Question

In the equation: $N{O}_2^{-}+H_{2}O\to N{O}_3^{-}+2H^{+}+n{e}^{-}$, $n$ stands for :

Answer 1

Given a reaction:
$N{O}_2^{-}+H_{2}O\to N{O}_3^{-}+2H^{+}+2e^{-}$
+3 +5
The change in oxidation state for $N$ is $+2$.
So, $n$ is $2$.