Electrochemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

Efficiency of a fuel cell $$\left( \phi \right) = \frac{{\Delta G}}{{\Delta H}} \times 100$$
Fuel cells are expected to have an efficiency of 100% .

The electrolyte in a mercury cell is a paste of $$KOH$$  and $$ZnO.$$  The electrode reactions for the cell are
At anode : $$Zn\left( {Hg} \right) + 2O{H^ - } \to $$    $$Zn{O_{\left( s \right)}} + {H_2}O + 2{e^ - }$$
At cathode : $$HgO + {H_2}O + 2{e^ - } \to $$    $$H{g_{\left( l \right)}} + 2O{H^ - }$$

Oxidation potential of $$M$$  is more than $$Ni$$  and less than $$Mn.$$  Hence reducing power $$Mn > M > Ni$$

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

$$Pb{O_2}$$  plates act as cathode while $$Pb$$  plates act as anode. Dilute (38%) $${H_2}S{O_4}$$  acts as an electrolyte.

$$P{b_{\left( s \right)}} + Pb{O_{2\left( s \right)}} + 2{H_2}S{O_{4\left( {aq} \right)}} \to $$       $$2PbS{O_{4\left( s \right)}} + 2{H_2}{O_{\left( l \right)}}$$

$$\eqalign{ & {\text{We know that, standard Gibbs energy,}} \cr & \Delta {G^ \circ } = - nFE_{cell}^ \circ \cr & {\text{For the cell reaction,}} \cr & 2\,A{g^ + } + Cu \to C{u^{2 + }} + 2\,Ag \cr & \Delta E_{cell}^ \circ = + 0.46\,V \cr & \,\,\,\Delta {G^ \circ } = - nF\,E_{cell}^ \circ \cr & \,\,\,\,\,\,\,\,\,\,n = 2 \cr & \Delta {G^ \circ } = - 2 \times 96500 \times 0.46 \cr & = - 88780\,J \cr & = - 88.7\,kJ \approx - 89.0\,kJ \cr} $$

$$\mathop {MnO_4^ - }\limits_{1\,mole} + \mathop {5{e^ - }}\limits_{5\,moles} \to \mathop {M{n^{2 + }}}\limits_{1\,mole} $$
$$5\,moles$$  of electrons are needed for reduction of $$1\,mole$$  of $$MnO_4^ - $$  to $$M{n^{2 + }}$$
$$5\,moles$$  of electrons = 5 Faradays
Quantity of charge required
= 5 × 96500
= 4.825 × 10⁵ Coulombs

$$\eqalign{ & OC{l^ - } \to C{l^ - };{E^ \circ } = 0.94\,V\left( {\text{i}} \right) \cr & C{l^ - } \to \frac{1}{2}C{l_2} + {e^ - };{E^ \circ } = 1.36\,V\left( {{\text{ii}}} \right) \cr} $$
$${\text{Add}}\,\left( {\text{i}} \right) + \left( {{\text{ii}}} \right)\,\,OC{l^ - } \to \frac{1}{2}C{l_2};$$       $${E^ \circ } = 0.94 - 1.36$$
$$ = - 0.42\,V$$

The reduction potential of $$NO_3^ - \,ion$$   is more than $${H^ + }\,ion.$$