Question

Let a fully charged lead storage battery contains $1.5L$ of $5M\text{H}_{2}SO_{4}$ . What will be the concentration of $H_{2}SO_{4}$ in the battery after $2.5$ ampere current is drawn from the battery for $6$ hour?

• A. 4.626 M
• B. 0.1865 M
• C. 0.373 M
• D. 9.627 M

Answer 1

Answer: (A)

Number of moles of $H_{2}SO_{4}$ before electrolysis
$=\frac{M V}{1000}$
$M\times V\left(m L\right)=5\times 1.5=7.5$
Write cell rection $\Rightarrow \underset{\left(0\right)}{P b}+\underset{\left(+ 4\right)}{P b O_{2}}+2H_{2}SO_{4} \rightarrow \underset{\left(+ 2\right)}{2 P b S O_{4}}+2H_{2}O$
$n=2$
For 2 moles of electrons
2 moles of $H_{2}SO_{4}$ are used
Moles of electrons used $=\frac{I \left(i n \, A\right) \times t \left(i n \, s e c \right)}{96500}=\frac{2.5 \times 6 \times 3600}{96500}=0.56$
Hence, moles of $H_{2}SO_{4}$ used $=0.56$
Remaining mol of $H_{2}SO_{4}=7.5-0.56=6.94$
Final molarity $=\frac{n}{V}\times 1000=\frac{6.94}{1.5 \times 1000}\times 1000$
$=4.626$