- A$Pb{O_2}$ is dissolved
- B$PbS{O_4}$ is deposited on lead electrode
- ✓${H_2}S{O_4}$ is regenerated
- DLead is deposited on lead electrode
$P b_{(s)}+P b O_{2(s)}+2 H_{2} S O_{4(a q)} \rightarrow 2 P b S O_{4(s)}+2 H_{2} O_{(l)}$
$\rightarrow$ During discharge, $H_{2} S O_{4}$ is consumed and its density decreases.
$\rightarrow$ Fully charged cell gives a potential of about $2 V$. When the cell potential reduces to about $1.8 V$ it should be recharged.
Net cell reaction during recharge:-
$2 P b S O_{4(s)}+2 H_{2} O_{(l)} \rightarrow P b_{(s)}+P b O_{2(s)}+2 H_{2} S O_{4(a q)}$
$\rightarrow$ The concentration of $H_{2} S O_{4}$ increases as it is generated.
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Statement$-I$ $: \alpha$ and $\beta$ forms of sulphur can change reversibly between themselves with slow heating or slow cooling.
Statement$-II :$ At room temperature the stable crystalline form of sulphur is monoclinic sulphur.
In the light of the above statements, choose the correct answer from the options given below:
$(I)$ $-OAc$ $(II)$ $-OMe$
$(III)$ $-O-SO_2-Me$ $(IV)$ $-O-SO_2-CF_3$
The order of leaving power is :
Reason : It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate.