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

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.
I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

\begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\ \ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction} \end{align}


In this, what about $\ce{CuO}$ in the product side? I have also learned that we need to balance the spectator ions 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.

  • $\begingroup$ 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? $\endgroup$ – Silvio Levy Jul 22 '14 at 7:05
  • $\begingroup$ 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: $\ce{1 CuS + x \cdot SO4^{2-} + \cdots}$) $\endgroup$ – Silvio Levy Jul 22 '14 at 7:11
  • $\begingroup$ @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 :-) $\endgroup$ – Freddy Jul 22 '14 at 7:19
  • $\begingroup$ @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. $\endgroup$ – Greg E. Jul 22 '14 at 7:25
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    $\begingroup$ @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. $\endgroup$ – Greg E. Jul 22 '14 at 7:27

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.

  1. Writing down half reactions: \begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2}&& \text{Oxidation}\\ \ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction} \end{align}

  2. Adding electrons in both reactions: \begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2 + 6e-}&& \text{Oxidation}\\ \ce{^{(6)}SO4^{2-} + 2e- &-> ^{(4)}SO2}&& \text{Reduction} \end{align}

  3. Balancing $\ce{Cu}$: \begin{align} \ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- + Cu^{+2}}&& \text{Oxidation} \end{align}

  4. Balancing charges of both half reactions: \begin{align} \ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\ \ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2}&& \text{Reduction} \end{align}

  5. Balancing oxygen and hydrogen: \begin{align} \ce{CuS^{(-2)} + 3H2O&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\ \ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2 + 2H2O}&& \text{Reduction} \end{align}

  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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