Identify from the reaction the reactants that undergo oxidation and reduction.
$F e + S \to F e S$

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Solution

The oxidation states of $F e$ and $S$ in $F e S$ are $+ 2$ and $- 2$ respectively. Therefore $F e$ is oxidized and $S$ is reduced to $F e S$ .
$F e - 2 e^{-} ⟶ F e^{2 +}$ [Oxidation]
$S + 2 e^{-} ⟶ S^{2 -}$ [Reduction]

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