Question

Write the complete balanced equation of $CuO + N{H_3} \to Cu + {N_2} + {H_2}O$ .

Answer 1

In order to answer this question, first we will do or solve the oxidation part or compound and then we will solve or balance the reduction part of the given equation. And then we will try to balance the equation by combining the oxidation and reduction parts.

Complete answer:
Oxidation:
${}^{3 - }N{H_3} \to {}^0{N_2} + 3{e^ - }$ ……(1)
Reduction:
${}^{2 + }CuO + 2{e^ - } \to {}^0Cu$ ……..(2)
Balancing $H$ atom in equation(1)-
$N{H_3} \to \dfrac{1}{2}{N_2} + 3{H^ + } + 3{e^ - }$ …………(3)
Balancing $O\,\& \,H$ atoms in equation(2)-
$CuO + 2{e^ - } + 2{H^ + } \to Cu + {H_2}O$ ………..(4)
Now, to balance the whole reaction: $eq(3) \times 2 + eq(4) \times 3$ :
$2N{H_3} + 3CuO \to {N_2} + 3Cu + 3{H_2}O$
The above reaction is the balanced equation.

Note:
Oxidation and Reduction regarding Electron Transfer. This is the most generally utilized meaning of oxidation and decrease and most broadly pertinent. For this situation, Oxidation is the deficiency of electrons and Reduction is the addition of electrons.