Question

$N{O}_2^{\ominus }+H_{2}O⟶N{O}_3^{\ominus }+2H^{\oplus }+n{e}^{-}$
$n$ stands for :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$N{O}_2^{\ominus }+H_{2}O⟶N{O}_3^{\ominus }+2H^{\oplus }+2e^{-}$

Here total 2 positive charge in whole reaction and no negative charge as the negative charge of reactant cancel the negative charge of product.

Thus,
On balancing charge on both sides, we get $n=2$.