ELECTROCHEMISTRY
1.1 Electrolysis
Electrolysis is the breaking down of a compound into its elements using
an electric current
Electrolyte is the compound that is broken down during electrolysis and it
is either a molten ionic compound or a concentrated aqueous solution of
ions
The positive electrode is anode – attracts anions (negatively charged ions)
where oxidation occurs
The negative electrode is cathode – attracts cations (positively charged ions) where reduction occurs
Electrolysis of molten electrolytes
Cations move to the cathode where they gain electrons and
get reduced – if a metal is formed it is deposited on the
cathode, if hydrogen forms bubbles are seen
Anions move to the anode where they lose electrons and get
oxidised
Electrolysis of aqueous solutions
Water contributes H+ and OH- ions to the solution, so the ions that are discharged during
electrolysis will depend on:
o The standard electrode potential (when standard half-cell is connected to standard
hydrogen cell with 0.00V) – in the cathode the cation with most positive Eꝋ will be reduced
easily and discharged (strong oxidising agent). In the anode the anion with most negative
Eꝋ will be oxidised easily and discharged. conditions to compare:
o Ion concentration of 1.00 mol dm-3
o A temperature of 298 K
o A pressure of 1 atm
o The concentration of the ions – ions that are present in higher concentrations are more
likely to be discharged
, 1.2 Faraday’s law and Avogadro’s constant
One faraday is the amount of electric charge carried by 1 mole of electrons or 1 mole of singly
charged ions - 1 faraday is 96 500 C mol-1
The Avogadro’s constant (L) is the number of particles in one mole. L = 6.02 x 1023 mol-1.
Finding Avogadro’s constant experimentally
Keep the amount of copper deposited is 0.13 g (2.0 x 10-3 mol). To deposit this 0.17 A current is
required and took 40 minutes.
Q = I x t = 0.17 x (60 x 40) = 408 C
The half-equation shows that 2 mol of electrons are needed to deposit one mol of copper:
Cu2+(aq) + 2e- → Cu(s)
So, the charge on 1 mol of electrons is:
Q = 408/(2.0 x 10-3 x 2) = 99 646 C
Given that the charge on one electron is 1.60 x 10-19 C, then L equals:
99 646/1.60 × 10-19 = 6.23 x 1023 mol-1
The experimentally determined value for L of 6.23 x 1023 mol-1 is very close to the theoretical value
of 6.02 x 1023 mol-1
1.1 Electrolysis
Electrolysis is the breaking down of a compound into its elements using
an electric current
Electrolyte is the compound that is broken down during electrolysis and it
is either a molten ionic compound or a concentrated aqueous solution of
ions
The positive electrode is anode – attracts anions (negatively charged ions)
where oxidation occurs
The negative electrode is cathode – attracts cations (positively charged ions) where reduction occurs
Electrolysis of molten electrolytes
Cations move to the cathode where they gain electrons and
get reduced – if a metal is formed it is deposited on the
cathode, if hydrogen forms bubbles are seen
Anions move to the anode where they lose electrons and get
oxidised
Electrolysis of aqueous solutions
Water contributes H+ and OH- ions to the solution, so the ions that are discharged during
electrolysis will depend on:
o The standard electrode potential (when standard half-cell is connected to standard
hydrogen cell with 0.00V) – in the cathode the cation with most positive Eꝋ will be reduced
easily and discharged (strong oxidising agent). In the anode the anion with most negative
Eꝋ will be oxidised easily and discharged. conditions to compare:
o Ion concentration of 1.00 mol dm-3
o A temperature of 298 K
o A pressure of 1 atm
o The concentration of the ions – ions that are present in higher concentrations are more
likely to be discharged
, 1.2 Faraday’s law and Avogadro’s constant
One faraday is the amount of electric charge carried by 1 mole of electrons or 1 mole of singly
charged ions - 1 faraday is 96 500 C mol-1
The Avogadro’s constant (L) is the number of particles in one mole. L = 6.02 x 1023 mol-1.
Finding Avogadro’s constant experimentally
Keep the amount of copper deposited is 0.13 g (2.0 x 10-3 mol). To deposit this 0.17 A current is
required and took 40 minutes.
Q = I x t = 0.17 x (60 x 40) = 408 C
The half-equation shows that 2 mol of electrons are needed to deposit one mol of copper:
Cu2+(aq) + 2e- → Cu(s)
So, the charge on 1 mol of electrons is:
Q = 408/(2.0 x 10-3 x 2) = 99 646 C
Given that the charge on one electron is 1.60 x 10-19 C, then L equals:
99 646/1.60 × 10-19 = 6.23 x 1023 mol-1
The experimentally determined value for L of 6.23 x 1023 mol-1 is very close to the theoretical value
of 6.02 x 1023 mol-1
