The oxidation of any element, by itself, is zero.
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
Cu
Yes, it is. It shows two oxidation states: Cu(I) and Cu(II).
Hydrogen's oxidation number is +1.Chlorin's oxidation number is +1.Oxygen's oxidation number is -2.
Both oxidation and reduction
Cu = +1
The oxidation number tells you the "combining power" of that element. For example, if Cu has a 1+ oxidation number then it will combine with Cl in a 1:1 ratio, and result in CuCl. If Cu has an oxidation number of 2+, then it will combine with Cl in a 1:2 ratio, and result in CuCl2.
Cu is +2 and F is -1
Oxidation number is the charge per atom in a compound. Cl2= 2- (Cl= 1-) Cu would have to be 2+ to balance the compound, because there is only one copper atom.
If the question is Cu2, then it is equivalent to Cu and the oxidation number for any element is zero. If the question is Cu2+, then the oxidation number is +2.
cu =1+ fe=3+ s=2-
+2 for Cu, +4 for C, -2 for O
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
Cu
+1 for Cu, +1 for H, +6 for S, -2 for each O
Cu ----> Cu 2+ + 2e-
Cl2 + 2Cu --> 2CuCl Oxidation reaction is Cu --> Cu+ + 1e Reduction reaction is Cl + 1e --> Cl- Redox reaction is Cu + Cl --> Cu+ + Cl-