2Cu + O2 -> 2CuO
Now you have an ionic bond because the oxygen has taken two electrons from copper into it's valance shell. Thus, copper is oxidized; loses electrons.
Zinc is oxidized more easily than copper.
Cu is being reduced and Zn is being oxidized, hence Zn + CuSO4 --> ZnSO4 + Cu
Cu(s)
-3.27V
-3.27V
Zinc is oxidized more easily than copper.
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 is being reduced and Zn is being oxidized, hence Zn + CuSO4 --> ZnSO4 + Cu
being oxidized by O2
Cu(s)
The final answer would be 2Cu O2---> 2CuO
The chemical reactions are:4 Cu(I) + O2 = 2 Cu2O2 Cu(II) + O2 = 2 CuO
oxidized
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Cu(s)
-3.27V
-3.27V