Question

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

(a) \({\bf{C}}{{\bf{l}}^{\bf{ - }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{C}}{{\bf{l}}_{\bf{2}}}\)

(b) \({\bf{M}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}\)

(c) \({{\bf{H}}_{\bf{2}}} \to {{\bf{H}}^{\bf{ + }}}\)

(d) \({\bf{NO}}_{\bf{3}}^{\bf{ - }} \to {\bf{NO}}\)

Short answer

1. Loss of electron is oxidation thus, \({\rm{C}}{{\rm{l}}^ - }\)loses two electrons and gets reduced to\({\rm{C}}{{\rm{l}}_2}\).
2. Loss of electrons is oxidation thus, \({\rm{M}}{{\rm{n}}^{2 + }}\) loses two electrons and gets oxidized to \({\rm{Mn}}{{\rm{O}}_2}\).
3. Loss of electrons is oxidation thus, \({{\rm{H}}_2}\) loses two electrons and gets oxidized to \({{\rm{H}}^ + }\).
4. Gain of electron is reduction thus, \({\rm{NO}}_3^ - \) gains three electrons and gets reduced to NO.

Step-by-step solution

Define oxidation and reduction

• 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\({{\bf{A}}^{\bf{ + }}}{\bf{(aq)}}\)with release of 1 electron. The oxidation state of A increases from 0 to 1.

• Similarly,

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

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.

a) Determine whether oxidation or reduction is occurring

The given reaction is as follows:

\({\rm{C}}{{\rm{l}}^ - } \to {\rm{C}}{{\rm{l}}_2}\)

Give coefficient 2 to \({\rm{C}}{{\rm{l}}^ - }\)to balance the number of chlorine atom and add 2 electrons to the right to balance the charge. Thus, balanced chemical reaction will be:\(2{\rm{C}}{{\rm{l}}^ - } \to {\rm{C}}{{\rm{l}}_2} + 2{e^ - }\)

The oxidation state of \({\rm{Cl}}\) in the reactant is \( - 1\) and on the product side is zero. Thus, the oxidation state of Cl increases in the above reaction.

Thus, \({\rm{C}}{{\rm{l}}^ - }\)gets oxidized to \({\rm{C}}{{\rm{l}}_2}\) in the reaction and in the above reaction, oxidation occurs.

Also, the loss of electron is oxidation thus, \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) loses two electrons and gets reduced to \({\rm{C}}{{\rm{l}}_2}\).

b) Determine whether an oxidation or reduction is occurring

The given reaction is as follows:

\({\rm{M}}{{\rm{n}}^{2 + }} \to {\rm{Mn}}{{\rm{O}}_2}\)

Balance the oxygen atom by adding two water molecules on the left side of the reaction arrow.

\({\rm{M}}{{\rm{n}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}} \to {\rm{Mn}}{{\rm{O}}_2}\)

Now, add four hydrogen ions to balance the hydrogen atoms.

\({\rm{M}}{{\rm{n}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}} \to {\rm{Mn}}{{\rm{O}}_2} + 4{{\rm{H}}^ + }\)

Now, balance the charge by adding two electrons to the right side of the reaction arrow.

\({\rm{M}}{{\rm{n}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}} \to {\rm{Mn}}{{\rm{O}}_2} + 4{{\rm{H}}^ + } + 2{e^ - }\)

The oxidation state of Mn in the reactant side is \( + 2\) and on product side it is \( + 4\).

Thus, the oxidation state of \({\rm{Mn}}\) increases from \( + 2\) to \( + 4\) in the above reaction.

Thus, \({\rm{M}}{{\rm{n}}^{2 + }}\) gets oxidized to \({\rm{M}}{{\rm{n}}^{4 + }}\) in the reaction and in the above reaction oxidation is occurring.

Also, loss of electrons is oxidation thus, \({\rm{M}}{{\rm{n}}^{2 + }}\) loses two electrons and gets oxidized to \({\rm{Mn}}{{\rm{O}}_2}\).

c) Determine whether oxidation or reduction occurs

The given reaction is as follows:

\({{\rm{H}}_2} \to {{\rm{H}}^ + }\)

To balance the above reaction, give coefficient 2 to \({{\rm{H}}^ + }\)on the product side and add 2 electrons to balance the charge. Thus, the balanced chemical reaction will be:

\({{\rm{H}}_2} \to 2{{\rm{H}}^ + } + 2{e^ - }\)

The oxidation state of \({\rm{H}}\) in the reactant side is zero and the product side is \( + 1\). Thus, the oxidation state of \({\rm{H}}\) increases from 0 to \( + 1\) in the above reaction.

Thus, \({{\rm{H}}_2}\) gets oxidized to \({{\rm{H}}^ + }\)in the reaction and in the above reaction, oxidation occurs.

Also, loss of electrons is oxidation thus, \({{\rm{H}}_2}\) loses two electrons and gets oxidized to \({{\rm{H}}^ + }\).

d) Determine whether an oxidation or reduction is occurring

The given reaction is as follows:

\({\rm{NO}}_3^ - \to {\rm{NO}}\)

First balance the number of oxygen atoms by adding two molecules of \({{\rm{H}}_2}{\rm{O}}\) on the right side of the reaction arrow.

\({\rm{NO}}_3^ - \to {\rm{NO}} + 2{{\rm{H}}_2}{\rm{O}}\)

Now, balance the hydrogen atoms by adding four hydrogen ions on the reactant side as follows:

\({\rm{NO}}_3^ - + 4{{\rm{H}}^ + } \to {\rm{NO}} + 2{{\rm{H}}_2}{\rm{O}}\)

To balance the charge, add three electrons on the left side as follows:

\({\rm{NO}}_3^ - + 4{{\rm{H}}^ + } + 3{e^ - } \to {\rm{NO}} + 2{{\rm{H}}_2}{\rm{O}}\)

The oxidation state of \(N\) in the reactant is \( + 5\) and on product side is \( + 2\). Thus, the oxidation state of \(N\) decreases in the above reaction.

Thus, \({\rm{NO}}_3^ - \)gets reduced to \({\rm{NO}}\) in the reaction and in the above reaction, reduction occurs.

Also, the gain of electron is reduction thus, \({\rm{NO}}_3^ - \)gains three electrons and gets reduced to NO.