Question:
What is the change in oxidation number of Pb at positive electrode of lead accumulator acting as galvanic cell ?
What is the change in oxidation number of Pb at positive electrode of lead accumulator acting as galvanic cell ?
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The positive plate is PbO2, which is reduced to PbSO4.
Updated On: Oct 1, 2026
- increases by 1
- decreases by 1
- increases by 2
- decreases by 2
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
In a lead accumulator working as a galvanic cell (discharging), the positive electrode is the cathode, where reduction occurs. It is made of \(\text{PbO}_2\).
Step 2: Half reaction:
\[ \text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O} \]
Step 3: Oxidation numbers:
In \(\text{PbO}_2\), Pb is +4. In \(\text{PbSO}_4\), Pb is +2. So the oxidation number decreases by 2.
At the negative electrode Pb goes from 0 to +2, which is an increase, but that is not the electrode asked about.
Final Answer:
Pb changes from +4 to +2, a decrease of 2, option (D). \[ \boxed{\text{decreases by } 2} \]
In a lead accumulator working as a galvanic cell (discharging), the positive electrode is the cathode, where reduction occurs. It is made of \(\text{PbO}_2\).
Step 2: Half reaction:
\[ \text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O} \]
Step 3: Oxidation numbers:
In \(\text{PbO}_2\), Pb is +4. In \(\text{PbSO}_4\), Pb is +2. So the oxidation number decreases by 2.
At the negative electrode Pb goes from 0 to +2, which is an increase, but that is not the electrode asked about.
Final Answer:
Pb changes from +4 to +2, a decrease of 2, option (D). \[ \boxed{\text{decreases by } 2} \]
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