Question
During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.5 to 1.1 g/ml. Sulphuric acid of density 1.5 g/ml is 40% H2SO4, by weight, and that of density 1.1 g/ml is 10% H2SO4, by weight. The battery holds 3.6 L of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere-hours which the battery should have been used. The electrode reactions are:
Pb + SO42− → PbSO4 + 2e
PbO2 + 4H+ + SO42− + 2e → PbSO4 + 2H2O
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Solution
Initial mass of H2SO4,
$w_{1} = 3600 \times 1.5 \times \frac{40}{100} = 2160\text{ gm}$
Final mass of H2SO4,
$w_{2} = 3600 \times 1.1 \times \frac{10}{100} = 396\text{ gm}$
∴ Moles of H2SO4 consumed = $\frac{w_{1} - w_{2}}{98} = 18$
Now, $n_{eq}H_{2}SO_{4} = \frac{Q}{F} \Rightarrow 18 \times 1 = \frac{\left( \text{amp-hr} \right) \times 3600}{96500}$
∴ Number of ampere-hr = 482.5
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