Q47 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
What happens in the oxidation number of Mn during working of dry cell.
- AIncreases by 1
- Correct answerBdecreases by 1
- CIncreases by 2
- Ddecreases by 2
Solution
In a dry cell, the reaction at the cathode involves the reduction of manganese dioxide ($MnO_2$).
The cathode reaction is:
$MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$
The oxidation state of Mn in $MnO_2$ is $+4$.
The oxidation state of Mn in $MnO(OH)$ is $+3$.
The oxidation number of Mn changes from $+4$ to $+3$. Thus, the oxidation number of Mn decreases by $1$.
Answer: decreases by 1
The cathode reaction is:
$MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$
The oxidation state of Mn in $MnO_2$ is $+4$.
The oxidation state of Mn in $MnO(OH)$ is $+3$.
The oxidation number of Mn changes from $+4$ to $+3$. Thus, the oxidation number of Mn decreases by $1$.
Answer: decreases by 1
Hence, the correct answer is option: 'B'.