Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{Pb}}_{(\mathrm{s})}$ is oxidised to ${\mathrm{Pb}}_{(\text{aq. }.}^{2+}$

${\mathrm{Pb}}_{\text{(aq.) }}^{2+}$ is oxidised to ${\mathrm{PbO}}_{2(\mathrm{s})}$

$\mathrm{Pb}{\mathrm{O}}_{2(\mathrm{s})}$ is reduced to ${\mathrm{Pb}}_{(\text{aq. })}^{2+}$

${\mathrm{Pb}}_{\text{(aq.) }}^{2+}$ is reduced to ${\mathrm{Pb}}_{(\mathrm{s})}$