Question

During the charging and discharging of a lead-acid battery (a $\text{Pb}$ anode, a grid of $\text{Pb}$ packed with $\text{PbO}_2$ as cathode, and an aqueous solution of $\text{H}_2\text{SO}_4$ as an electrolyte), which of the following redox reactions does NOT occur?

• A. $\text{Pb}^{4+} + 4\text{e}^- \rightarrow \text{Pb}$
• B. $\text{Pb}^{2+} \rightarrow \text{Pb}^{4+} + 2\text{e}^-$
• C. $\text{Pb} \rightarrow \text{Pb}^{2+} + 2\text{e}^-$
• D. $2\text{Pb}^{2+} \rightarrow \text{Pb}^{4+} + \text{Pb}$

Hint:

In both charging and discharging of a lead-acid battery, the stable end-product of both electrode reactions is always solid insoluble $\text{PbSO}_4$ (where lead is in the $+2$ state).
Therefore, only 2-electron transfer steps involving $\text{Pb}^{0} \leftrightarrow \text{Pb}^{2+}$ and $\text{Pb}^{2+} \leftrightarrow \text{Pb}^{4+}$ occur.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to identify which of the given redox half-reactions or overall reactions does not take place during the operation (either charging or discharging) of a lead-acid storage battery.

Step 2 : Key Formulas and Approach:
We will write down the chemical reactions that occur at both electrodes of the lead-acid battery during discharging and charging.
By tracking the oxidation states of lead ($\text{Pb}$ in $0$, $+2$, and $+4$ states), we can determine which electron transfer steps actually occur.

Step 3 : Detailed Explanation:
Let us analyze the chemistry of the lead-acid battery:

• During Discharging (Galvanic Cell Mode):
• At Anode (Oxidation): Metallic lead ($\text{Pb}$) is oxidized to lead(II) ions:
\[ \text{Pb} \, (\text{s}) + \text{SO}_4^{2-} \, (\text{aq}) \rightarrow \text{PbSO}_4 \, (\text{s}) + 2\text{e}^- \]
This corresponds to: $\text{Pb} \rightarrow \text{Pb}^{2+} + 2\text{e}^-$ (Reaction C occurs).

• At Cathode (Reduction): Lead dioxide ($\text{PbO}_2$, where lead is in $+4$ state) is reduced to lead(II) ions:
\[ \text{PbO}_2 \, (\text{s}) + \text{SO}_4^{2-} \, (\text{aq}) + 4\text{H}^+ \, (\text{aq}) + 2\text{e}^- \rightarrow \text{PbSO}_4 \, (\text{s}) + 2\text{H}_2\text{O} \]
This corresponds to the reduction: $\text{Pb}^{4+} + 2\text{e}^- \rightarrow \text{Pb}^{2+}$.

• During Charging (Electrolytic Cell Mode):
The reverse reactions occur under an applied external potential:
• At anode: $\text{Pb}^{2+} \rightarrow \text{Pb}^{4+} + 2\text{e}^-$ (Reaction B occurs).

• At cathode: $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}^0$.

• Net Cell Reaction of Charging:
If we combine both charging half-reactions, we obtain:
\[ 2\text{Pb}^{2+} \rightarrow \text{Pb}^{4+} + \text{Pb} \]
This is a disproportionation-like redox change, showing that Reaction D also occurs.

• There is no step in either the charging or discharging process where $\text{Pb}^{4+}$ (from $\text{PbO}_2$) is directly reduced to metallic $\text{Pb}^0$ in a single 4-electron transfer step. Thus, Reaction (A) does not occur.

Step 4 : Final Answer:
The reaction $\text{Pb}^{4+} + 4\text{e}^- \rightarrow \text{Pb}$ does not occur in a lead-acid battery.
This corresponds to Option (A).