For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

(a) \({\bf{F}}{{\bf{e}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{Fe}}\)

(b) \({\bf{Cr}} \to {\bf{C}}{{\bf{r}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}}\)

(c) \({\bf{MnO}}_{\bf{4}}^{{\bf{2 - }}} \to {\bf{MnO}}_{\bf{4}}^{\bf{ - }}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}\)

(d) \({\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}} \to {\bf{Li}}\)

• Oxidation and reduction can be identified by the change in oxidation state. If the oxidation state of an atom of an element increases, it undergoes oxidation, and if it decreases, it undergoes reduction.
• The general oxidation and reduction half reactions are as follows:

\({\rm{A}}(s) \to {{\rm{A}}^ + }(aq) + e\)

Here, A(s) undergoes oxidation to form\({{\bf{A}}^{\bf{ + }}}{\bf{(aq)}}\)by releasing 1 electron. The oxidation state of A increases from 0 to 1.

\({{\rm{B}}^ + }(aq) + e \to {\rm{B}}(s)\)

Here,\({{\bf{B}}^{\bf{ + }}}{\bf{(aq)}}\)undergoes reduction to form B(s) with addition of 1 electron. The oxidation state of A decreases from 1 to 0.

The given reaction is as follows:

\({\rm{F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + 3}}{{\rm{e}}^{\rm{ - }}} \to {\rm{Fe}}\)

The oxidation state of Fe in the reactant is \( + 3\) and on the product side is zero. Thus, the oxidation state of \({\rm{Fe}}\) decreases in the above reaction.

Thus, \({\rm{F}}{{\rm{e}}^{3 + }}\) is getting reduced to \({\rm{Fe}}\) in the reaction and in the above reaction, reduction occurs.

Also, the gain of electron is reduction. Thus, \({\rm{F}}{{\rm{e}}^{3 + }}\) gains 3 electrons and gets reduced to \({\rm{Fe}}\).

The given reaction is as follows:

\({\rm{Cr}} \to {\rm{C}}{{\rm{r}}^{3 + }} + 3{e^ - }\)

The oxidation state of \({\rm{Cr}}\) in the reactant side is zero and on the product side is \( + 3\). Thus, the oxidation state of \({\rm{Cr}}\) increases from zero to \( + 3\) in the above reaction.

Thus, \({\rm{Cr}}\) is getting oxidized to \({\rm{C}}{{\rm{r}}^{3 + }}\) in the reaction, and in the above reaction, oxidation is occurs.

Also, loss of electrons is oxidation thus, \({\rm{Cr}}\) loses 3 electrons and gets oxidized to\({\rm{C}}{{\rm{r}}^{3 + }}\).

The given reaction is as follows:

\({\rm{MnO}}_4^{2 - } \to {\rm{MnO}}_4^ - + {e^ - }\)

The oxidation state of \({\rm{MnO}}_4^{2 - }\) in the reactant side is \( + 4\) and on the product side is \( + 7\). Thus, the oxidation state of Mn increases from \( + 4\) to \( + 7\) in the above reaction.

Thus, \({\rm{M}}{{\rm{n}}^{4 + }}\) is getting oxidized to \({\rm{M}}{{\rm{n}}^{7 + }}\) in the reaction and in the above reaction oxidation occurs.

Also, loss of electrons is oxidation. Thus, \({\rm{MnO}}_4^{2 - }\) loses 1 electron and gets oxidized to \({\rm{MnO}}_4^ - \).

The given reaction is as follows:

\({\rm{L}}{{\rm{i}}^ + } + {e^ - } \to {\rm{Li}}\)

The oxidation state of Li in a reactant is \( + 1\) and on the product side is zero. Thus, the oxidation state of Li decreases in the above reaction.

Thus, \({\rm{L}}{{\rm{i}}^ + }\)gets reduced to \({\rm{Li}}\) in the reaction and in the above reaction, reduction occurs.

Also, the gain of electron is reduction thus, \({\rm{L}}{{\rm{i}}^ + }\)gains 1 electron and gets reduced to Li.