Question

Let a fully charged lead storage battery contains $1.5L$ of $5M{H}_{2}S{O}_4$. 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. $4.626M$
• B. $0.1865M$
• C. $0.373M$
• D. $9.627M$

Answer 1

Answer: (A)

$4.626M$

From Faraday's First law of electrolysis,

$w=Z\times I\times t$

where, $Z=\frac{Equivalentweight}{F}$
no. of equivalents of $H_2=\frac{It}{F}$

given, $I=2.5A$

$t=6$ hours

$\Rightarrow$ Equivalent of $H_2=\frac{It}{F}=\frac{\left(2.5\right)\left(60\right)\left(60\right)\left(6\right)}{96500}=0.56$

Moles of $H^{+}=\frac{0.56}{2}=0.28M$

Molarity of $H^{+}=\left[H^{+}\right]=\frac{0.28}{1.5}=0.186M$

$H_2{SO}_4\to {2H}^{+}+{SO}_4^{2-}$

$\left[H_2{SO}_4\right]=5-2\left[H^{+}\right]=5-(2\times 0.186)=4.626M$

Final concentration of $H_2{SO}_4$ is $4.626M$.