A) \[{{\operatorname{SO}}_{2}} is evolved\]
B) Lead sulphate is consumed
C) Lead is formed
D) Sulphuric acid is consumed
Correct Answer: D
Solution :
Sulphuric acid is consumed In discharged state, both +ve and -ve plates become lead (II) sulfate\[\left( PbS{{O}_{4}} \right)\]. The electrolyte loses much of its dissolved sulphuric acid and becomes primarily water. The discharge process is driven by the conduction of electrons from negative plate back into the cell at positive plate in the external circuit. At?ve plate, \[\operatorname{Pb} \left( S \right) + HS{{O}^{-}}_{4}\left( aq \right)\Rightarrow PbS{{O}_{4}} \left( S \right)+ {{H}^{+}}\left( aq. \right)+ 2{{e}^{-}}\]At +ve plate, \[{{\operatorname{PbO}}_{2}}\left( S \right) + HS{{O}^{-}}_{4}\left( aq. \right) +3{{H}^{+}}\left( aq. \right) + 2{{e}^{-}}\to \] \[{{\operatorname{PbSO}}_{4}}\left( S \right)+2{{H}_{2}}O\left( \ell \right)\] On combining \[\operatorname{Pb} \left( S \right) + Pb{{O}_{2}} \left( S \right) + 2{{H}^{+}}\left( aq. \right) + 2HS{{O}^{-}}_{4}\left( aq. \right)\to \] \[2{{\operatorname{PbSO}}_{4}}\left( S \right)+2{{H}_{2}}O\left( \ell \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
