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? $\endgroup$ Jul 22, 2014 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: $\ce{1 CuS + x \cdot SO4^{2-} + \cdots}$) $\endgroup$ Jul 22, 2014 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, 2014 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, 2014 at 7:25
  • 1
    $\begingroup$ @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, 2014 at 7:27

1 Answer 1


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.