JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Redox Reactions and Electrochemistry

On charging, the cell reaction runs in reverse, $2\mathrm{PbS}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Pb}+\mathrm{Pb}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$. At the lead plate the lead in $\mathrm{PbS}{\mathrm{O}}_4$ is reduced from $+2$ to $0$, while at the lead-dioxide plate the lead in $\mathrm{PbS}{\mathrm{O}}_4$ is oxidized from $+2$ to $+4$. Reading the oxidation states in the order anode then cathode, they are $+2,0,+2,+4$.