What is the final, balanced equation that is formed by combining these two half reactions?
Cu-Cu2+ + 2e-
NO3 + 2e + 2H* —>N0Z+H20
Cu2+ + NO3 + 4e- + 2H+ -->Cu +
u + NO3 + H20
Cu + NO3
+ 2HCu2+ + NO2 + H20
2Cu + NO3 +
+ 2H+ ->20u2+ +
+ NO 2 + H20
Cu+ NO3 +
+ 2H+
Cu2+ + NO + + 2H20

Cu + 2NO3- + 4H+ > Cu2+ + 2NO2 + 2H20

Explanation:

The half reactions are;Cu > Cu2+ + 2e-

NO3 + 2e + 2H* —> N02 + H20

It is worthy to note that oxidation is the an increase in oxidation number while reduction is a decrease in oxidation number.

In the first half reaction involving Copper, copper has been oxidized as its oxidation number increased from 0 to +2.

IN the reduction half reaction, NO3- has been reduced as its oxidation number decreased from +5 to +4.

In balancing half reactions, the following must be observed;

1. split the reactions into half equations, one for oxidation and the other for reduction.

Cu > Cu2+ + 2e-   ( oxidation)

NO3- + 2e- > NO2   (reduction).

2. balance on half with respect to atoms and charges.

3. balance the other half with respect to atoms and charges

4. balance by adding H+ ions in acidic solution

NO3- + e- > NO2

(-1 )       (-1)             (0)

NO3- + 2H+ +e- > NO2

(-1)       (+2)     (-1)            (0)

5. balance hydrogen by adding water molecules

NO3 - + 2H+ + e - >  NO2 + H2O

6. combine both equations

Cu ---> Cu2+ + 2e-                                  *1

NO3- + 2H+ + e- > NO2 + H2O      *2

= Cu > Cu2+ + 2e-

2NO3- +4H+ + 2e- > 2 NO2 + 2H20

Combined equation  = Cu + 2NO3- + 4H+ > Cu2+ + 2NO2 + 2H20