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Structural Isomerism

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Question

Write Chemical equation when $N_{2} H_{4}$ is oxidised to $N_{2}$ in acidic medium and to $N_{3}^{-}$ in basic medium.

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Solution

0. In basic medium, $N_{2} H_{4}$ oxidized to $N_{2}$
$N_{2} H_{4} + 4 O H^{-} \to N_{2} + 4 H_{2} O + 4 e^{-}$

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Similar questions

N_{2} H_{4}

N_{2}

N_{2} H_{4}

N_{2} H_{4}

I O_{3}^{-}

I^{+}

N_{2} H_{4} + I O_{3}^{-} + H^{+} ⟶ N_{2} + I^{+} + H_{2} O

N_{2} H_{4}

N_{2}

N -

N_{2}

N_{2} H_{4}

C r O_{4}^{2 -}

C r_{2} O_{7}^{2 -}

C r_{2} O_{7}^{2 -}

C r O_{4}^{2 -}

R - N = C

N a_{2} S_{2} O_{3} \to N a_{2} S O_{4}

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