`=>` Products of electrolysis depend on :
(i) The nature of material being electrolysed and
(ii) The type of electrodes being used
`=>` If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons.
`=>` If the electrode is reactive, it participates in the electrode reaction.
`=>` Therefore, the products of electrolysis may be different for reactive and inert electrodes.
`=>` The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials.
`=>` Some of the electrochemical processes are feasible but very slow kinetically that at lower voltages these don’t seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur.
`color{red}text(Examples) :` (i) If we use molten `NaCl`, the products of electrolysis are sodium metal and `Cl_2` gas.
Here, we have only one cation (`Na^+`) which is reduced at the cathode (`Na^+ + e^(-) → Na`) and one anion `(Cl^(-))` which is oxidised at the anode (`color{red}(Cl^(-) → ½Cl_2+e^(-))`).
(ii) During the electrolysis of aqueous sodium chloride solution, the products are `NaOH`, `Cl_2` and `H_2`.
In this case besides `Na^+` and `Cl^-` ions, we also have `H^+` and `OH^-` ions along with the solvent molecules, `H_2O`.
`->` At the cathode, there is competition between the following reduction reactions :
`color{red}(Na^(+) (aq) + e^(-) → Na (s) \ \ \ \ \ E_text(cell)^(⊖) = -2.71V)`
`color{red}(H^(+) (aq) + e^(-) → 1/2 H_2 (g) \ \ \ \ \ \ \ E_text(cell)^(⊖) = 0.00V)`
`color{green}text(Note) :` The reaction with higher value of `E^(⊖)` is preferred and, therefore, the reaction at the cathode during electrolysis is :
`color{red}(H^(+) (aq) +e^(-) → 1/2 H_2 (g))` ......................(6).
But `H^+ (aq)` is produced by the dissociation of `H_2O ` i.e.
`color{red}(H_2O (l) → H^(+) (aq) + OH^(-) (aq))`................(7).
Therefore, the net reaction at the cathode may be written as the sum of (3.33) and (3.34) and we have
`color{red}(H_2O(l) + e^(-) → 1/2 H_2 (g) + OH^-)` ...................(8).
`->` At the anode, the following oxidation reactions are possible :
`color{red}(Cl^(-) (aq) → 1/2 Cl_2 (g) + e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.36 V)` ............(9).
`color{red}(2H_2O(l) → O_2 (g) + 4 H^(+) (aq) +4 H^(+) (aq) +4e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.23 V)` ..........(10).
`color{green}text(Note) :` The reaction at anode with lower value of `E^(⊖)` is preferred and therefore, water should get oxidised in preference to `Cl^(-) (aq)`. However, on account of overpotential of oxygen, reaction (9) is preferred.
Therefore, the net reactions may be summarised as :
`color{red}(NaCl (aq) overset(H_2O)→ Na^+ (aq) + Cl^(-) (aq)) `
`color{green}("Cathode")` : `color{red}(H_2O (l) + e^(-) → 1/2 H_2 (g) + OH^(-) (aq))`
`color{green}("Anode")` : `color{red}(Cl^(-) (aq) → 1/2 Cl_2 (g) + e^(-))`.
`color{green}("Net reaction ")`: `color{red}(NaCl (aq) + H_2O (l) → Na^(+) (aq) + OH^(-) (aq) +1/2 H_2 (g) +1/2Cl_2 (g))`.
`=>` The standard electrode potentials are replaced by electrode potentials given by Nernst equation to take into account the concentration effects.
`color{red}text(Example) :` During the electrolysis of sulphuric acid, the following processes are possible at the anode :
`color{red}(2H_2O(l) → O_2 (g) +4H^(+) (aq) +4 e^(-) \ \ \ \ E_text(cell)^(⊖) = +1.23 V)` ................(11).
`color{red}(2SO_4^(2-) (aq) → S_2 O_8^(2-) (aq) +2 e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.96 V) ` ...............(12).
For dilute sulphuric acid, reaction (11) is preferred but at higher concentrations of `H_2SO_4` process, reaction (12) is preferred.
