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Faraday's Second Law

NH 3 is oxidi...

Question

${N H}_{3}$ is oxidised to $N O$ by $O_{2}$ (air) in basic medium. Number of equivalents of ${N H}_{3}$ oxidised by $1$ mole of $O_{2}$ is :

4

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5

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6

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7

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Solution

The correct option is A $4$
Reaction:
$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$
-3 +2
From balanced reaction, it can be seen that one mole of $O_{2}$ oxidises $\frac{4}{5}$ moles of $N H_{3} = \frac{4 \times 5}{5} = 4$ equivalents of $N H_{3}$ (as change in oxidation state of N is 5).

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