The secondary reactions which takes place in dry cell are-1- MnO-2-NH-4-e- MnO-OH-NH-3-2- 2NH-4Cl-2-OH- -160- 2NH-3-2Cl- 2H-2O-3- -160- Zn-2MnO-2- -160- -Zn-NH-3-2-2OH-2MnO-OH-4- -160- Zn-2-2NH-3-2Cl- -160- -Zn-NH-3-2-Cl-2 or ZnCl-2-2NH-3-Correct statements are-1-2-41-2-3-42-4Only 4

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Dry Cell

The secondary...

Question

The secondary reactions which takes place in dry cell are:
1) $M n O_{2} + N H_{4}^{+} + e^{-} \to M n O (O H) + N H_{3}$
2) $2 N H_{4} C l + 2 (O H^{-}) \to$ $2 N H_{3} + 2 C l^{-} + 2 H_{2} O$
3) $Z n + 2 M n O_{2} \to$ $[Z n (N H_{3})_{2}] + 2 O H^{-} + 2 M n O (O H)$
4) $Z n^{2 +} 2 N H_{3} + 2 C l^{-} \to$ $[Z n (N H_{3})_{2}] C l_{2} or Z n C l_{2} .2 N H_{3}$
Correct statements are:

1,2,4

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1,2,3,4

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2,4

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Only 4

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In the reaction Zn +2 $H^{+}$ +2 $C l^{-}$ $⟶$ $Z n^{+} 2$ + 2 $C l^{-}$ + $H_{2}$ the spectator ion

$Z n + 2 H^{+} + 2 C l^{-} - - - - - - - > Z n^{2 +} + 2 C l^{-} + H_{2}$

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{\circ} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q), E^{\circ} = 0.34 V

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

△ G^{\circ}

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

E_{A g^{+} / A g}^{\circ} = + 0.80 V

E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V

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