# How to balance the redox reaction between bismuth(III) sulfide and nitrate anion?

I have to solve the following redox reaction which occurs in an acidic environment, using the half-equation method.

$$\ce {Bi2S3_{(s)} + NO3-_{(aq)} -> BiO3-_{(aq)} + SO4^2-_{(aq)} + NO_{(g)}}$$

There seems to be a lot elements that change their oxidation numbers, except for oxygen:

• $\ce{Bi}$ from $\ce{+III}$ to $\ce{+V}$
• $\ce{S}$ from $\ce{-II}$ to $\ce{+VI}$
• $\ce{N}$ from $\ce{+V}$ to $\ce{+II}$

What should I begin with? How would one go about grouping three redox pairs?

• If possible, please try not to use MathJax in titles on Chemistry.SE. Please see this meta discussion for more details. Apr 8 '17 at 18:30
• There might not be a unique answer. Multiple redoxes are difficult because you cannot necessarily add two reduction potentials or two oxidation potentials...
– Zhe
Apr 8 '17 at 18:39
• N is from +5, not +3. Apr 9 '17 at 5:30

The key is to note that both elements in $\ce{Bi2S3}$ are oxidized, and hence we have a stoichiometric relationship between the two oxidation half-reactions. This suggests we write the oxidation half-reaction as follows: $$\ce{18H2O + Bi2S3 -> 2BiO3- + 3SO4^{2-} + 36H+ + 28e-,}$$ and we can proceed with the usual algorithm for balancing half-reactions.