Oxidant and Reductanct in the following reactions are respectively

$10 H^{+} (a q) + 4 Z n (s) + N O_{3}^{-} (a q) \to 4 Z n^{2 +} (a q) + N H_{4}^{+} (a q) + 3 H_{2} O (l)$

Zn , N

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N, Zn

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Zn, O

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O, Zn

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Solution

The correct option is B
N, Zn

First let us assign oxidation numbers in the equation.

$10^{+} (a q) + 4 0 Z n (s) + + 5 N - 2 O_{3}^{-} (a q) \to 4 + 2 Z n^{2 +} (a q) + - 3 N + 1 H_{4}^{+} (a q) + 3 {+ 1 H}_{2} - 2 O (l)$

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

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

Oxidant : $\to N$

Reductant: $\to Z n$

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

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Similar questions

Oxidant and Reductanct in the following reactions are respectively

$10 H^{+} (a q) + 4 Z n (s) + N O_{3}^{-} (a q) \to 4 Z n^{2 +} (a q) + N H_{4}^{+} (a q) + 3 H_{2} O (l)$

Z n (s) + N O_{3}^{-} (a q) + H^{+} (a q) ⟶ {Z n}^{2 +} (a q) + {N H}_{4}^{+} (a q)

Z n (s)

N O_{3}^{-} (a q)

{C r}_{2} O_{7}^{2 -} (a q) + H {N O}_{2} (a q) ⟶ {C r}^{3 +} (a q) + {N O}_{3}^{-} (a q)

{N H}_{4}^{+} (a q) + {C N}^{-} (a q) ⇌ H C N (a q) + {N H}_{3} (a q)

2 N O_{2} (g) + 2 O H^{-} (a q) \to N O_{3}^{-} (a q) + H_{2} O (l) + N O_{2}^{-}

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