Question

The anodic half-cell of lead-acid battery is recharged unsing electricity of $0.05$ Faraday. The amount of $ {PbSO_4}$ electrolyzed in g during the process in : (Molar mass of $ {PbSO_4 = 303 \; g \; mol^{-1}}$)

• A. 22.8
• B. 15.2
• C. 7.6
• D. 11.4

Answer 1

The Correct Option is B

To determine the amount of \(PbSO_4\) electrolyzed in a lead-acid battery when recharged with \(0.05\) Faraday of electricity, we can use Faraday's laws of electrolysis. According to Faraday's first law:

\(m = \dfrac{Q \cdot M}{F \cdot n}\)

where:

• \(m\) is the mass of the substance (in grams) that is deposited or dissolved at an electrode.
• \(Q\) is the charge in coulombs. Since \(1\) Faraday \(= 96500 \, \text{C}\), \(0.05\) Faraday \(= 0.05 \times 96500 \, \text{C} = 4825 \, \text{C}\).
• \(M\) is the molar mass of \(PbSO_4\), which is \(303 \, \text{g/mol}\).
• \(F\) is Faraday's constant, \(96500 \, \text{C/mol}\).
• \(n\) is the number of electrons transferred per molecule of \(PbSO_4\) during the reaction. In the electrochemical reaction involving \(PbSO_4\), \(n = 2\).

Substituting the known values into the formula:

\(m = \dfrac{4825 \times 303}{96500 \times 2} = \dfrac{1463175}{193000} \approx 15.2 \, \text{g}\)

Therefore, the amount of \(PbSO_4\) electrolyzed is approximately 15.2 g.

Hence, the correct answer is 15.2.