Electrochemistry - Result Question 48
####47. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to $1.139 g / mL$. Sulphuric acid of density $1.294 g / mL$ is $39 % H _2 SO _4$ by weight and that of density $1.139 g / mL$ is $20 % H _2 SO _4$ by weight. The battery holds $3.5 L$ of the acid and the volume remained practically constant during the discharge.
Calculate the number of ampere-hours for which the battery must have been used. The charging and discharging reactions are
$$ \begin{aligned} & Pb+SO _4^{2-}=PbSO _4+2 e^{-} \text {(charging) } \ & PbO _2+4 H^{+}+SO _4^{2-}+2 e^{-} \ & \quad=PbSO _4+2 H _2 O \text { (discharging) } \end{aligned} $$
(1986, 5M)
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Solution:
- For $1.0 L H _2 SO _4$ :
Initial mass of $H _2 SO _4=1294 \times \frac{39}{100}=504.66 g$
Final mass of $H _2 SO _4=1139 \times \frac{20}{100}=227.80 g$
$\Rightarrow H _2 SO _4$ consumed/litre $=504.66-227.80=276.86 g$ $\Rightarrow$ Total $H _2 SO _4$ used up $=276.86 \times 3.5=969.01 g$
$$ =\frac{969.01}{98} mol=9.888 mol $$
$\because 1$ mole of $H _2 SO _4$ is associated with transfer of 1.0 mole of electrons, total of 9.888 moles of electron transfer has occurred.
Coulomb produced $=9.888 \times 96500$
$$ \text { Ampere-hour }=\frac{9.888 \times 96500}{3600}=265 Ah $$
