# How do I write a balanced redox reaction for the oxidation of chalcocite [(Cu2S) (s)] by O2 (g) to form Cu2+ (aq) and SO42- (aq) [closed]

I'm not exactly sure how to get the oxidation and reduction half reactions. Any help will be much appreciated!

• Do you know what is the oxidation states of Cu and S in the reagents? Apr 4 at 16:12
• Apparently I'm supposed to arrive at this answer: Cu2S +5/2O2 +(2H+) gives 2Cu2+ + SO4^2- +H2O Apr 4 at 18:18
• I'm just not sure how to write the reduction and oxidation half reactions to arrive at the above given answer. Its a question from the book ''Soil and water chemistry'' by Michael Essington Apr 4 at 18:20

I think that the half-reaction method for this reaction would look something like this:

• $$\ce{Cu2S + O2 + H2SO4 -> CuSO4 + H2O}$$

• $$\ce{2Cu^+ -2e^- -> 2Cu^2+}$$ (oxidation)
• $$\ce{S^2- + 4H2O - 8e^- -> SO4^- + 8H^+}$$ (oxidation)
• $$\ce{O2 + 4H^+ +4e^- -> 2H2O}$$ (reduction)

• $$\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$$

I think that the following method would look parallel to the method mentioned in the above post:

• $$\ce{4H2O + Cu2S + H2SO4 -> 2CuSO4 + 10H+ + 10e^- }$$ (oxidation)
• $$5(\ce{Cu2S + 3O2 + H2SO4 + 2H+ + 2e^- ->2CuSO4 + 2H2O})$$ (reduction)

• $$\ce{6Cu2S + 15O2 + 6H2SO4 -> 12CuSO4 + 6H2O}$$

• $$\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$$