Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half reactions in an acidic solution.

(a) \({\bf{Ca}} \to {\bf{C}}{{\bf{a}}^{{\bf{2 + }}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {{\bf{F}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2\;}}{{\bf{F}}^{\bf{ - }}}\)

(b) \({\bf{Li}} \to {\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2C}}{{\bf{l}}^{\bf{ - }}}\)

(c) \({\bf{Fe}} \to {\bf{F}}{{\bf{e}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2B}}{{\bf{r}}^{\bf{ - }}}\)

(d) \({\bf{Ag}} \to {\bf{A}}{{\bf{g}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{MnO}}_{\bf{4}}^{\bf{ - }}{\bf{ + 4}}{{\bf{H}}^{\bf{ + }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{ + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\)

• 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 the half equation method.
• The oxidation and reduction can be identified by change in the 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 A(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 half reactions are as follows:

\(\begin{array}{l}{\rm{Ca}} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2{e^ - }\\{{\rm{F}}_2} + 2{e^ - } \to 2\;{{\rm{F}}^ - }\end{array}\)

The overall reaction can be obtained by adding the two half reactions. Thus, the balanced reaction will be:

\(\begin{array}{l}{\rm{Ca}} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2{e^ - }\\{{\rm{F}}_2} + 2{e^ - } \to 2\;{{\rm{F}}^ - }\\\overline {{\rm{Ca}} + {{\rm{F}}_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2\;{{\rm{F}}^ - }} \end{array}\)

Therefore, the balanced chemical reaction is as follows:

\({\rm{Ca}} + {{\rm{F}}_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2\;{{\rm{F}}^ - }\)

The given half reactions are as follows:

\(\begin{array}{l}{\rm{Li}} \to {\rm{L}}{{\rm{i}}^ + } + {e^ - } \ldots ...{\rm{ (1) }}\\{\rm{C}}{{\rm{l}}_2} + 2{e^ - } \to 2{\rm{C}}{{\rm{l}}^ - } \ldots ...(2)\end{array}\)

The overall reaction can be obtained by adding the two half reactions. To balance the number of electrons, multiply equation (1) by 2.

\(2{\rm{Li}} \to 2{\rm{L}}{{\rm{i}}^ + } + 2{e^ - }\)

Thus, the balanced reaction will be:

\(\begin{array}{l}2{\rm{Li}} \to 2{\rm{L}}{{\rm{i}}^ + } + 2{e^ - }\\\frac{{{\rm{C}}{{\rm{l}}_2} + 2{e^ - } \to 2{\rm{C}}{{\rm{l}}^ - }}}{{2{\rm{Li}} + {\rm{C}}{{\rm{l}}_2} \to 2{\rm{L}}{{\rm{i}}^ + } + 2{\rm{C}}{{\rm{l}}^ - }}}\end{array}\)

Therefore, the balanced chemical reaction is as follows:

\(2{\rm{Li}} + {\rm{B}}{{\rm{r}}_2} \to 2{\rm{L}}{{\rm{i}}^ + } + 2{\rm{B}}{{\rm{r}}^ - }\)

The given half reactions are as follows:

\(\begin{array}{l}{\rm{Fe}} \to {\rm{F}}{{\rm{e}}^{3 + }} + 3{e^ - } \ldots \ldots (1)\\{\rm{B}}{{\rm{r}}_2} + 2{e^ - } \to 2{\rm{B}}{{\rm{r}}^ - } \ldots \ldots (2)\end{array}\)

The overall reaction can be obtained by adding the two half reactions. To balance the number of electrons, multiply equation (1) by 2 .

\(2{\rm{Fe}} \to 2{\rm{F}}{{\rm{e}}^{3 + }} + 6{e^ - }\)

Also, multiply equation (2) by 3,

\(3{\rm{B}}{{\rm{r}}_2} + 6{e^ - } \to 6{\rm{B}}{{\rm{r}}^ - }\)

Thus, the balanced reaction will be:

\(\begin{array}{l}2{\rm{Fe}} \to 2{\rm{F}}{{\rm{e}}^{3 + }} + 6{e^ - }\\3{\rm{B}}{{\rm{r}}_2} + 6{e^ - } \to 6{\rm{B}}{{\rm{r}}^ - }\\\overline {2{\rm{Fe}} + 3{\rm{B}}{{\rm{r}}_2} \to 2{\rm{F}}{{\rm{e}}^{3 + }} + 6{\rm{B}}{{\rm{r}}^ - }} \end{array}\)

Therefore, the balanced chemical reaction is as follows:

\(2{\rm{Fe}} + 3{\rm{B}}{{\rm{r}}_2} \to 2{\rm{F}}{{\rm{e}}^{3 + }} + 6{\rm{B}}{{\rm{r}}^ - }\)

The given half reactions are as follows:

\(\begin{array}{l}{\rm{Ag}} \to {\rm{A}}{{\rm{g}}^ + } + {e^ - } \ldots \ldots (1)\\{\rm{MnO}}_4^ - + 4{{\rm{H}}^ + } + 3{e^ - } \to {\rm{Mn}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \ldots \ldots (2)\end{array}\)

The overall reaction can be obtained by adding the two half reactions. To balance the number of electrons, multiply equation (1) by 3.

\(3{\rm{Ag}} \to 3{\rm{A}}{{\rm{g}}^ + } + 3{e^ - }\)

Thus, the balanced reaction will be:

\(\begin{array}{l}3{\rm{Ag}} \to 3{\rm{A}}{{\rm{g}}^ + } + 3{e^ - }\\{\rm{MnO}}_4^ - + 4{{\rm{H}}^ + } + 3{e^ - } \to {\rm{Mn}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}\\\overline {{\rm{MnO}}_4^ - + 4{{\rm{H}}^ + } + 3{\rm{Ag}} \to 3{\rm{A}}{{\rm{g}}^ + } + {\rm{Mn}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}} \end{array}\)

Therefore, the balanced chemical reaction is as follows:

\({\rm{MnO}}_4^ - + 4{{\rm{H}}^ + } + 3{\rm{Ag}} \to 3{\rm{A}}{{\rm{g}}^ + } + {\rm{Mn}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}\)