Question

Let a fully charged lead storage battery contains $1.5L$ of $5M{\text{H}}_{2}S{O}_4$ . What will be the concentration of $H_{2}S{O}_4$ 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 $H_{2}S{O}_4$ before electrolysis
$=\frac{MV}{1000}$
$M\times V\left(mL\right)=5\times 1.5=7.5$
Write cell rection $\Rightarrow \underset{\left(0\right)}{\mathrm{Pb}}+\underset{\left(+4\right)}{Pb{O}_2}+2H_{2}S{O}_4\to \underset{\left(+2\right)}{2PbS{O}_4}+2H_{2}O$
$n=2$
For 2 moles of electrons
2 moles of $H_{2}S{O}_4$ are used
Moles of electrons used $=\frac{I\left(inA\right)\times t\left(insec\right)}{96500}=\frac{2.5\times 6\times 3600}{96500}=0.56$
Hence, moles of $H_{2}S{O}_4$ used $=0.56$
Remaining mol of $H_{2}S{O}_4=7.5-0.56=6.94$
Final molarity $=\frac{n}{V}\times 1000=\frac{6.94}{1.5\times 1000}\times 1000$
$=4.626$