Ammonia released during the reduction at cathode of Leclanche cell combines with zinc ions to formA. `[Zn(NH_(3))_(4)]^(2+)`B. `[Zn(NH_(3))_(2)]^(2+)`C. `[Zn(NH_(3))_(6)]^(2+)`D. `[Zn(NH_(3))_(5)]^(2+)`

Ammonia released during the reduction at cathode of Leclanche cell com

1.	Ammonia released during the reduction at cathode of Leclanche cell combines with zinc ions to formA. `[Zn(NH_(3))_(4)]^(2+)`B. `[Zn(NH_(3))_(2)]^(2+)`C. `[Zn(NH_(3))_(6)]^(2+)`D. `[Zn(NH_(3))_(5)]^(2+)`
Answer» Correct Answer - A Leclanche produces about `1.5V` but drops to about `0.8 V` as the reaction proceeds. This drop is because the reaction products cannot diffuse aways from the electrolyte. If such a cell is left unused for a short time, the voltage may rise back again to about `1.3 V`. The cell apears to be dead, because large excess of `[Zn(NH_(3))_(4)]Cl_(2)` formed crystallizes out around the anode. The electrolyte becomes unable to conduct electricity effectively. A little warming of the cell may restore the voltage again due to the diffusion of the complex.

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Ammonia released during the reduction at cathode of Leclanche cell combines with zinc ions to formA. `[Zn(NH_(3))_(4)]^(2+)`B. `[Zn(NH_(3))_(2)]^(2+)`C. `[Zn(NH_(3))_(6)]^(2+)`D. `[Zn(NH_(3))_(5)]^(2+)`

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