[bookworm]

An acidic solution of MnO4- is used to remove S2- from waste water. The redox reaction forms sulphur (S) and the purple solution turns colourless because of the formation of Mn2+.

how do I balance the redox solution?

[Astrokel]

It is pretty straightforward since products are given to you but still you need to show your attempts.

[Borek]

If your textbook doesn't help, there are plenty of tutorials on the web. Like this one: balancing redox reactions.

[bookworm]

i can only get

[MnO4-] + [8H+] + [5e-] -> [Mn2+] + [4H2O]   ----- (eq1)

[S2-]  ->  + [2e-] --- (eq2)

[Borek]

First equation is OK.

You are told what is product in the second half reaction, reread the question.

[bookworm]

S2- -> S + 2e-   -- (eq2)

[Borek]

So where is the problem now? Just combine them so that electrons cancel out

[bookworm]

 
sorry, i got confused with my lecturer's answer.
from his answer script it stated : H2S -> S
and i was wondering why is it H2S when the question only states S2- for reactant and S for product.
thanks anyway

[Astrokel]

If you are considering redox reaction, S^2- is good enough as the hydrogen in hydrogen sulphide is not the one undergoing oxidation nor reduction. However if you were to write out the full equation, you do need to use H₂S. I have no idea why does your lecturer put H₂S because there are many sulphide compound, nevertheless, reaction between hydrogen sulphide and potassium permanganate is pretty well known.

[Borek]

This reaction goes in acidic condition, with pK_a1 around 7 S^2- has to be completely protonated.