will deposit $$\frac{{27}}{3} \times \frac{1}{2} = 4.5\,g$$    of $$Al$$

$$\eqalign{ & {\text{At anode}}:Fe \to F{e^{2 + }}\left( {0.001\,M} \right) + 2{e^ - } \cr & \underline {{\text{At cathode}}:2{H^ + }\left( {1\,M} \right) + 2{e^ - } \to {H_2}\left( {1\,atm} \right)} \cr & \underline {{\text{Net reaction:}}\,Fe + 2{H^ + } \to F{e^{2 + }} + {H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & {\text{Nernst equation for the given cell,}} \cr & {E_{{\text{cell}}}} = E_{{\text{cell}}}^ \circ - \frac{{0.0591}}{2}\log \frac{{\left[ {F{e^{2 + }}} \right]\left[ {{H_2}} \right]}}{{\left[ {Fe} \right]{{\left[ {{H^ + }} \right]}^2}}} \cr} $$

$$\eqalign{ & {H_2}O \to 2{H^ + } + \frac{1}{2}{O_2} + 2{e^ - }{\text{(Oxidation)}} \cr & \underline {C{u^{2 + }} + 2{e^ - } \to Cu\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(Reduction)}}} \cr & \underline {C{u^{2 + }} + {H_2}O \to Cu + 2{H^ + } + \frac{1}{2}{O_2}\,\,\,\,\,\,\,\,} \cr} $$

$$\frac{{63.5}}{2} = 31.75\,g\,\,{\text{of}}\,\,C{u^{2 + }}$$      $$ \equiv \frac{1}{2} \times 22400 = 11200\,cc\,\,{\text{of}}\,\,{O_2}$$
$$31.75\,g\,\,{\text{of}}\,\,C{u^{2 + }}\,\,{\text{evolve}}$$      $${\text{11200}}\,{\text{cc}}\,\,{\text{of}}\,\,{O_2}\,\,{\text{at}}\,\,NTP$$
$$0.4\,g\,\,{\text{of}}\,\,C{u^{2 + }}\,\,{\text{will}}\,\,{\text{evolve}}$$      $$\frac{{11200}}{{31.75}} \times 0.4 = 141\,cc\,\,{\text{of}}\,\,{O_2}$$

$$\frac{{{W_1}}}{{{W_2}}} = \frac{{{E_1}}}{{{E_2}}} = \frac{{{Z_1}\,{\text{it}}}}{{{Z_2}\,{\text{it}}}}\,\,\,\,\,\,\,\therefore \,\,\frac{{{Z_1}}}{{{Z_2}}} = \frac{{{E_1}}}{{{E_2}}}$$
Here$${E_1}$$ & $${E_2}$$ are equivalent weights of the ions.

$$\eqalign{ & Zn\left( s \right) + 2{H^ + }\left( {aq} \right) \rightleftharpoons Z{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right) \cr & {E_{cell}} = E_{cell}^ \circ - \frac{{0.059}}{2}\log \frac{{\left[ {Z{n^{2 + }}} \right]\left[ {{H_2}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} \cr} $$
Addition of $${H_2}S{O_4}$$  will increase $$\left[ {{H^ + }} \right]$$  and $${E_{cell}}$$  will also increase and the equilibrium will shift towards $$RHS$$