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

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    $\begingroup$ Do you know what is the oxidation states of Cu and S in the reagents? $\endgroup$ – Ivan Neretin Apr 4 at 16:12
  • $\begingroup$ Apparently I'm supposed to arrive at this answer: Cu2S +5/2O2 +(2H+) gives 2Cu2+ + SO4^2- +H2O $\endgroup$ – Annie Apr 4 at 18:18
  • $\begingroup$ 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 $\endgroup$ – Annie Apr 4 at 18:20
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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)

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

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

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

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