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Question: Predict the products of electrolysis in each of the following: (i) An aqueous solution of \(AgN...

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3AgN{{O}_{3}} with silver electrodes.
(ii) An aqueous solution of AgNO3AgN{{O}_{3}} with platinum electrodes.
(iii) A dilute solution of H2SO4{{H}_{2}}S{{O}_{4}} with platinum electrodes.
(iv) An aqueous solution of CuCl2CuC{{l}_{2}} with platinum electrodes.

Explanation

Solution

Write the ions that will go to the electrodes i.e. cathode and anode. We have to consider the electrode potential for the respective half reactions. Remember to write the oxidation and reduction of water molecules along with the ions of the salt.

Complete step-by-step answer:
As suggested in the hint we will write the reactions taking place at anode and cathode for each electrolysis process and then predict the possible products of electrolysis.
(i) An aqueous solution of AgNO3AgN{{O}_{3}} with silver electrodes.
Electrolyte : AgNO3AgN{{O}_{3}}
Cathode (Silver) :
Ag+ + e− → Ag\text{A}{{\text{g}}^{+}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Ag}
This is because silver ions have lower discharge potential than hydrogen ions.
Anode (Silver):
Ag → Ag+ + e−\text{Ag }\to \text{ A}{{\text{g}}^{+}}\text{ + }{{\text{e}}^{-}}
Product : Ag
(ii) An aqueous solution of AgNO3AgN{{O}_{3}} with platinum electrodes.
Electrolyte: AgNO3AgN{{O}_{3}}
Cathode(inert) :
Ag+ + e− → Ag\text{A}{{\text{g}}^{+}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Ag}
This is because silver ions have lower discharge potential than hydrogen ions.
Anode(inert) :
4OH− → 2H2O + O2 + 4e−4\text{O}{{\text{H}}^{-}}\text{ }\to \text{ 2}{{\text{H}}_{2}}\text{O + }{{\text{O}}_{2}}\text{ + 4}{{\text{e}}^{-}}
This is because hydroxide ions have a lower discharge potential than nitrate ions.
Products: Ag, O2 {{\text{O}}_{2}}\text{ }
(iii) A dilute solution of H2SO4{{H}_{2}}S{{O}_{4}} with platinum electrodes.
Electrolyte: H2SO4{{H}_{2}}S{{O}_{4}}
Cathode(inert) :
2H+ + 2e− → H2\text{2}{{\text{H}}^{+}}\text{ + 2}{{\text{e}}^{-}}\text{ }\to \text{ }{{\text{H}}_{2}}
Anode(inert) :
4OH− → 2H2O + O2 + 4e−4\text{O}{{\text{H}}^{-}}\text{ }\to \text{ 2}{{\text{H}}_{2}}\text{O + }{{\text{O}}_{2}}\text{ + 4}{{\text{e}}^{-}}
This is because hydroxide ions have a lower discharge potential than sulphate ions.
Products: H2{{\text{H}}_{2}}, O2{{\text{O}}_{2}}
(iv) An aqueous solution of CuCl2CuC{{l}_{2}} with platinum electrodes.
Electrolyte: CuCl2CuC{{l}_{2}}
Cathode(inert) :
Cu2+ + 2e− → Cu\text{C}{{\text{u}}^{2+}}\text{ + 2}{{\text{e}}^{-}}\text{ }\to \text{ Cu}
This is because copper(II) ions have lower discharge potential than hydrogen ions.
Anode(inert) :
2Cl− → Cl2 + 2e−\text{2C}{{\text{l}}^{-}}\text{ }\to \text{ C}{{\text{l}}_{2}}\text{ + 2}{{\text{e}}^{-}}
Chloride ions are oxidised in preference to hydroxide ions.
Products: Cu, Cl2\text{C}{{\text{l}}_{2}}

Note: It is important to know that oxidation potential of chlorine ions is lower than hydroxide ions. However, in the electrolysis aqueous solution of CuCl2CuC{{l}_{2}} with platinum electrodes the product obtained at anode is chlorine. This is because the formation of chlorine molecules happens at a faster rate than formation of oxygen molecules. In electrochemistry we give preference to chemical kinetics over thermodynamics.