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Oxidation Number Method

For the react...

Question

For the reaction, ${N H}_{3} + {O C l}^{-} \to N_{2} H_{4} + {C l}^{-}$ occurring in basic medium, the coefficient of $N_{2} H_{4}$ in the balanced equation will be :

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Solution

The correct option is A 1
Reaction is
$N H_{3} + O C l^{-} ⟶ N_{2} H_{4} + {C l}^{-}$

For ammonia,
$2 N H_{3} ⟶ N_{2} H_{4} + 2 H^{+} + 2 e^{-}$
Since, it is in basic media

For $O C l^{-}$ ,
$O C l^{-} + 2 H^{+} + 2 e^{-} ⟶ C l^{-} + H_{2} O$

Adding both equations we get,
$2 N H_{3} + O C l^{-} ⟶ N_{2} H_{4} + {C l}^{-} + H_{2} O$

Coefficient of $N_{2} H_{4}$ is 1.

Suggest Corrections

Similar questions

B r O_{3}^{-} + N_{2} H_{4} ⟶ {B r}^{-} + N_{2}

{N O}_{3}^{-} + B i (s) \to {B i}^{3 +} + {N O}_{2}

F e {(O H)}_{3} (s) + H_{2} O_{2} \to F e {(O H)}_{2} (s) + H_{2} O + O_{2}

{C r}_{2} O_{7}^{2 -} + C_{2} H_{4} O \to C_{2} H_{4} O_{2} + {C r}^{3 +}

N_{2} H_{4} + H_{2} O ⇌ N_{2} H_{5}^{+} + {O H}^{-}

25^{o} C

2 \times 10^{- 6}

N_{2} H_{4}

0.01 M

[N_{2} H_{5}^{+}]

Δ {H_{f}}^{o} (N_{2} H_{4}, l) = 50 k J / m o l, Δ {H_{f}}^{o} ({N H}_{3}, g) = - 46 k J / m o l

B . E . (N - H) = 393 k J / m o l, B . E . (H - H) = 436 k J / m o l

Δ_{v a p} H (N_{2} H_{4}, l) = 18 k J / m o l

N - N

N_{2} H_{4}

0.16 g

N_{2} H_{4}

500 m L

N_{2} H_{4}

K_{b}

N_{2} H_{4}

4.0 \times 10^{- 6} M

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