write oxidation and reduction halfreaction of 3CuO+2NH3>3Cu+N2+3H2O
Cu^2+ ==> Cu you balance.
NH3 ==> N2 you balance
To determine the oxidation and reduction half-reactions in the chemical equation 3CuO + 2NH3 → 3Cu + N2 + 3H2O, we need to assign oxidation numbers to each element in the reaction:
- Copper (Cu) has an oxidation number of +2 in CuO and 0 in Cu.
- Nitrogen (N) has an oxidation number of -3 in NH3 and 0 in N2.
- Oxygen (O) has an oxidation number of -2 in CuO and -2 in H2O.
- Hydrogen (H) has an oxidation number of +1 in NH3 and +1 in H2O.
Based on the changes in oxidation numbers, the half-reactions can be determined as follows:
Oxidation Half-Reaction:
CuO → Cu
The oxidation half-reaction shows that copper in CuO goes from an oxidation state of +2 to 0, which means it loses electrons. Since copper loses electrons, it undergoes oxidation.
Reduction Half-Reaction:
2NH3 → N2 + 3H2O
The reduction half-reaction shows that nitrogen in NH3 goes from an oxidation state of -3 to 0, which means it gains electrons. Since nitrogen gains electrons, it undergoes reduction.
Remember that in a balanced chemical equation, both the oxidation and reduction half-reactions must have an equal number of electrons. The balanced equation for the overall reaction is obtained by adding the two half-reactions together after multiplying them by suitable coefficients.