Chemistry - EDEXCEL IGCSE - Electrolysis

ELECTROLYSIS:

An electric current is the flow of ions (e.g. electrons)

Covalent compounds cannot conduct electricity as they have no ions meaning there can be no potential difference. Ionic compounds conduct electricity when molten or dissolved as it means the ions in the compound can now move around freely which means that it is possible for current to flow as the ions can now carry the current.

Experiment to distinguish between electrolytes and non-electrolytes: 

If when the circuit is turned on (the switch is closed) the lamp turns on, then the substance is an electrolyte.

Electrolysis of molten salts (e.g.PbBr₂)

1. Bubbling (effervescence) at the anode (positive) of Br gas
2. Lead fired at the cathode (negative)

Electrolysis of an aqueous solution with carbon electrodes:

Sodium Chloride:

The experiment is set up as before except that it is flipped and two test tubes are filled with water placed above the electrodes. This is to collect the gas.

At Cathode:

For positive ions, the lower down in the reactivity series the more easily it accepts electrons. Therefore H+ in the water discharges rather than Na+. Therefore, hydrogen gas is released.

2H⁺ + 2e^- —> H₂(g)

The sodium reacts with the remaining OH- in the water to form an alkaline sodium hydroxide solution  - Na⁺(aq) + OH^-(aq) —> NaOH(aq)

At anode:

The chlorine and hydroxide ions are at a similar position in the reactivity series. But there are more chlorine ions in the solution so it is mainly these that are discharged.

2Cl^- —> Cl₂(g) + 2e^-

Copper Sulphate:

At cathode:

Copper is lower in the reactivity series compared to hydrogen so copper is formed.

Cu²⁺ + 2e^- —> Cu

At anode:

The hydroxide in the water discharges, releasing oxygen.

4OH^- —> 2H₂O(l) + O₂(g) + 4e^-

Sulphuric Acid:

At cathode:

Hydrogen ions discharges.

2⁺ + 2e^- —> H₂(g)

At anode:

Once again the hydroxide ions discharge, releasing oxygen.

4OH^- —> 2H₂O(l) + O₂(g) + 4e^-

Calculations involving electrolysis:

1 Faraday = 1 Mole of electrons = 96500 Coulombs

Coulomb = current * time

A current of 0.2 amp is passed through copper sulphate for 2 hours, how much copper is formed?

Cu²⁺ + 2e^- —> Cu

2 hours = 2 * 60 * 60 = 7200

7200 * 0.2 = 1440

Therefore, there are 1440 coulombs

1440/96500 = 0.015

Therefore, there is 0.015 mole

0.015 / 2 = Cu²⁺

0.0075 = Cu²⁺

Ar of Cu = 63.5

63.5 * 0.0075 = 0.48

Therefore, answer is 0.48g