Question

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and thereducing agent in each of the reactions of the previous problem.

\(\begin{array}{l}{\bf{(a) H_2O_2 + S}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{H_2O + S}}{{\bf{n}}^{{\bf{4 + }}}}\\{\bf{(b) PbO_2 + Hg}} \to {\bf{Hg}}{{\bf{2}}^{{\bf{2 + }}}}{\bf{ + P}}{{\bf{b}}^{{\bf{2 + }}}}\\{\bf{(c) Al + Cr_2O}}{{\bf{7}}^{{\bf{2 - }}}} \to {\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{ + C}}{{\bf{r}}^{{\bf{3 + }}}}\end{array}\)

Short answer

1. \({{\rm{H}}_2}{{\rm{O}}_2}\)is an oxidizing agent,\({\rm{S}}{{\rm{n}}^{2 + }}\) is a reducing agent.
2. \({\rm{Pb}}{{\rm{O}}_2}\)is an oxidizing agent, Hg is a reducing agent.
3. \({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }\) is an oxidizing agent, Al is a reducing agent.

Step-by-step solution

Define oxidation and reduction

The oxidation-reduction reaction is also known as a redox reaction.

In this reaction, one reactant is oxidized and the other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and the other is reduction reaction.

The balancing of a redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The oxidation and reduction can be identified by 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:

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

Here, A(s) undergoes oxidation to form A with the release of 1 electron. The oxidation state of A increases from 0 to 1.

Similarly,

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

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

 Step 2: a) Determine the balanced reaction for each pair of half reactions in an acidic solution

The given reaction is as follows:

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

In the above reaction, the two half reactions are as follows:

\({{\rm{H}}_2}{{\rm{O}}_2} \to {{\rm{H}}_2}{\rm{O}} \ldots \ldots {\rm{ (1) }}\)

And,

\({\rm{S}}{{\rm{n}}^{2 + }} \to {\rm{S}}{{\rm{n}}^{4 + }} \ldots ...(2)\)

To balance the reaction (1), add 1 water molecule to the right side of the reaction arrow:

\({{\rm{H}}_2}{{\rm{O}}_2} \to 2{{\rm{H}}_2}{\rm{O}}\)

Now, add two hydrogen ions to the left,

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

Now, add two electrons to the left to balance the charge:

\({{\rm{H}}_2}{{\rm{O}}_2} + 2{{\rm{H}}^ + } + 2{e^ - } \to 2{{\rm{H}}_2}{\rm{O}} \ldots ...{\rm{ (3) }}\)

Similarly, to balance the reaction (2), add two electrons to the right side of the reaction arrow thus,

\({\rm{S}}{{\rm{n}}^{2 + }} \to {\rm{S}}{{\rm{n}}^{4 + }} + 2{e^ - } \ldots \ldots {\rm{ (4) }}\)

Now, reduction means gain of electrons thus, in reaction (3), \({{\rm{H}}_2}{{\rm{O}}_2}\) gets reduced to \({{\rm{H}}_2}{{\rm{O}}_2}\) by gain of two electrons.

And the species that oxidizes others and itself gets reduced are oxidizing agents thus, \({{\rm{H}}_2}{{\rm{O}}_2}\) is an oxidizing agent.

Similarly, oxidation means loss of electrons thus, in reaction (4), \({\rm{S}}{{\rm{n}}^{2 + }}\) gets oxidized to \({\rm{S}}{{\rm{n}}^{4 + }}\) by the loss of two electrons.

And the species that reduces others and itself gets oxidized are reducing agents thus, \({\rm{S}}{{\rm{n}}^{2 + }}\) is reducing agent.

b) Determine the balanced reaction for each pair of half reactions in an acidic solution.

