The balanced half-equations for the reaction 2Cu + O2 = 2CuO are the oxidation half-reaction Cu(s) → Cu2+(aq) + 2e- and the reduction half-reaction O2(g) + 4e- → 2O2-(aq). These reactions illustrate the electron transfer that constitutes the redox process.
The main answer in 2 lines to write two balanced half-equations for the redox reaction 2Cu + O2 = 2CuO would be:
- Oxidation half-reaction: Cu(s) → Cu2+(aq) + 2e-
- Reduction half-reaction: O2(g) + 4e- → 2O2-(aq)
In a redox reaction, oxidation and reduction processes occur simultaneously. Oxidation involves the loss of electrons, and reduction involves the gain of electrons. In the given reaction, copper (Cu) loses electrons and is oxidized to copper(II) oxide (CuO), while molecular oxygen (O2) gains electrons and is reduced. The half-reactions must be balanced for mass and charge. For the oxidation of copper, two electrons are lost per copper atom to form Cu2+. For the reduction half, each oxygen molecule gains four electrons to form two oxide ions (O2-). These half-reactions reflect the electron transfer process in the overall reaction.
balancing the half-reactions involves accounting for both the number of atoms and charges involved in the redox process. The balanced half-reactions represent the oxidation of copper and the reduction of oxygen in the formation of copper(II) oxide.