Dry cell has a small life time because of the corrosion of Zn container to form .

N H_{4}^{+} (a q)

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C l^{-} i o n s

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M n_{2} O_{3} (s)

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[Z n (N H_{3})_{4}] C l_{2}

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Solution

The correct option is D $[Z n (N H_{3})_{4}] C l_{2}$
Dry cells reactions

At anode

$Z n (s) ⟶ Z n^{2 +} (a q) + 2 e^{-}$

At cathode

$M n O_{2} (s) + 2 N H_{4} C l + 2 e^{-} ⟶ M n_{2} O_{3} (s) + 2 N H_{3} + H_{2} O (l)$

Overall reaction

$Z n (s) + M n O_{2} (s) + 2 N H_{4} C l ⟶ Z n^{2 +} (a q) + M n_{2} O_{3} (s) + 2 N H_{3} + H_{2} O (l)$
Zinc and ammonia react to form the complex given in the answer.

Suggest Corrections

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N H_{4} C l

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{Z n}^{+ 2} + 4 {N H}_{3} ⇌ {[Z n {({N H}_{3})}_{4}]}^{+ 2}

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E_{{[Z n {({N H}_{3})}_{4}]}^{+ 2} / Z n + 4 {N H}_{3}}^{o} = - 1.03 V

E_{Z n / {Z n}^{+ 2}}

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