The given reaction is as follows:

\({\rm{Pb}}{{\rm{O}}_2} + {\rm{Hg}} \to {\rm{Hg}}_2^{2 + } + {\rm{P}}{{\rm{b}}^{2 + }}\)

In the above reaction, the two half reactions are as follows:

\({\rm{Pb}}{{\rm{O}}_2} \to {\rm{P}}{{\rm{b}}^{2 + }} \ldots \ldots {\rm{ (1) }}\)

And,

\({\rm{Hg}} \to {\rm{Hg}}_2^{2 + } \ldots \ldots {\rm{ (2) }}\)

To balance reaction (1), add two water molecules to the right side of the reaction arrow:

\({\rm{Pb}}{{\rm{O}}_2} \to {\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}}\)

Now, add four hydrogen ions to the left,

\({\rm{Pb}}{{\rm{O}}_2} + 4{{\rm{H}}^ + } \to {\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}}\)

Now, add two electrons to the left to balance the charge:

\({\rm{Pb}}{{\rm{O}}_2} + 4{{\rm{H}}^ + } + 2{e^ - } \to {\rm{P}}{{\rm{b}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}} \ldots \ldots {\rm{ (3) }}\)

Similarly, to balance reaction (2), give coefficient 2 to \({\rm{Hg}}\) on left side of the reaction arrow:

\(2{\rm{Hg}} \to {\rm{Hg}}_2^{2 + }\)

Now, add two electrons to the right side of the reaction arrow,

\(2{\rm{Hg}} \to {\rm{Hg}}_2^{2 + } + 2{e^ - } \ldots \ldots {\rm{ (4) }}\)

Reduction means the gain of electrons. Thus, in reaction (3), \({\rm{Pb}}{{\rm{O}}_2}\) reduces to \({\rm{P}}{{\rm{b}}^{2 + }}\) by gain of two electrons.

And the species that oxidize others and itself gets reduced are oxidizing agents. Thus, \({\rm{Pb}}{{\rm{O}}_2}\) is an oxidizing agent.

Similarly, oxidation means the loss of electrons. So, in reaction (4), Hg oxidizes to \({\rm{Hg}}_2^{2 + }\) by loss of two electrons.

The species that reduces others and itself gets oxidized are reducing agents. Thus, Hg is a reducing agent.

c) Determine the balanced reaction for each pair of half reactions in an acidic solution.

The given reaction is as follows:

\({\rm{Al}} + {\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } \to {\rm{A}}{{\rm{l}}^{3 + }} + {\rm{C}}{{\rm{r}}^{3 + }}\)

In the above reaction, the two half reactions are as follows:

\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } \to {\rm{C}}{{\rm{r}}^{3 + }} \ldots \ldots {\rm{ (1) }}\)

And,\({\rm{Al}} \to {\rm{A}}{{\rm{l}}^{3 + }} \ldots ..(2)\)

To balance the reaction (1), give coefficient 2 to \({\rm{C}}{{\rm{r}}^{3 + }}\) thus,

\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } \to 2{\rm{C}}{{\rm{r}}^{3 + }}\)

Now, add seven water molecules to the right side of the reaction arrow:

\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } \to 2{\rm{C}}{{\rm{r}}^{3 + }} + 7{{\rm{H}}_2}{\rm{O}}\)

Now, add 14 hydrogen ions to the left,

\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } + 14{{\rm{H}}^ + } \to 2{\rm{C}}{{\rm{r}}^{3 + }} + 7{{\rm{H}}_2}{\rm{O}}\)

Now, add six electrons to the left to balance the charge:

\({\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}^{2 - } + 14{{\rm{H}}^ + } + 6{e^ - } \to 2{\rm{C}}{{\rm{r}}^{3 + }} + 7{{\rm{H}}_2}{\rm{O}} \ldots ..{\rm{ (3) }}\)

Similarly, to balance reaction (2), add 3 electrons to the right side of the reaction arrow,

\({\rm{Al}} \to {\rm{A}}{{\rm{l}}^{3 + }} + 3{e^ - } \ldots \ldots (4)\)

Now, reduction means gain of electrons. Thus, in reaction \((3){\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }\) gets reduced to \({\rm{C}}{{\rm{r}}^{3 + }}\) by gain of six electrons.

The species that oxidizes others and itself gets reduced are oxidizing agents. So, \({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }\) is an oxidizing agent.

Similarly, oxidation means the loss of electrons. Thus, in reaction (4) Al gets oxidized to \({\rm{A}}{{\rm{l}}^{3 + }}\) by loss of 3 electrons.

The species that reduces others and itself gets oxidized are reducing agents. So, Al is a reducing agent.