Question

In a lead-acid battery, if 1 A current is passed to charge the battery for 1 h, what is the amount of PbSO$_4$ converted to PbO$_2$? (Given data: 1 F = 96500 C mol$^{-1}$)

• A. 0.0373 moles
• B. 0.0186 moles
• C. 0.0093 moles
• D. 0.0268 moles
• E. 0.0400 moles

Hint:

Faraday's first law: Amount of substance $\propto$ Total charge. Remember to check the oxidation state change to find '$n$' (for Pb, it changes from +2 to +4, so $n=2$).

Answer 1

The Correct Option is B

Concept: During charging of a lead-acid battery, the reaction at the anode is: $\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^-$. This indicates that 2 moles of electrons are required to convert 1 mole of $\text{PbSO}_4$ to $\text{PbO}_2$.

Step 1: {Calculate the total charge (Q) passed.} Current ($I$) = 1 A, Time ($t$) = 1 hour = 3600 s. $$Q = I \times t$$ $$Q = 1 \text{ A} \times 3600 \text{ s} = 3600 \text{ C}$$

Step 2: {Calculate the number of Faradays passed.} Number of Faradays ($F_{passed}$) = $\frac{Q}{96500}$. $$F_{passed} = \frac{3600}{96500} \approx 0.0373 \text{ F}$$

Step 3: {Determine the moles of PbSO$_4$ converted.} From the stoichiometry, $n = 2$ electrons per mole of $\text{PbSO}_4$. $$\text{Moles} = \frac{F_{passed}}{n} = \frac{0.0373}{2}$$ $$\text{Moles} = 0.01865 \text{ moles}$$