`=>` Products of electrolysis depend on :
(i) The nature of material being electrolysed and
(ii) The type of electrodes being used
`=>` If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons.
`=>` If the electrode is reactive, it participates in the electrode reaction.
`=>` Therefore, the products of electrolysis may be different for reactive and inert electrodes.
`=>` The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials.
`=>` Some of the electrochemical processes are feasible but very slow kinetically that at lower voltages these don’t seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur.
`color{red}text(Examples) :` (i) If we use molten `NaCl`, the products of electrolysis are sodium metal and `Cl_2` gas.
Here, we have only one cation (`Na^+`) which is reduced at the cathode (`Na^+ + e^(-) → Na`) and one anion `(Cl^(-))` which is oxidised at the anode (`color{red}(Cl^(-) → ½Cl_2+e^(-))`).
(ii) During the electrolysis of aqueous sodium chloride solution, the products are `NaOH`, `Cl_2` and `H_2`.
In this case besides `Na^+` and `Cl^-` ions, we also have `H^+` and `OH^-` ions along with the solvent molecules, `H_2O`.
`->` At the cathode, there is competition between the following reduction reactions :
`color{red}(Na^(+) (aq) + e^(-) → Na (s) \ \ \ \ \ E_text(cell)^(⊖) = -2.71V)`
`color{red}(H^(+) (aq) + e^(-) → 1/2 H_2 (g) \ \ \ \ \ \ \ E_text(cell)^(⊖) = 0.00V)`
`color{green}text(Note) :` The reaction with higher value of `E^(⊖)` is preferred and, therefore, the reaction at the cathode during electrolysis is :
`color{red}(H^(+) (aq) +e^(-) → 1/2 H_2 (g))` ......................(6).
But `H^+ (aq)` is produced by the dissociation of `H_2O ` i.e.
`color{red}(H_2O (l) → H^(+) (aq) + OH^(-) (aq))`................(7).
Therefore, the net reaction at the cathode may be written as the sum of (3.33) and (3.34) and we have
`color{red}(H_2O(l) + e^(-) → 1/2 H_2 (g) + OH^-)` ...................(8).
`->` At the anode, the following oxidation reactions are possible :
`color{red}(Cl^(-) (aq) → 1/2 Cl_2 (g) + e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.36 V)` ............(9).
`color{red}(2H_2O(l) → O_2 (g) + 4 H^(+) (aq) +4 H^(+) (aq) +4e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.23 V)` ..........(10).
`color{green}text(Note) :` The reaction at anode with lower value of `E^(⊖)` is preferred and therefore, water should get oxidised in preference to `Cl^(-) (aq)`. However, on account of overpotential of oxygen, reaction (9) is preferred.
Therefore, the net reactions may be summarised as :
`color{red}(NaCl (aq) overset(H_2O)→ Na^+ (aq) + Cl^(-) (aq)) `
`color{green}("Cathode")` : `color{red}(H_2O (l) + e^(-) → 1/2 H_2 (g) + OH^(-) (aq))`
`color{green}("Anode")` : `color{red}(Cl^(-) (aq) → 1/2 Cl_2 (g) + e^(-))`.
`color{green}("Net reaction ")`: `color{red}(NaCl (aq) + H_2O (l) → Na^(+) (aq) + OH^(-) (aq) +1/2 H_2 (g) +1/2Cl_2 (g))`.
`=>` The standard electrode potentials are replaced by electrode potentials given by Nernst equation to take into account the concentration effects.
`color{red}text(Example) :` During the electrolysis of sulphuric acid, the following processes are possible at the anode :
`color{red}(2H_2O(l) → O_2 (g) +4H^(+) (aq) +4 e^(-) \ \ \ \ E_text(cell)^(⊖) = +1.23 V)` ................(11).
`color{red}(2SO_4^(2-) (aq) → S_2 O_8^(2-) (aq) +2 e^(-) \ \ \ \ \ E_text(cell)^(⊖) = 1.96 V) ` ...............(12).
For dilute sulphuric acid, reaction (11) is preferred but at higher concentrations of `H_2SO_4` process, reaction (12) is preferred.