Electrolysis

Always put state symbols

Electrolysis is the chemical change caused by passing an electric current through a compound which is either molten or in solution. 

1.48- Electricity is a flow of ions or electrons

1.49- Understand that covalent compounds do not conduct electricity. 

• Covalent compounds do not have any free electrons in their outer shells.
• They don't have free ions either as they are covalently bonded
• So they do not conduct electricity

1.50- Understand why ionic compounds conduct electricity only when molten or in solution 

• The ions must be free to move to allow the ionic compound to conduct electricity. 
• This happens only when the ionic compounds are molten or in solution

1.51- Describe experiments to distinguish between electrolytes and nonelectrolytes

• Set up an electric circuit with a battery and a bulb 
• Make a break in the wire
• And put both ends of the wire into the electrolyte
• If the bulb glows then that means the substance is an electrolyte. If it does not glow, that means the substance is a non-electrolyte

1.52-Understand that electrolysis involves the formation of new substances when ionic compounds conduct electricity

• Electrolysis is carried out on electrolytes which are ionic substances. 
• The ionic substances being made of cations and anions separate and get attracted to the cathodes and anodes respectively.
• At the cathode, cations gain electrons to become neutral atoms
• At the anode, anions lose electrons to become neutral atoms.

1.53- Describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the products 

• The electrolyte in this case is the lead bromide. 
• The inert electrodes can be made out of platinum.

1. Dissolve the lead bromide in water or heat it until it becomes molten.
2. Set up a circuit and create a break in it. Place both ends of the wire in the molten/aqueous substance.
3. If bubbles are seen (gas is being produced,) collect it over water, upwards into a test tube, or downwards into a test tube where appropriate. 
4. Test the substances produced.

^{Products produced- Lead  and Bromine}


Electrolysis of water 

H₂O(l)ßà H⁺(aq) + OH^- (aq)

1.54- Describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the products

In an aqueous solution if the cation has a high reactivity, then hydrogen is most likely to be produced. If the ion is lower than hydrogen, then the metal is going to be produced.

Water electrolyses to form H⁺ ions and OH^- ions.  SInce the H

Electrolysis of NaCl(brine):(Platinum electrodes)

Since the hydrogen ions are attracted to the cathode, there is a buildup of hydroxide ionn. So we can say that the hydroxide ions make the region around the cathode alkaline. We can also describe electrolysis as the forming of sodium hydroxide solution.  

At the cathode, the hydrogen ions are selectively discharged over sodium ions.

2H^{+   (aq)}+ 2e^-  à H_{2 (g)          Reduction}

At the anode, the chlorine ions are selectively discharged over OH- ions. (Even though it is easier for the OH- ion to reduce, there are a lot more chlotide ions in brine which makes chloride ions discharge)

2Cl^-(aq) à Cl₂(g)+2e^{-             Oxidation}

^{Electrolysis of copper(11) sulfate: (Carbon electrodes)}

At the cathode: Copper(11) ions are selectivily discharged and reduced over Hydrogen ions. This is because copper is more unstable than hydrogen ion s it is easier for copper to gain an electron.

Cu²⁺(aq)+2e^- à Cu(s) Reduction

At the anode: Hydroxide ions are dischared and oxidised over Sulfate ions. This is because Sulfate ions are more stable and so it is easier for hydroxide ions to lose an electron.

4OH^-(aq) à O₂ (g)+ 2H₂O(l) +4e^{- Oxidation}

If this electrolysis is carried out for a long time, the colour of the solution will fade from blue to colourless as all the copper ions would be used up. All the hydroxide ions would get used up as well.
This would lead to dilute sulfuric acid becoming an electrolyte.

Electrolysis of dilute sulfuric acid 

At the cathode: Hydrogen ions from both the water and sulfuric acid are reduced.

2H^{+   (aq)}+ 2e^-  à H_{2 (g)          Reduction}

At the anode: hydroxide ions will again be selectively discharged over sulfate ions. This is because the sulfate ions are too stable to be discharged.

4OH^-(aq) à O₂ (g)+ 2H₂O(l) +4e^{- Oxidation}

Twice as much hydrogen molecules are produced as oxygen molecules.

H₂:O₂ = 4:2 = 2:1

This is because 1 oxygen molecule consists of 2 oxgen molecules. Both the atoms, will oxidise to give away 2 electrons each. The 2 electrons recieved by the positive anode, will move around the circuit and reach the negative cathode. At the negative cathode these 4 electrons will be gained by 4 hydrogen atoms due to reduction. 4 atoms of hydrogen atoms would result in 2 molecules of hydrogen. Therefore twice the number of hydrogen molecules  are produced compared to number of oxygen molecules.Al

ONE FARADAY REPRESENTS 1 MOLE OF ELECTRONS

^{If you have reasonably concentrated solutions of the halide then the halide is produced at the anode.} 


5.1 Explain how methods of extraction of this section are related to their positions in the reactivity series.

If the element is below carbon in the reactivity series then it can be displaced from its ore by carbon.

5.2 describe and explain the extraction of aluminium from purified aluminium oxide by electrolysis, including: i the use of molten cryolite as a solvent and to decrease the required operating temperature ii the need to replace the positive electrodes iii the cost of the electricity as a major factor


Bauxite is purified to aluminum oxide.
This is then dissolved in molten cryolite to bring down the melting point, This decreases the required operating temperature.
The walls of the tank are the cathode; here the aluminum ions are discharged.
The aluminum sinks to the bottom and is then tapped off.

Oxygen is formed at the positive electrode. It reacts with graphite to form carbon dioxide.

So the anodes need to be changed often.
It is very expensive to supplay this electricity for electrolysis.

5.3 describe and explain the main reactions involved in the extraction of iron from iron ore (haematite), using coke, limestone and air in a blast furnace