I'm supposed to balance a redox equation. I arrive at $$ \ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} $$
The "official" solution is $$ 4 \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow 4\ce{NH3} + 2\ce{S} + 2\ce{H2O} $$
There is no information about if this is occuring in basic or acidic medium. But IMHO in the first equation the charges and all atoms are balanced, so are these equations equivalent?
Let's start with the question: How do we get from your answer to the official answer? We can transform your answer into the official answer by first adding 3 ammonium ions to each side:
$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\ + 3 \ce{NH3}+3\ce{H^+} & \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\ \ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-} \end{aligned}$
Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:
$\begin{aligned} \ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\ -\ce{H2O} & \rightarrow - \ce{H2O}\\ 4\ce{NH4+}+2\ce{S^2-}+\ce{O2} & \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O} \end{aligned}$
Producing what is the 'official' result. Now we can ask: are the two answers equivalent? Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:
$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$
Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation. I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.
