Question

Please solve the question.

$CU_{2} + HNO_{3} ------\ \textgreater \ Cu(NO_{3})_{2} + CuSO_{4} + NO_{2} + H_{2}O$

Answer 1

Answer:

Cu0+H+1N+5O3−2→Cu+2(N+5O3−2)2+N+4O2−2+H2+1O−2

Change in Ox. no. has occurred in copper and nitrogen.

Cu0→Cu+2(NO3)2

HN+5O3→N+4O2

Increase in Ox. no. of copper =2 units per molecule Cu.

Decrease in Ox. no. of nitrogen=1 unit per molecule HNO3.

To make increase and decrease equal, eq. (ii) is multiplied by 2

Cu+2HNO3→Cu(NO3)2+2NO2+H2O

Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained:

Cu+4HNO3→Cu(NO3)2+2NO2+2H2O

This is the balanced equation.

Answer 2

Answer:

$\sf\fbox\red{Answer:-}$

CuO+H+1N+503-2-Cu+2(N+503-2)2+N+

402-2+H2+10-2

Change in Ox. no. has occurred in copper and nitrogen.

CuO Cu+2(NO3)2 HN+503➡N+402

Increase in Ox. no. of copper =2 units per molecule Cu.

Decrease in Ox. no. of nitrogen=1 unit per molecule HNO3.

To make increase and decrease equal, eq. (ii) is multiplied by 2

Cu+2HNO3 Cu(NO3)2+2NO2+H2O

Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained:

Cu+4HNO3➡Cu(NO3)2+2NO2+2H2O