Copper atoms are oxidised at the positive electrode to Cu2+ ions, as shown in the half equation.
Cu(s) Cu2+(aq) + 2e−
(i) How does the half equation show that copper atoms are oxidised?
Oxidation state changes reflect charges of atoms Cu2+(aq) has an oxidation state of 2, while Cu(s) has an oxidation state of 0. So, Cu(s)=Cu2+(aq)+2e->>>(2+(-2))=Cu
