Question

Oxidant and Reductanct in the following reactions are respectively

$10H^{+}\left(aq\right)+4Zn\left(s\right)+N{O}_3^{-}\left(aq\right)\to 4Z{n}^{2+}\left(aq\right)+N{H}_4^{+}\left(aq\right)+3H_{2}O\left(l\right)$

• A. Zn , N
• B. N, Zn
• C. Zn, O
• D. O, Zn

Answer 1

The correct option is B

N, Zn

First let us assign oxidation numbers in the equation.

$10\stackrel{+1}{H^{+}}\left(aq\right)+4\stackrel{0}{Z}n\left(s\right)+\stackrel{+5}{N}\stackrel{-2}{O_3^{-}}\left(aq\right)\to 4\stackrel{+2}{Z}{n}^{2+}\left(aq\right)+\stackrel{-3}{N}\stackrel{+1}{H_4^{+}}\left(aq\right)+3{\stackrel{+1}{H}}_2\stackrel{-2}{O}\left(l\right)$

Oxidation No. of Zn is increasing $\Rightarrow$ Oxidising $\Rightarrow$ Reductant

Oxidation No. of N is decreasing$\Rightarrow$ reducing $\Rightarrow$ Oxidant

Oxidant : $\to N$

Reductant: $\to Zn$

Note $\Rightarrow$ Oxidation is increase in oxidation in oxidation state while reduction is decrease in oxidation state.