Question

Balance the redox reaction in acidic solution:
Cu(s)+NO−3(aq)→Cu2+(aq)+NO2(g)

Answer 1

Oxidation-reduction reaction or redox reaction:

A reaction in which one reactant is oxidized and another is reduced.

Oxidation:

Oxidation is a process in which either 1 or all following changes occurs:

1. Gaining of oxygen atoms
2. Loss of electrons
3. Loss of hydrogen atom.
4. Increasing oxidation number.

Reduction:

Reduction is a process in which either 1 or all following changes occurs:

1. Loss of oxygen atoms
2. Gaining of electrons
3. Gaining of hydrogen atom.
4. Decreasing oxidation number.

Oxidation and reduction process are reverse process.

The given reaction is as follows:

Cu(s) + NO3(aq)---> Cu2+ (aq) + NO2(g)

First we determine oxidation and reduction reaction in the given reaction:

Oxidation: Cu(s) -----> Cu2+(aq) + 2e- ----1

Due to increase of oxidation number.

Reduction: NO31-(aq) + e- -----> NO2(g) ---2

Due to decrease of oxidation number

(2)Balance each half-reaction with respect to atoms and electrons.
Oxidation: Cu(s) -----> Cu2+(aq) + 2e-
Reduction: 2NO3-(aq) + 2 e- -----> 2NO2(g)

(3)

Balance reduction reaction by H+, and water molecules.

Reduction: 2NO3-(aq) + 2 e- + 4H+(aq) -----> 2NO2(g) + 2H2O(l)

                               

(4)Add the two half-reactions :

Cu(s) -----> Cu2+(aq) + 2e-

2NO3-(aq) + 2 e- + 4H+(aq) -----> 2NO2(g) + 2H2O(l)

Cu(s) + 2NO3-(aq) + 4H+(aq) -----> Cu2+(aq) + 2NO2(g) + 2H2O(l)