Chemistry - Electrochemistry Question with Solution | TestHub

$\underset{0.05/2}{\left(\text{A}\right){\text{PbSO}}_4\left(\text{s}\right)}+2{\text{OH}}^{-}\to {\text{PbO}}_2+{\text{H}}_2{\text{SO}}_4+\underset{0.05\text{F}}{2{\text{e}}^{-}}$
$\underset{0.05/2}{\left(\text{C}\right){\text{PbSO}}_4\left(\text{s}\right)}+\underset{0.05\text{F}}{2{\text{e}}^{-}}+2{\text{H}}^{+}\to \text{Pb}\left(\text{s}\right)+{\text{H}}_2{\text{SO}}_4$
${\text{n}}_{{\text{PbSO}}_4}=\frac{0.05}{2}\text{mole}$
${\text{w}}_{{\text{PbSO}}_4}=\frac{0.05}{2}\times 303=7.6\text{gm}$