# Sulfate as product of Cu2S oxidation and reaction medium [closed]

The question asks for a balanced reaction for oxidation of $$\ce{Cu2S}$$ to ferric sulfate in sulfuric acid solution to form cupric sulfate.

I know the first half reaction is $$\ce{Fe^3+ +e- -> Fe^2+}$$

I also know $$\ce{H+}$$ and $$\ce{SO4^2-}$$ are spectators.

But what is the equation for $$\ce{Cu2S}$$ oxidizing?

$$\ce{Cu2S -> 2Cu^2+ + SO4^2- + 4e- (?)}$$

I know $$\ce{SO4^2-}$$ is a spectator so I can count it in the end. But because of $$\ce{Cu2S}$$ my final equation does not balance due to an extra sulfur.

• Spectator ions are neither produced neither consumed by a reaction. – Poutnik Sep 16 at 5:39
• Also, chemical reactions conserve atoms and charges. – Poutnik Sep 16 at 5:42
• Your half-reaction is unbalanced in O and e-. – Ivan Neretin Sep 16 at 6:25

Starting out with

$$\ce{Cu2S->2Cu^{2+} + SO4^{2-}}$$

where you balance the elements other than hydrogen and oxygen (Step 1), is a good first step. From here you need to do the following:

Step 2) Compare the combined oxidation states of the elements you just balanced. The electrons given or taken will match the difference between the combined oxidation states, not what it takes to balance charges. That comes later.

Here the two copper and one sulfur atom add up to a combined oxidation state of $$0$$ in the neutral compound $$\ce{Cu2S}$$, but $$+10$$ in the products on the right -- $$+2$$ for each copper atom and $$+6$$ for the sulfur. So you place not four but ten electrons with the products:

$$\ce{Cu2S->2Cu^{2+} + SO4^{2-} + 10e-}$$

Step 3) Now comes the time to balance charges. In basic solution reactions you would use $$\ce{OH-}$$, in acidic solutions such as this case you use $$\ce{H+}$$ (understanding, of course, that these ions are not bare but in their heavily water-solvated states). It is not hard to see that this particular reaction then requires eight hydrogen ions among the products:

$$\ce{Cu2S->2Cu^{2+} + SO4^{2-} + 8 H+ + 10e-}$$

Step 4) Finally, you must balance the hydrogen and oxygen atoms. That is done with water molecules. If you balanced the other elements and tracked the oxidation states correctly in Steps 1 and 2, adding the water will neatly get both elements balanced in one blow. Clearly there are four water molecules among the reactants to balance the hydrogen, in which case you can check that the oxygen is indeed balanced as well to give

$$\ce{Cu2S + 4H2O->2Cu^{2+} + SO4^{2-} + 8 H+ + 10e-}$$

This should be good enough for a class setting. However, there is one final, perhaps overly pedantic point when you have sulfate in acid solution. Sulfuric acid is a strong acid, but only for dissociating one hydrogen ion unless the acid is so dilute that the pH is 2 or higher. Typically in sulfuric acid reactions the acid is more concentrated than that, so the sulfate ion should be rendered as bisulfate, $$\ce{HSO4-}$$, by using one of the hydrogen ions. So the more precise half-reaction is

$$\ce{Cu2S + 4H2O->2Cu^{2+} + HSO4- + 7 H+ + 10e-}$$