Electrochemistry 1 Question 6
6. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of $\mathrm{PbSO}_4$ electrolysed in ‘g’ during the process is (Molar mass of $\mathrm{PbSO}_4=303 \mathrm{g} \mathrm{mol}^{-1}$ )
(2019 Main, 9 Jan I)
(a) $11.4$
(b) $7.6$
(c) $15.2$
(d) $22.8$
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Answer:
Correct Answer: 6. (b)
Solution:
Key Idea: This question is based upon Faraday’s first law which states that “Mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”
During charging:
$ \mathrm{Pb}+\mathrm{SO}_4^{2-} \longrightarrow \mathrm{PbSO}_4+2 e^{-} $
$ \begin{aligned} \Rightarrow \quad 1 \mathrm{F} & \equiv 1 \text { g-equiv. of } \mathrm{PbSO}_4 \\ & =\frac{1}{2} \text { mol of } \mathrm{PbSO}_4 \Rightarrow \frac{303}{2} \mathrm{g} \mathrm{PbSO}_4 \\ \therefore \quad 0.05 \mathrm{F} & \equiv \frac{303}{2} \times 0.05 \mathrm{g} \text { of } \mathrm{PbSO}_4 \\ & =7.575 \mathrm{g} \text { of } \mathrm{PbSO}_4 \end{aligned} $
BETA
