What is the change in oxidation number of Pb at positive electrode of lead accumulator acting as galvanic cell ?
Show Hint
- increases by 1
- decreases by 1
- increases by 2
- decreases by 2
The Correct Option is D
Solution and Explanation
Positive electrode of a galvanic cell = cathode = site of reduction. Reduction lowers oxidation number.
Step 2: Compute:
$\text{PbO}_2$ (Pb = +4) becomes $\text{PbSO}_4$ (Pb = +2). Change = 2 - 4 = -2, so it decreases by 2 (D).
Final Answer:
Decrease by 2. \[ \boxed{\text{(D)}} \]
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