A) \[PbS{{O}_{4}}\]on cathode is reduced to \[Pb\]
B) \[PbS{{O}_{4}}\] on anode is oxidized to \[Pb{{O}_{2}}\]
C) \[PbS{{O}_{4}}\] on anode is reduced to \[Pb\]
D) \[PbS{{O}_{4}}\] on cathode is oxidized to \[Pb\].
Correct Answer: C
Solution :
: Reactions occurring during charging are: \[\underset{\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\mathop{2PbS{{O}_{4(s)}}+2{{H}_{2}}O\xrightarrow{{}}P{{b}_{(s)}}+Pb{{O}_{2(s)}}+Pb{{O}_{2(s)}}+2{{H}_{2}}S{{O}_{4(aq)}}}}\,}{\overset{{}}{\mathop{\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\mathop{\begin{matrix} Anode:\,\,\,\,\,PbS{{O}_{4(s)}}+2{{e}^{-}}\xrightarrow{{}}P{{b}_{(s)}}+SO_{4(aq)}^{2-} \\ \begin{align} & Cathode:\,\,PbS{{O}_{4(s)}}+2{{H}_{2}}O\xrightarrow{{}}Pb{{O}_{2(s)}}+SO_{4(aq)}^{2-} \\ & +4H_{(aq)}^{+}+2{{e}^{-}} \\ \end{align} \\ \end{matrix}}}\,}}}\,\]
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