Electrochemistry 1 Question 48
47. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to $1.139 \mathrm{g} / \mathrm{mL}$. Sulphuric acid of density $1.294 \mathrm{g} / \mathrm{mL}$ is $39 % \mathrm{H} _{2} \mathrm{SO} _{4}$ by weight and that of density $1.139 \mathrm{g} / \mathrm{mL}$ is $20 % \mathrm{H} _{2} \mathrm{SO} _{4}$ by weight. The battery holds $3.5 \mathrm{L}$ 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} & \mathrm{Pb}+\mathrm{SO} _{4}^{2-}=\mathrm{PbSO} _{4}+2 e^{-} \text {(charging) } \\ & \mathrm{PbO} _{2}+4 \mathrm{H}^{+}+\mathrm{SO} _{4}^{2-}+2 e^{-} \\ & \quad=\mathrm{PbSO} _{4}+2 \mathrm{H} _{2} \mathrm{O} \text { (discharging) } \end{aligned} $
(1986, 5M)
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Answer:
Correct Answer: 47. $265 \mathrm{Ah}$
Solution:
- For $1.0 \mathrm{L} \mathrm{H} _{2} \mathrm{SO} _{4}$ :
Initial mass of $\mathrm{H} _{2} \mathrm{SO} _{4}=1294 \times \frac{39}{100}=504.66 \mathrm{g}$
Final mass of $\mathrm{H} _{2} \mathrm{SO} _{4}=1139 \times \frac{20}{100}=227.80 \mathrm{g}$
$\Rightarrow \mathrm{H} _{2} \mathrm{SO} _{4}$ consumed/litre $=504.66-227.80=276.86 \mathrm{g}$ $\Rightarrow$ Total $\mathrm{H} _{2} \mathrm{SO} _{4}$ used up $=276.86 \times 3.5=969.01 \mathrm{g}$
$ =\frac{969.01}{98} \mathrm{mol}=9.888 \mathrm{mol} $
$\because 1$ mole of $\mathrm{H} _{2} \mathrm{SO} _{4}$ 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 \mathrm{Ah} $
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