The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount

Q: Why does the anode undergo oxidation during recharge? I thought the anode was the negative terminal!

In an electrolytic cell (which is what the battery becomes during recharge), the external power source forces electrons into the negative terminal (making it the cathode, where reduction happens) and pulls electrons from the positive terminal (making it the anode, where oxidation happens). Oxidation always occurs at the anode, regardless of the cell type!

Q: What is the reaction at the cathodic half-cell during recharge?

At the cathode during recharge, lead sulfate is reduced to pure lead: $\text{PbSO}_4(s) + 2e^- \rightarrow \text{Pb}(s) + \text{SO}_4^{2-}(aq)$. Notice that it also involves 2 electrons per mole of $\text{PbSO}_4$!

Q: How did we know the n-factor is 2?

By looking at the balanced half-reaction: $\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \text{PbO}_2 + \text{SO}_4^{2-} + 4\text{H}^+ + 2e^-$. For every 1 mole of $\text{PbSO}_4$ consumed, 2 moles of electrons are produced. Hence, the n-factor is 2.

Q: What if the question asked for the amount of $\text{PbSO}_4$ electrolysed at the cathode?

It would be exactly the same! The cathodic reaction ($\text{PbSO}_4 + 2e^- \rightarrow \text{Pb} + \text{SO}_4^{2-}$) also involves 2 electrons per mole of $\text{PbSO}_4$. The symmetry of the lead-acid battery is quite elegant.

Q: Why do we use 303 as the molar mass instead of calculating it ourselves?

The problem explicitly provides the molar mass of $\text{PbSO}_4$ as $303 \text{ g mol}^{-1}$. It's always best to use the given values to avoid minor discrepancies in atomic weights that could lead to a slightly different decimal answer.

Question

The anodic half-cell of lead-acid battery is recharged using electricity of $0.05$ Faraday. The amount of $	ext{PbSO}_4$ electrolysed in g during the process is (Molar mass of $	ext{PbSO}_4 = 303 	ext{ g mol}^{-1}$)

Options:

(A) 11.4

(B) 7.6

(C) 15.2

(D) 22.8

Sigma Percentile · Accepted Answer

### The Dual Life of a Lead-Acid Battery

Imagine the heavy lead-acid battery sitting under the hood of a car. It leads a fascinating double life. When you start the engine, it acts as a **galvanic cell**, spontaneously converting chemical energy into electrical energy to crank the motor. But what happens when the battery is drained? We pump electricity back into it using an alternator. In this recharging phase, the battery transforms into an **electrolytic cell**. The external voltage forces the non-spontaneous chemical reactions to occur in reverse.

### The Anodic Half-Cell During Recharge

During the discharge phase, the positive terminal (cathode) was made of lead dioxide ($	ext{PbO}_2$), and the negative terminal (anode) was pure lead ($	ext{Pb}$). Both electrodes became coated with lead sulfate ($	ext{PbSO}_4$) as the battery drained.

When we recharge the battery, we connect the positive terminal of our external power source to the positive electrode of the battery. This forces the positive electrode to become the **anode** of our new electrolytic setup. Why? Because the external source pulls electrons away from it, forcing an oxidation reaction.

The solid lead sulfate on this electrode is forced to oxidize back into lead dioxide:

$$	ext{PbSO}_4(s) + 2	ext{H}_2	ext{O}(l) ightarrow 	ext{PbO}_2(s) + 	ext{SO}_4^{2-}(aq) + 4	ext{H}^+(aq) + 2e^-$$

### Faraday's Law in Action

Look closely at the stoichiometry of that half-cell reaction. For every **$1$ mole** of $	ext{PbSO}_4$ that is electrolysed, exactly **$2$ moles** of electrons are released into the circuit.

Michael Faraday taught us that one mole of electrons carries exactly one Faraday ($1 	ext{ F}$) of electrical charge. Therefore, it takes $2 	ext{ F}$ of charge to electrolyse $1$ mole of $	ext{PbSO}_4$.

The problem states that we are passing $0.05 	ext{ F}$ of electricity through the cell. If $2 	ext{ F}$ corresponds to $1$ mole, then $0.05 	ext{ F}$ will correspond to:

$$n = \frac{0.05 	ext{ F}}{2 	ext{ F/mol}} = 0.025 	ext{ moles of } 	ext{PbSO}_4$$

### The Final Calculation

We now know exactly how many moles of lead sulfate were electrolysed. The final step is to convert this molar quantity into a tangible mass. The problem kindly provides the molar mass of $	ext{PbSO}_4$ as $303 	ext{ g mol}^{-1}$.

Using the fundamental relation $	ext{Mass} = 	ext{moles} 	imes 	ext{Molar Mass}$:

$$	ext{Mass} = 0.025 	ext{ mol} 	imes 303 	ext{ g mol}^{-1} = 7.575 	ext{ g}$$

Looking at our multiple-choice options, they are all rounded to one decimal place. When we round $7.575 	ext{ g}$ to one decimal place, the $7$ in the hundredths place forces us to round up, giving us **$7.6 	ext{ g}$**. This matches perfectly with option (b). A truly elegant application of electrochemistry!

Animated Solution for Chemistry - Electrochemistry: The anodic half-cell of lead-acid battery is recharged using electricity of $0.05$ Faraday. The amount of $PbSO_{4}$ electrolysed in g during the process is (Molar mass of $PbSO_{4} = 303 g mol^{- 1}$ )

Select Answer:

Visualized Solution

\text{The Recharging Process}

\text{Faraday's Law of Electrolysis}

\text{Calculating Moles Electrolysed}

\text{Converting Moles to Mass}

\text{Final Conclusion}

The Sigma Insight: Electrochemical Cells

The Dual Life of a Lead-Acid Battery

The Anodic Half-Cell During Recharge

Faraday's Law in Action

The Final Calculation

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