Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.

(a) \({\rm{Mg}}(s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {\rm{M}}{{\rm{g}}^{2 + }}(aq) + {\rm{Ni}}(s)\)

(b) \(2{\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Ag}}(s)\)

(c) \({\rm{Mn}}(s) + {\rm{Sn}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) \to {\rm{Mn}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

(d)\(3{\rm{CuN}}{{\rm{O}}_3}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) \to 3{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

• The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
• The balancing of 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 given balanced reaction is as follows:

\({\rm{Mg}}(s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {\rm{M}}{{\rm{g}}^{2 + }}(aq) + {\rm{Ni}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{Mg}}(s) \to + {\rm{M}}{{\rm{g}}^{2 + }}(aq) + 2{e^ - }\\{\rm{N}}{{\rm{i}}^{2 + }}(aq) + 2{e^ - } \to {\rm{Ni}}(s)\end{array}\)

Here, magnesium atom gets oxidized to magnesium ion by release of two electrons and nickel ion gained 2 electrons to form neutral nickel atom which is a reduction reaction. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Mg}\nolimits} (s)\left| {M{g^{2 + }}(aq)||{\rm{N}}{{\rm{i}}^{2 + }}(aq)} \right|{\rm{Ni}}(s)\)

The given balanced reaction is as follows:

\(2{\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Ag}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{e^ - }\\2{\rm{A}}{{\rm{g}}^ + }(aq) + 2{e^ - } \to 2{\rm{Ag}}(s)\end{array}\)

Here, copper atom gets oxidized to copper ion by release of two electrons and silver ion gained 2 electrons to form neutral silver atom which is a reduction reaction. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines. Thus, balanced reaction with cell notation will be:\({\rm{Cu}}(s)\left| {{\rm{C}}{{\rm{u}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)

The given balanced reaction is as follows:

\({\rm{Mn}}(s) + {\mathop{\rm Sn}\nolimits} {\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) \to {\mathop{\rm Mn}\nolimits} {\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\mathop{\rm Sn}\nolimits} (s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\mathop{\rm Mn}\nolimits} (s) \to {{\mathop{\rm Mn}\nolimits} ^{2 + }}(aq) + 2{e^ - }\\{{\mathop{\rm Sn}\nolimits} ^{2 + }}(aq) + 2{e^ - } \to {\mathop{\rm Sn}\nolimits} (s)\end{array}\)

Here, manganese atom gets oxidized to manganese ion by release of two electrons and lead ion gained 2 electrons to form neutral lead atom which is a reduction reaction.

In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Mn}\nolimits} (s)\left| {{{{\mathop{\rm Mn}\nolimits} }^{2 + }}(aq) || {{{\mathop{\rm Sn}\nolimits} }^ + }(aq)} \right|{\mathop{\rm Sn}\nolimits} (s)\)

The given balanced reaction should be written using the cell notation.

The given balanced reaction is as follows:

\(3{\rm{CuN}}{{\rm{O}}_3}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) \to 3{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}(s)\)

The above reaction can be split into two half reactions as follows:

\(\begin{array}{l}{\rm{C}}{{\rm{u}}^ + }(aq) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + {e^ - }\\{\rm{A}}{{\rm{u}}^{3 + }}(aq) + 3{e^ - } \to {\rm{Au}}(s)\end{array}\)

Here, \({\rm{Cu}}\) (I) gets oxidized to \({\rm{Cu}}\) (II) ion by release of one electron and gold(III) ion gained 3 electrons to form neutral gold atom which is a reduction reaction. Here, Pt electrode is used as an inert electrode. In a cell reaction, first oxidation is shown which is separated from the reduction via a salt bridge represented as two vertical lines.

Thus, balanced reaction with cell notation will be: \({\mathop{\rm Pt}\nolimits} (s)\left| {{\rm{C}}{{\rm{u}}^ + }(aq),{\rm{C}}{{\rm{u}}^{2 + }}(aq) || A{u^{3 + }}(aq)} \right|Au(s)\)