In a half reaction, nitrate is reduced by 6e- reduction to x as follows 7H++NO3 -+6e- arrow 2H2O+x . The 'x' in the reaction is

Question

In a half reaction, nitrate is reduced by 6e- reduction to x as follows 7H++NO3 -+6e- arrow 2H2O+x . The 'x' in the reaction is (A) NO (B) NH2NH2 (C) NH3 (D) NH2OH

Tardigrade Admin · Accepted Answer

overset+ 5NO3 - arrow overset- 1NH2OH , Change in oxidation number is 6

Q. In a half reaction, nitrate is reduced by $6 e^{-}$ reduction to $x$ as follows $7 H^{+} + N O_{3}^{-} + 6 e^{-} \to 2 H_{2} O + x$ . The $^{'} x^{'}$ in the reaction is

$NO$

$N H_{2} N H_{2}$

$N H_{3}$

$N H_{2} O H$

$N + 5 O_{3}^{-} \to N - 1 H_{2} O H$ ,
Change in oxidation number is $6$

Q. In a half reaction, nitrate is reduced by 6e− reduction to x as follows 7H++NO3−​+6e−→2H2​O+x . The ′x′ in the reaction is

NO

NH2​NH2​

NH3​

NH2​OH