# Are these two reactions equivalent?

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?

\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:
$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$