Question #5a2d7

May 18, 2015

You can identify a redox reaction by determining whether or not the oxidation state of an atom changes during a chemical reaction.

For a given atom, if the oxidation state decreases, that means the atom has gained electrons, or has been reduced; likewise, if the oxidation state increases, that means the atom lost electrons, or has been oxidized.

So, let's take the equations one by one and try to see if the oxidation state of any of the atoms changes

$\stackrel{\textcolor{b l u e}{+ 2}}{M g} \stackrel{\textcolor{b l u e}{- 2}}{O} + \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{O} \to \stackrel{\textcolor{b l u e}{+ 2}}{M g} {\left(\stackrel{\textcolor{b l u e}{- 2}}{O} \stackrel{\textcolor{b l u e}{+ 1}}{H}\right)}_{2}$

Notice that the oxidation states of all the atoms involved in the reaction do not change. This means that this reaction is not a redox reaction.

$2 \stackrel{\textcolor{b l u e}{0}}{{O}_{3}} \to 3 \stackrel{\textcolor{b l u e}{0}}{{O}_{2}}$

Once again, the oxidation state of the oxygen atoms was not changed by the reaction, so you're not dealing with a redox reaction.

$\stackrel{\textcolor{b l u e}{0}}{Z n} + \stackrel{\textcolor{b l u e}{+ 2}}{C u} \stackrel{\textcolor{b l u e}{+ 6}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z n} \stackrel{\textcolor{b l u e}{+ 6}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}} + \stackrel{\textcolor{b l u e}{0}}{C u}$

Notice that the oxidation state of zinc went from 0 on the reactants' side, to +2 on the products' side, which means that it has been oxidized.

At the same time, the oxidation state of copper went from +2 on the reactants' side, to 0 on the products' side, which means that it has been reduced.

Zinc metal lost two electrons to become $Z {n}^{2 +}$, while $C {u}^{2 +}$ gained two electrons to become copper metal.

One species was oxidized and another was reduced, so this qualifies as a redox reaction.

FInally,

$\stackrel{\textcolor{b l u e}{+ 2}}{F e} \stackrel{\textcolor{b l u e}{- 2}}{S} + 2 \stackrel{\textcolor{b l u e}{+ 1}}{H} \stackrel{\textcolor{b l u e}{- 1}}{C l} \to \stackrel{\textcolor{b l u e}{+ 2}}{F e} \stackrel{\textcolor{b l u e}{- 1}}{C {l}_{2}} + \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{S}$

The oxidation states of all the atoms remain unchanged by the reaction, so this cannot be considered a redox reaction.