The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of $$PbS{O_4}$$ electrolyzed in $$g$$ during the process is : ( Molar mass of $$PbS{O_4} = 303\,g\,mo{l^{ - 1}}$$ )
A.
22.8
B.
15.2
C.
7.6
D.
11.4
Answer :
7.6
Solution :
Half cell reaction : $$PbS{O_4} \to P{b^{4 + }} + 2{e^ - }$$
According to the reaction :
$$PbS{O_4} \to P{b^{4 + }} + 2{e^ - }$$
We require $$2F$$ for the electrolysis of 1 mol or $$303 g$$ of $$PbS{O_4}$$
∴ Amount of $$PbS{O_4}$$ electrolysed by $$0.05F$$
$$\eqalign{
& = \frac{{303}}{2} \times .05 \cr
& = 7.575\,g \approx 7.6\,g \cr} $$
Releted MCQ Question on Physical Chemistry >> Electrochemistry
Releted Question 1
The standard reduction potentials at $$298 K$$ for the following half reactions are given against each
$$\eqalign{
& Z{n^{2 + }}\left( {aq} \right) + 2e \rightleftharpoons Zn\left( s \right)\,\,\,\,\,\,\,\,\, - 0.762 \cr
& C{r^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons Cr\left( s \right)\,\,\,\,\,\,\,\,\, - 0.740 \cr
& 2{H^ + }\left( {aq} \right) + 2e \rightleftharpoons {H_2}\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.000 \cr
& F{e^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons F{e^{2 + }}\left( {aq} \right)\,\,\,\,\,\,\,\,0.770 \cr} $$
which is the strongest reducing agent ?
A solution containing one mole per litre of each $$Cu{\left( {N{O_3}} \right)_2};AgN{O_3};H{g_2}{\left( {N{O_3}} \right)_2};$$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are :
$$\eqalign{
& Ag/A{g^ + } = + 0.80,\,\,2Hg/H{g_2}^{ + + } = + 0.79 \cr
& Cu/C{u^{ + + }} = + 0.34,\,Mg/M{g^{ + + }} = - 2.37 \cr} $$
With increasing voltage, the sequence of deposition of metals on the cathode will be :