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Problem:

Balance the following reaction using the ion-electron method:

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

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?
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    $\begingroup$ Your book is wrong. And you are right. In acidic solution $\ce{H2S}$ is not dissociated, and not deprotonized. It reacts as $\ce{H2S}$. Anyway the ion $\ce{S^{2-}}$ is nearly inexistant even in rather basic solution. The main ion produced by $\ce{H2S}$ in basic solution is $\ce{HS-}$ $\endgroup$
    – Maurice
    May 10 at 16:18

1 Answer 1

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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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