By:Chemistry Lecturer

School of Allied Health SciencesCity University College of Science and Technology

Electrolytes: Chemical compound that can conduct electricity in the molten state or in aqueous solution◦ Electricity conducted by free moving ions

Non electrolytes: Chemical compound that cannot conduct electricity in any state

ELECTROLYTES

NON-ELECTROLYTES

• Electrolysis is a process whereby a compound is decomposed by electric current

• Electrolytic cell consists of two electrodesi. Anode – Positive terminal Anions (-ve charged ions) attracted to anode

ii. Cathode – Negative terminal Cations (+ve charged ions) attracted to

cathode

• Generally a molten compound electrolytes AnBm produces Am+ cations and Bn- anions

AnBm Am+ Bn-

• Examples:i. PbBr2 Pb2+ + 2Br-

ii. NaCl Na+ + Cl-

• Gives the anion and cation for the electrolytes:

i. PbCl2 -ii. AgCl -iii.CuCl2 -iv.CuBr2 -

• Gives the anion and cation for the electrolytes:

i. PbCl2 - Pb2+ , Cl-ii. AgCl - Ag+ , Cl-iii.CuCl2 - Cu2+ , Cl-iv.CuBr2 - Cu2+ , Br-

• Two steps occur during electrolysisi. Movement of ions to the electrodes

Cations (+ve) move towards the cathode Anions (-ve) move towards the anode

ii. Discharge of ions at the electrodes Cations discharged by receiving electrons

(losing positive charge to become neutral)An+ + ne- A

Anions discharged by releasing electrons (losing negative charge to become neutral)Bn- B + ne-

Steps in writing half equation

Anode Cathode

Step 1 Ion to neutral Br- Br2 Cu2+ Cu

Step 2 Balance number of atoms

2Br- Br2 Cu2+ Cu

Step 3 Balance the charge by adding

the electrons

2Br- Br2 + 2e

Cu2+ + 2e Cu

Balance the number of electrons from the half equation

Example:Half equationAt the Anode: 2Cl- Cl2 + 2eAt the cathode: Na+ + e Na

Overall equation2Cl- Cl2 + 2e(Na+ + e Na) x 2 = 2Na+ + 2e 2Na

= 2Cl- + 2Na+ Cl2 + 2Na

In the electrolysis of a molten compound,◦ The metal component of the compound is formed

at the cathode

◦ The non-metal component is formed at the anode

Example: NaCl molten electrolytes

NaCl Na+ + Cl-Anode: Cl-Cathode: Na+

Half equationAt the Anode: 2Cl- Cl2 + 2eAt the cathode: Na+ + e Na

Overall equation2Cl- Cl2 + 2e(Na+ + e Na) x 2 = 2Na+ + 2e 2Na2Cl- + 2Na+ Cl2 + 2Na

ProductAnode: Cl2 gasCathode: Na metal

Example: PbBr2 molten electrolytesSolution

Equation PbBr2 Pb2+ + Br-

Anode: Br-

Cathode: Pb2+

Half equationAnode: 2Br- Br2 + 2eCathode: Pb2+ +2e Pb

Overall equation2Br- Br2 + 2ePb2+ +2e Pb2Br- + Pb2+ Br2 +Pb

ProductsAnode: Br2 gasCathode: Pb metal

Aqueos solution consists of TWO types of cations and anions; H+ and OH-

Example: Aqueos sodium chloride (NaCl) solutionCations: Na+ and H+

Anions: Cl- and OH-

However only ONE type of cation and anion will be discharged at each electrode

1. Positions of ions in the electrochemical series

The tendency of ions to be selectively discharged depends on their positions in a series known as electrochemical series (ES).

The lower the position of the ion in the ES, the easier the ion will be discharged.

Example: Aqueos sodium chloride (NaCl) solution electrolytesCations: Na+ and H+

Anions: Cl- and OH-

Half equationAnode: 4OH- 2H2O + O2 + 4eCathode: 2H+ + 2e H2

Overall equation 4OH- 2H2O + O2 + 4e (2H+ + 2e H2 ) x 2 = 4H+ + 4e 2H2 4OH- + 4H+ 2H2O + O2 + 2H2

ProductsAnode: oxygen gasCathode: hydrogen gas

Example 2: Aqueos sodium sulphate (Na2SO4) solution electrolytesCations: Na+ and H+

Anions: SO42- and OH-

Half equationAnode: 4OH- 2H2O + O2 + 4eCathode: 2H+ + 2e H2

Overall equation 4OH- 2H2O + O2 + 4e (2H+ + 2e H2 ) x 2 = 4H+ + 4e 2H2 4OH- + 4H+ 2H2O + O2 + 2H2

ProductsAnode: oxygen gasCathode: hydrogen gas

2. Effect of ion concentration When the concentration of a particular

type of ion is higher, ion will more likely to be discharged in electrolysis

Usually concentrated halide (Cl- / Br- / I- ions)

3. Effect of types of electrodes used The types of electrodes used can determined

the type of ions discharged in electrolysis Using metal electrodes at anode, ions are

not discharged instead the metal dissolves by releasing electrons to form metal ions.

Example: using copper (Cu) electrodes.Cu Cu2+ + 2eHence mass of anode decrease.

1. Gives the diagram, anion, cation, half equation, overall equation, products and observation using electrolytes

i. PbBr2 moltenii. PbCl2 molteniii. CuSO4 solution

THANK YOU…