Electrochemistry - Result Question 99
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^{-} \\ & =PbSO _4+2 H _2 O \text { (discharging) } \end{aligned} $
(1986, 5M)
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Answer:
Correct Answer: 47. ($265$ Ah)
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
BETA
