Why did my book write S2- if hydrogen sulfide isn't ionic?

**Problem:**


Balance the following reaction using the [ion-electron method](https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.02%3A_Balancing_Redox_Reactions#:%7E:text=systematic%20method.%20The-,ion%2Delectron%20method,-allows%20one%20to):


$$\ce{KMnO4 + H2SO4 + H2S -> MnSO4 + S + K2SO4 +H2O}$$


**My book's solution:**


Oxidation half-reaction:


$$\ce{S^{2-}->S + 2e^{-}}\tag{1}$$


$$...$$


**My question:**


1. Is writing $(1)$ correct? I'm asking this because $\ce{H2S}$ is not an ionic compound. How can my book write $\ce{S^2-}$ then?




It is indeed better to render (1) as


$\ce{H2S -> S +2H^+ + 2e^-}.$


Since you did not ask, I assume you know the permanganate reduction equation. When you combine this reduction with the correctly written reaction for hydrogen sulfide oxidation in acid you should get


$\ce{5 H2S + 2 MnO4^- + 6H^+ -> 5 S + 2 Mn^{2+} + 8 H2O,}$


where the net hydrogen ions on the left side are interpreted as coming from the sulfuric acid (which is not oxidized or reduced, so does not appear directly in the half-equations). You should then infer that the balanced ratio of reactants is $\ce{5H2S:2KMnO4:3H2SO4}$.


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