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.
reduced
Both oxidation and reduction
2 Fe + 3 CuO -> Fe2O3 + 3 Cu
CuO
The chemical formula of cupric oxide is CuO.
Yes, that is correct. When copper carbonate (CuCO3) is heated, it decomposes into copper(II) oxide (CuO) and carbon dioxide (CO2).
Both oxidation and reduction
cuo
CuO + 2HCL - CuCl2 + H2O
cuo
heating of copper nitrate at 200 degree yields copper oxide
When a metal oxide (such as Copper oxide, CuO) is reduced, it is separated from its oxygen and usually becomes a solid metal again. It usually needs the help of an acid (such as hydrochloric acid or nitric acid) to make this happen. In combination with the oxygen (CuO), the metal has a positive charge (it is a cation). When it is reduced, it gains electrons from a more reactive metal (such as zinc) and its charge drops to 0. Copper is reduced from Cu+2 to Cu(s). The other metal is oxidized. The zinc, in this case, looses electrons and becomes Zn+2. Here is the balanced reaction. CuO(s) +2 HCl (aq) à CuCl2(aq) + H2O The acid strips the Copper from the Oxygen, but the charges are the same. CuCl2(aq) + Zn(s) à Cu(s) + ZnCl2(aq) The copper is reduced and the zinc is oxidized.
2 Fe + 3 CuO -> Fe2O3 + 3 Cu
CuO + 2HCl --> CuCl2 + H2O This equation is a double displacement; because Cu (Copper) has pushed out O (oxygen), so that Cu can be with Cl (chlorine). Also, H (Hydrogen) has pushed out Cl to be with O. Both elements have pushed out their partner, therefore a double displacement.
copper oxide.......CuO
CuO
no reaction
light blue color