Balance the equation in aqueous basic solution:
As2S3(s) + H2O2(aq) → AsO43-(aq) + SO42-(aq)
I normally understand how to balance redox equations, but this one confuses me because I would normally think that the As2S3 is being reduced to AsO4^3-, but since the As2S3 has a "S", do i also include the SO4^2-? but even when i do that, and try to balance the electrons for the half rxn, it seems to change into oxidation... I'm really confused!
2 answers
I usually attack this type this way. Yes, both As and S are changing, As from +6(total of 2 atoms As) to +10(total of 2 atoms As). S is changing from -6 (total of 3 S atoms to +18(total of 3 atoms S). Most of the time it works out if one simply combines the As and S into one kind of reaction (add electrons algebraically). What has been omitted is the reaction of the H2O2. You'll see what to do when that is included.
Here’s a hint. The half reactions that must be balanced are As₂S₃ —> AsO₄³⁻ + SO₄²⁻ (oxidation) and H₂O₂ —> ? (reduction). You can determine which ions to group by examining oxidation numbers. Note that since H₂O₂ consists of only hydrogen and oxygen, you can instantly balance the half reaction by adding water and H⁺/OH⁻ ions as you would for any half reaction once the non-hydrogen/non-oxygen species have been balanced by inspection. This explains why there is no obvious product in the half reaction (the product is water which is added in only when balancing).
2 answers