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Q.

Balance the half reaction, ${NO}_{2}^{-} \to_{}^{H^{+}} {NO}_{3}^{-}$ and calculate the product of number of electrons and $H^{+}$ ions involved in the balanced half reaction

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a

2

b

8

c

6

d

4

answer is B.

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Detailed Solution

$\begin{array}{l} {NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} \\ {NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} + 2 H^{+} \\ {NO}_{2}^{-} + H_{2} O \to {NO}_{3}^{-} + 2 H^{+} + 2 \bar{e} \\ Product of number of electrons and H^{+} ions = 2 \times 2 = 4 \end{array}$

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