The reaction which take place at cathode in Leclanché cell is

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$H g O + H_{2} O + 2 e^{-} \to H g + 2 O H^{-}$

$P b O_{2} + S O_{4}^{2 -} + 4 H^{+} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O$

$M n O_{2} + N H_{4}^{+} + e^{-} \to M n O (O H) + N H_{3}$

$Z n \to Z n^{2 +} + 2 e^{-}$

answer is C.

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In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state. Ammonia produced in the reaction forms a complex with Zn²⁺ to give [Zn(NH₃)₄ ]²⁺.

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