Question

During the discharge of a lead storage battery the density of sulphuric acid fell from $1.294$ to $1.139{\text{g mL}}^{-1}$ . $H_{2}S{O}_4$ of density $1.294{\text{g mL}}^{-1}$ is 39% and that of density $1.139{\text{g mL}}^{-1}$ is 20% by weight. The battery holds $3.5\text{L}$ of acid and the volume practically remains constant during the discharge. Calculate the number of Ampere hours for which the battery must have been used. The discharging reactions are:
$Pb+S{O}_4^{2-}\to PbS{O}_4+2e^{-}\left(\text{anode}\right)$
$Pb{O}_2+4H^{+}+S{O}_4^{2-}+2e^{-}\to PbS{O}_4+2H_{2}O\left(\text{cathode}\right)$

• A. $190$
• B. $530$
• C. $265$
• D. $132.5$

Answer 1

The correct option is C $265$

$Pb+Pb{O}_2+4H^{+}+2S{O}_4^{2-}\to 2PbS{O}_4+2H_{2}O$
Here normality and molarity of $H_{2}S{O}_4$ are equal.
After electrolysis,molarity of $H_S{O}_4$ will be,
$=\frac{20\times 1.139\times 1000}{98\times 100}=2.325$
The number of moles = $2.325\times 3.5=8.1375$
Before electrolysis,molarity of $H_S{O}_4$ will be,
$=\frac{39\times 1.294\times 1000}{98\times 100}=5.15$
The number of moles = $5.15\times 3.5=18.025$
Equivalents of $H_{2}S{O}_4$ used = $18.025-8.1375=9.8875m=ZIt$
$I\times t=9.8875\times 96500÷3600=265.04$ ampere hr