Chemistry Electrolytic Cell, Electrolysis and Faraday's Law of Electrolysis
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Topics Covered :

● Electrolytic Cell
● Electrolysis
● Faraday's Law of Electrolysis
● Products of Electrolysis

Electrolytic Cells and Electrolysis :

`color{green}text(Electrolytic Cell) :` In an electrolytic cell external source of voltage is used to bring about a chemical reaction. The electrochemical processes are of great importance in the laboratory and the chemical industry.

`color{green}text(Components of Electrolytic Cell) :` One of the simplest electrolytic cell consists of two copper strips dipping in an aqueous solution of copper sulphate. If a DC voltage is applied to the two electrodes,

`=>` At the cathode (negatively charged), `Cu^(2+)` ions discharge and the following reaction takes place :

`color{red}(Cu^(2+) (aq) +2 e^(-) → Cu(s)) `.......(1).

Copper metal is deposited on the cathode.

`=>` At the anode, copper is converted into `Cu^(2+)` ions by the reaction :

`color{red}(Cu(s) → Cu^(2+) (s) +2 e^(-))` .............(2).

Thus, copper is dissolved (oxidised) at anode and deposited (reduced) at cathode.

`color{green}text(Note) :` (i) This is the basis for an industrial process in which impure copper is converted into copper of high purity.

(ii) The impure copper is made an anode that dissolves on passing current and pure copper is deposited at the cathode.

`color{green}text(Application) :` (i) Many metals like `Na, Mg, Al`, etc. are produced on large scale by electrochemical reduction of their respective cations where no suitable chemical reducing agents are available for this purpose.

(ii) Sodium and magnesium metals are produced by the electrolysis of their fused chlorides and aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.

Quantitative Aspects of Electrolysis :

Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Faraday's Law are given as :

Faraday’s Laws of Electrolysis :

After his deep research on electrolysis of solutions and melts of electrolytes, Faraday published the following well known Faraday’s two laws of electrolysis :

`color{green}text(First Law) :` The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

`color{green}text(Equivalent Weight) = text(Atomic Mass of Metal) / text(Number of electrons required to reduce the cation)`

`=>` There were no constant current sources available during Faraday’s times.

`=>` So, coulometer (a standard electrolytic cell) was being used for determining the quantity of electricity passed from the amount of metal (generally `Ag` or `Cu`) deposited or consumed.

`=>` Now, coulometers are obsolete and we have constant current `(I)` sources available and the quantity of electricity `Q`, passed is given by

`Q = It`

`Q` is in coloumbs when `I` is in ampere and `t` is in second.

`=>` The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction :

`color{red}(Ag^(+) (aq) + e^(-) → Ag (s))` ................(3).

One mole of the electron is required for the reduction of one mole of silver ions.

Since charge on one electron is equal to `1.6021× 10^(–19)C`. Therefore, the charge on one mole of electrons is equal to :

`color{red}(N_A xx 1.6021 xx 10^(-19) \ \ \ \ C = 6.02xx10^(23) mol^(-1) xx 1.6021xx 10^(-19))`

`C = 96487 C mol^(-1)`.

● This quantity of electricity is called `text(Faraday)` and is represented by the symbol `F`.

● For approximate calculations we use `1F ≃ 96500 C mol^(-1)`.

`=>` For the electrode reactions : `color{red}(Mg^(2+) (I) +2e^(-) → Mg(s))` ..................(4).

`color{red}(Al^(3+) (I) +3e^(-) → Al (s))` .....................(5).

It is obvious that one mole of `Mg^(2+)` and `Al^(3+)` require 2 mol of electrons `(2F)` and 3 mol of electrons `(3F)` respectively.

`=>` The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds.

`=>` In commercial production of metals, current as high as 50,000 amperes are used that amounts to about `0.518 F` per second.

`color{green}text(Second Law) :` The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

Q 2987345287

A solution of `CuSO_4` is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?


`t = 600 s` charge = current × time = `1.5 A xx 600 s = 900 C`

According to the reaction:

`Cu^(2+) (aq) +2 e^(-) = Cu (s)`

We require `2F` or `2 xx 96487 C` to deposit `1 mol` or `63 g` of `Cu.`

For `900 C`, the mass of `Cu` deposited `= ( 63 g mol^(-1) xx 900 C)/(2xx96487 C mol^(-1))`

` = 0.2938 g`.

Products of Electrolysis :

`=>` 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.