Page 107 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

Electrochemistry II: Electrolytic Cells 91
+ 2 mol of electrons reduce 1 mol of copper, CU(~), and 1 mol reduces
0.5 mol of copper.
=+ the mass of 1 chemical equivalent of copper is the mass of 0.5 mol.
This leads to the definition of the faraday. The faraday is defined as
the quantity of charge carried by 1 mol of electrons.
1F = 96500 C
When 1 F of electricity is passed through an electrolytic cell, 1 mole of
electrons passes through the cell and 1 chemical equivalent is deposited
at the cathode, i.e. since ‘CROA’ still applies in electrolytic cells, this
could be expressed as 1 equivalent of substance reduced at the cathode.
(1 mol) electrons = 96 500 C I
I =. 1 F = charge carried by 6.022 x
Examples:

(a) Consider the half-reaction: Ag + (as) + e -+ Ago(s)
1 F --+ 1 mol Ag; 96 500 C -+ 1 mol Ag; 96 500 C -+ 107.868 g of
silver, produced at the cathode during electrolysis, since 1 mol
of Ag contains 107.868 g.
(b) Consider the half-reaction: Mg2+(aq) + 2e + Mgo(s)
2 F -+ 1 mol Mg; 1 F -+ 0.5 mol Mg; 96 500 C -+ 0.5 mol Mg;
96 500 C -+ 0.5 x 24.305 g = 12.1525 g of magnesium produced
at the cathode during electrolysis, since 1 mol of Mg contains
24.305 g.
(c) Aluminium is produced by the electrolysis of aluminium oxide,
A1203 dissolved in molten cryolite, Na3AlF6. Determine the
mass (in kg) of aluminium produced in 12 hours, in an electro-
lytic cell operating at 95 kA, given that the molar mass (M) of
A1 is 26.982 g mol-’ and 1 F = 96 500 C.

First, the oxidation state of A1 in the oxide has to be calculated:
2x + 3( -11) = 0; 2x = 6; x = 111, i.e. Allll(aq). Then, aluminium will
be deposited at the cathode (reduction), according to Faraday’s
Second Law of Electrolysis, i.e. A13+(aq) + 3e + AlO,,)
3F4 1 molofAl; 1 F -+~molofAl;96500C--+~molofAl.
But, Q = It + Q = 95000 x (12 x 60 x 60), since 12 hours =
12 x 60 x 60 s
=+ Q = 4.104 x lo9 C
96 500 C + f (26.982) g = 8.994 g of A1
Hence, 1 C -+ (8.994/96 500) g of A1
4.104 x lo9 C + (8.994/96500)(4.104 x lo9) = 382501.31 g
Answer: 382.5 kg of aluminium