Here's the equation to balance:

Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Here are the half-reactions the wrong answer:

Sb2S3 + 6e¯ ---> 2Sb + 3S2¯
C + H2O ---> CO + 2H+ + 2e¯
4H+ + CO32¯ + 2e¯ ---> CO + 2H2O

Multiply the second half-reaction through by 4:

Sb2S3 + 6e¯ ---> 2Sb + 3S2¯
4C + H2O ---> 4CO + 8H+ + 8e¯
4H+ + CO32¯ + 2e¯ ---> CO + 2H2O

Add the haf-reactions together and simplify to get the final answer (which, I emphasize, is wrong):

Sb2S3 + CO32¯ + 4C + 2H2O ---> 2Sb + 3S2¯ + 5CO + 4H+

Why is it wrong?

1) In going from the net ionic equation back to the molecular equation, there's no way to recover the sodium ion without introducing a new compound. Look to the left-hand side of the equation and you will see that two sodium ions are required to make one sodium carbonate. Now, look to the right-hand side of the equation. You will see that six sodium ions are required to make three sodium sulfides.

Where are the other four sodium ions needed on the left-hand side going to come from? They can only come from introducing a new compound, say four sodium chlorides. (The four chloride ions, incidently would combine with the four hydrogen ions to make four HCl. There is no loss of charge balance by introducing the chlorides.)

However, I maintain introducing a new compound, not written in the problem, is unacceptable! (Unless there is no other way.)

2) The above solution ignores the nature of chemical reactivity. The sulfide ions would not stay in solution in the presence of hydrogen ions. These two ions would immediately combine and there would be a production of H2S gas from the reaction.

However, no such formula is shown in the problem. For the above solution to be correct, we would have to act as if the original problem statement was incomplete.

Now, the problem might be deliberately incomplete and so there would be a chemical reason to introduce a new compound. That chemical reason does not exist in this problem.

By the way, there is a compound present that is often not shown in the original problem statement. That compound is water. It is present because almost every chemical reaction discussed in the early years of chemistry classes happens in water solution.

Interestingly, the reaction discussed above does NOT occur in aqueous solution. It occurs at high temperature (several hundred degree Celsius) so no aqueous solution can be involved, as occurred in the above wrong answer.

In the correct answer, all items (H+, H2O) used to help balance the equation disappeared in the final answer.