Question

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of \({\rm{Pt}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)forms Pt metal, while in the other half-cell, Cu metal goes into a \({\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)solution with all solute concentrations 1M.

(b) The cathode consists of a gold electrode in a \(0.55{\rm{MAu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\) solution and the anode is a magnesium electrode in \(0.75{\rm{MMg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) solution.

(c) One half-cell consists of a silver electrode in a \(1{\rm{MAgN}}{{\rm{O}}_3}\) solution, and in the other half-cell, a copper electrode in \(1M{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) is oxidized.

Short answer

1. The cell notation will be \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) ||P{t^{2 + }}(aq,1M)} \right|{\rm{Pt}}({\rm{s}})\) .
2. The cell notation will be \({\rm{Mg}}({\rm{s}})\left| {{\rm{M}}{{\rm{g}}^{2 + }}({\rm{aq}},0.75{\rm{M}})|| A{u^{3 + }}(aq,0.55M)} \right|{\rm{Au}}({\rm{s}})\).
3. The cell notation is \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) || A{g^ + }(aq,1M)} \right|{\rm{Ag}}({\rm{s}})\) .

Step-by-step solution

Define oxidation and reduction

• The oxidation-reduction reaction is also known as aredox 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 ashalf equation method.

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

The cell notation will be \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) ||P{t^{2 + }}(aq,1M)} \right|{\rm{Pt}}({\rm{s}})\) .

In one half cell, \({\rm{Pt}}\) metal is formed from \({\rm{P}}{{\rm{t}}^{2 + }}\), which means that reduction takes place here and this represents reduction half-cell or cathode.

In other half cell, \({\rm{Cu}}\) metal converts into \({\rm{C}}{{\rm{u}}^{2 + }}\), which means that oxidation takes place here and this represents reduction half-cell or anode.

Therefore, the cell notation is will be: \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) ||P{t^{2 + }}(aq,1M)} \right|{\rm{Pt}}({\rm{s}})\)

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

The cell notation will be \({\rm{Mg}}({\rm{s}})\left| {{\rm{M}}{{\rm{g}}^{2 + }}({\rm{aq}},0.75{\rm{M}})|| A{u^{3 + }}(aq,0.55M)} \right|{\rm{Au}}({\rm{s}})\). .

The cathode consists of a gold electrode in \(0.55{\rm{MAu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\) solution, which means \({\rm{A}}{{\rm{u}}^3} + \) reduce into \({\rm{Au}}\) here.

Therefore, the cell notation will be:\({\rm{Mg}}({\rm{s}})\left| {{\rm{M}}{{\rm{g}}^{2 + }}({\rm{aq}},0.75{\rm{M}})|| A{u^{3 + }}(aq,0.55M)} \right|{\rm{Au}}({\rm{s}})\).

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

The cell notation is \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) || A{g^ + }(aq,1M)} \right|{\rm{Ag}}({\rm{s}})\) .

In one half cell, Ag metal is formed from \({\rm{A}}{{\rm{g}}^ + }\), which means that reduction takes place here and this represents reduction half-cell or cathode.

In other half cell, \({\rm{Cu}}\) metal converts into \({\rm{C}}{{\rm{u}}^{2 + }}\), which means that oxidation takes place here and this represents reduction half-cell or anode.

Therefore, the cell notation is will be: \({\rm{Cu}}({\rm{s}})\left| {{\rm{C}}{{\rm{u}}^{2 + }}({\rm{aq}},1{\rm{M}}) || A{g^ + }(aq,1M)} \right|{\rm{Ag}}({\rm{s}})\)