A) \[S{{O}_{2}}\]is evolved
B) lead sulphate is consumed
C) lead is formed
D) \[{{H}_{2}}S{{O}_{4}}\]is consumed
Correct Answer: D
Solution :
\[\underline{\begin{align} & \text{At}\,\text{anode}:\,\,\,\,\,Pb+{{H}_{2}}S{{O}_{4}}\to PbS{{O}_{4}}+2{{H}^{+}}+2{{e}^{-}} \\ & \text{At}\,\text{cathode}:\,Pb{{O}_{2}}+{{H}_{2}}S{{O}_{4}}+2{{H}^{+}}+2{{e}^{-}}\to \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PbS{{O}_{4}}+2{{H}_{2}}O \\ \end{align}}\]Net discharging equation: \[Pb+Pb{{O}_{2}}+2{{H}_{2}}S{{O}_{4}}\to 2PbS{{O}_{4}}\to 2{{H}_{2}}O\] So, \[{{H}_{2}}S{{O}_{4}}\]is consumed during discharging of lead storage battery.
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