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Valency and Oxidation States

2NO3-+10e-+12...

Question

$2 {N O}_{3}^{-} + 10 e^{-} + 12 H^{+} \to N_{2} + 6 H_{2} O$
Which statement is true about the reaction shown above?

Oxygen is reduced

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Hydrogen is reduced

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Nitrogen is reduced

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Oxygen is oxidized

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Solution

The correct option is D Nitrogen is reduced
In $N O_{3}^{-}$ , $x + 3 (- 2) = - 1 ⟹ x = + 5.$
But in $N_{2}$ , oxidation state of $N$ is $0$ .
So, Nitrogen is reduced from $5$ to $0.$

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Similar questions

2 B r O_{3}^{-} + 12 H^{+} + 10 e \to B r_{2} + 6 H_{2} O

N_{2}

2 N H_{3} + 3 O_{2} \to 2 N O_{2}^{-} + 2 H^{+} + 2 H_{2} O

2 N O_{2}^{-} + O_{2} \to 2 N O_{3}^{-}

2 K^{+} + B r O^{3 -} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O + 2 K

K B r O_{3}

2 K^{+} B r O_{3}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O + 2 K^{+}

K B r O_{3}

=

K B r O_{3}

85.4 m L

0.672 N

{B r O_{3}}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O

2 {B r O_{3}}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O

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