Chemistry - Electrochemistry Question with Solution | TestHub

Number of moles of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$before electrolysis
$=\frac{\mathrm{M}\mathrm{V}}{1000}$
$\mathrm{M}\times \mathrm{V}\left(\mathrm{m}\mathrm{L}\right)=5\times 1.5=7.5$
Write cell rection$\Rightarrow \underset{\left(0\right)}{\mathrm{P}\mathrm{b}}+\underset{\left(+4\right)}{\mathrm{P}\mathrm{b}{\mathrm{O}}_2}+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \underset{\left(+2\right)}{2\mathrm{P}\mathrm{b}\mathrm{S}{\mathrm{O}}_4}+2{\mathrm{H}}_2\mathrm{O}$
$\mathrm{n}=2$
For 2 moles of electrons
2 moles of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$are used
Moles of electrons used$=\frac{\mathrm{I}\left(\mathrm{i}\mathrm{n }\mathrm{}\mathrm{A}\right)\times \mathrm{t}\left(\mathrm{i}\mathrm{n }\mathrm{}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{}\right)}{96500}=\frac{2.5\times 6\times 3600}{96500}=0.56$
Hence, moles of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$used$=0.56$
Remaining mol of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4=7.5-0.56=6.94$
Final molarity$=\frac{\mathrm{n}}{\mathrm{V}}\times 1000=\frac{6.94}{1.5\times 1000}\times 1000$
$=4.626$