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Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction


In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if the spectator ion is other than oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

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Your "half-reactions" involve the right oxidizer/reducer (oxygen in both cases), but a half-reaction needs to preserve atoms and charges, so you need to try again. In fact the original equation is just a shorthand -- notice that there are no charges on the left but there on the right, so something is missing. You'll need to add $\ce{H^+}$ on one side and $\ce{H_2O}$ on the other. Which goes on which side? – Silvio Levy Jul 22 '14 at 7:05
Also, does the problem say you have to use the half-reaction method? I ask because this method is not the easiest way to solve this problem (again, because O is being both oxidized and reduced). Completing the equation with $\ce{H^+}$ and water and then comparing coefficients for each type of atom and for the charges will be easier (you might want to introduce an unknown, x, on the left side: $\ce{1 CuS + x \cdot SO4^{2-} + \cdots}$) – Silvio Levy Jul 22 '14 at 7:11
@SilvioLevy i know this is not the good approach but my teacher told you should learn this so i have show him by solving this reaction only by this method :-) – Freddy Jul 22 '14 at 7:19
@SilvioLevy, I don't think oxygen changes oxidation states; it appears to be -2 in all products and reactants. Sulfur, however, is -2 in the sulfide, +6 in the sulfate, but +4 in sulfur dioxide. So, the sulfide is oxidized while the sulfate is reduced. – Greg E. Jul 22 '14 at 7:25
@Freddy, almost every sulfur redox reaction I've seen is done in acidic solution. Is that the case here? This is relevant, since the half-reaction methods of balancing equations for acidic and basic solution are different. – Greg E. Jul 22 '14 at 7:27
up vote 3 down vote accepted

1) Writing down half reactions:

(number written in brackets are oxidation number of S)

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

2) Adding electrons in both reactions

$\ce{CuS^{(-2)}-> ^{(4)}SO2 + 6e-}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} + 2e- -> ^{(4)}SO2}$ Reduction half reaction

3)Balancing $Cu$

$\ce{CuS^{(-2)}-> ^{(4)}SO2 + 6e- + Cu^{+2}}$ Oxidation half reaction

4) Balancing charges of both half reactions

$\ce{CuS^{(-2)}-> ^{(4)}SO2 + 6e- +CuO + 6H+}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} + 2e- +4H+ -> ^{(4)}SO2}$ Reduction half reaction

5) Balancing oxygen and hydrogen

$\ce{CuS^{(-2)} + 3H2O-> ^{(4)}SO2 + 6e- +CuO + 6H+}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} + 2e- +4H+ -> ^{(4)}SO2 + 2H2O}$ Reduction half reaction

6) Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction

$\ce{CuS + 3SO4^{2-} + 6H+ -> 4SO2 +CuO + 3H2O}$

(thanks to Silvio Levy for helping me to solve this problem)

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