A) voltage of cell increases
B) electrolyte of cell dilutes
C) resistance of cell increases
D) none of the above
Correct Answer: A
Solution :
The discharged acid cell (lead storage battery or cell) can be recharged by passing a direct current through it in opposite direction. During recharging of cell, the electrode material are restored in their original forms. Due to formation of \[{{\text{H}}^{\text{+}}}\]and \[\text{SO}_{4}^{2-}\]ions the specific gravity of sulphuric acid increases and it attains the original value \[(1.3\,g/c{{m}^{3}})\] and also its EMF (voltage) increases. During charging At negative electrode: \[PbS{{O}_{4}}(s)+2{{e}^{-}}\xrightarrow[current]{direct}Pb(s)+SO_{4}^{2-}(aq)\] At positive electrode: \[PbS{{O}_{4}}(s)+2{{H}_{2}}O\xrightarrow[current]{direct}\] \[Pb{{O}_{2}}(s)+4{{H}^{+}}(aq)+SO_{4}^{2-}(aq)+2{{e}^{-}}\] overall charging reaction: \[\underbrace{2PbS{{O}_{4}}(s)+2{{H}_{2}}O}_{\Pr esent\,of\,electrodes}\xrightarrow{Ch\arg ing}\underbrace{Pb(s)+Pb{{O}_{2}}(s)}_{\begin{smallmatrix} electrode\,materials \\ re\operatorname{cov}ered \end{smallmatrix}}\]\[+\underbrace{4\overset{+}{\mathop{H}}\,(aq)+SO_{4}^{2-}(aq)}_{{{H}_{2}}S{{O}_{4}}\,\,reformed}\]overall reaction during charging : \[Pb(s)+Pb{{O}_{2}}(s)+4{{H}^{+}}(aq)+2SO_{4}^{2-}(aq)\xrightarrow{{}}\] \[2PbS{{O}_{4}}(s)+2{{H}_{2}}O\]
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