Section 1 i) Specification

1.48 understand that an electric current is a flow of electrons or ions

Electric current is the movement of charged particles in a sort of stream. Usually it is in reference to electrons, but free ions can also conduct electricity/carry charge, as they are charged particles.

1.49 understand why covalent compounds do not conduct electricity

In covalent compounds, there are no free charged particles (no delocalised electrons, no ions) available to carry charge.

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

Ionic compounds, when solid, do not conduct electricity. This is because the ions are not free to move; they are held in place by strong ionic bonding. When molten or aqueous, the ions become free and can then carry electrical charge.

1.51 describe experiments to distinguish between electrolytes and nonelectrolytes

Make a simple circuit involving a buzzer or a lightbulb, then, in series with the chosen component, place a beaker containing the liquid in question with electrodes attached into the circuit. If the component works, it is an electrolyte, because it allowed current to flow through. If the component does not work, it is a nonelectrolyte as it has broken the connection of the circuit.

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

Electrolysis is a process designed to separate ionic compounds and re-form them back into their individual elements.

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

1. Place carbon electrodes into a beaker containing lead bromide, so they are submerged. 
2. Hook the electrodes up to a cell and allow current to flow for some time.
3. At the cathode (negative electrode), positive metal ions will be reduced, in this case lead will form. 
4. At the anode (positive electrode), negative non-metal ions will be attracted and oxidised, in this case bromine will form. 

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

Sodium chloride

1. Place carbon electrodes into a beaker containing lead bromide, so they are submerged. 
2. Hook the electrodes up to a cell and allow current to flow for some time.
3. At the cathode (negative electrode), positive hydrogen ions will be reduced, and bubbles of hydrogen gas will form
4. At the anode (positive electrode), negative halide ions will be attracted and oxidised, in this case chlorine will form. 

Copper (II) sulphate

1. Place carbon electrodes into a beaker containing lead bromide, so they are submerged. 
2. Hook the electrodes up to a cell and allow current to flow for some time.
3. At the cathode (negative electrode), positive copper ions will be reduced, and copper will coat the electrode.
4. At the anode (positive electrode), negative hydroxide ions will be attracted and oxidised, and bubbles of oxygen gas will form. 

Dilute sulphuric acid

1. Place carbon electrodes into a beaker containing lead bromide, so they are submerged. 
2. Hook the electrodes up to a cell and allow current to flow for some time.
3. At the cathode (negative electrode), positive hydrogen ions will be reduced, and bubbles of hydrogen gas will form
4. At the anode (positive electrode), negative hydroxide ions will be attracted and oxidised, and bubbles of oxygen gas will form.

1.55 write ionic half-equations representing the reactions at the electrodes during
electrolysis

Lead bromide
Cathode: Pb2+ + 2e- --> Pb
Anode: 2Br- --> Br2 + 2e-

Sodium chloride
Cathode: 2H+ + 2e- --> H2
Anode: 2Cl- --> Cl2 + 2e-

Copper (II) sulphate
Cathode: Cu2+ + 2e- --> Cu
Anode: 4OH- --> O2 + H2O + 4e-

Dilute Sulphuric acid
Cathode: 2H+ + 2e- --> H2
Anode: 4OH- --> O2 +H2O + 4e-

1.56 recall that one faraday represents one mole of electrons

One faraday = 96500 coulombs = 1 mole of electrons

1.57 calculate the amounts of the products of the electrolysis of molten
salts and aqueous solutions.

If a current of 1.5A is passed through lead bromide for 1 hour, how much lead will form?
Half equation: Pb2+ + 2e- --> Pb
Coulombs = Current (in amperes) x Time (in seconds) = 5400 C

Conversion into moles of electrons:
1 mole = 1 faraday = 96500 coulombs
5400 / 96500 = 54/965 mol

There are 2 moles of electrons for every mole of lead, so divide by 2 = 27/965 mol
Then multiply by Mr to find mass 27/965 x 207 = 5.79 g 


If a current of 2A is passed through sodium chloride for 2 hours, what volume of chlorine will form?
Half equation: 2Cl- --> Cl2 + 2e-
Q = I T = 14400 C

1 mole = 1 faraday - 96500 C
14400 / 96500 = 144/965 mol

2 electrons per atom, so divide by 2 = 72/965 mol
Then multiply by 24 000 to find volume in cm3 72/965 x 24000 = 1790.67 cm3


If a current of 4.2A is passed through copper (II) sulphate for 45 minutes, what mass of copper will be formed?
Half equation: Cu2+ + 2e- --> Cu
Q = I T = 11340C

11340 / 96500 = 567/4825 mol

2 electrons per atom, so divide by 2 = 567/9650 mol
Then multiply by Mr 567/9650 x 63.5 = 3.73 g


If a current of 0.2A is passed through dilute sulphuric acid for 5 hours and 45 minutes, what volume of hydrogen is produced?
Half equation: 2H+ + 2e- --> H2
Q = I T = (5 x 60 + 45) x 60 x 0.2 = 4140C

4140 / 96500 = 207/4825 mol

2 moles of electrons per mole of hydrogen, so divide by 2 = 207/9650 mol
Then multiply by 24 000 to get the volume in cubic centimetres 207/9650 x 24 000 = 514.82 cm3