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

CALCULATIONS:

- If 2.5 amps are passing through sodium chloride for 15 minutes. What mass of sodium do we get?

- 1. Write out the half-equation:
2H+ + 2e- -> H2

- Calculate number of faradays.
Q = IxT
Q = 2.5 x (15 x 60 = 900) = 2250 coulombs.
Faradays = Q / FQ
Faradays = 2250 / 96000 = 0.0234375 F

- Calculate number of moles of sodium produced.
Moles = Faradays / (Electrons in half-equation)
0.0234375 / 2 = 0.01171875 moles of sodium atoms.

- Calculate Mr to workout mass.
Mass of sodium = Mr x Moles
23 x 0.01171875 = 0.26953125 = 0.27 g

For better equations go to: (http://www.docbrown.info/page04/4_73calcs13pec.htm)

1.56 recall that one faraday represents one mole of electrons

FARADAYS:

- A faraday is the same as 96,485.3365 coulombs but for the exam you are allowed to know it as 96,000 coulombs.
- One faraday also contains one mole of electrons.

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

IONIC HALF-EQUATIONS:

- At the positive electrode, also known as the anode, electrons are lost.
2Br- -> Br2 + 2e-

- At the negative electrode, also known as the cathode, electrons are gained.
Pb^2+ + 2e- -> Pb

As http://hannahhelpchemistry.blogspot.co.uk/ has said:
- At the anode the half-equation should be equal on both sides: 1- -> 1-.
- At the cathode the half-equation should be equal on both sides: 0 -> 0.

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:

- Place electrodes into solution, then at the anode, negatives will form and vice versa with the cathode.

- NaCl contains Na+, Cl-, OH- and H+
Hydrogen is easier to accept electrons and therefore at the cathode hydrogen gas is produced.
Chloride is easier to lose electrons than hydroxide and so chlorine gas is produced.

- CuSO4 contains Cu2+, SO4^2-, H+ and OH
Copper accepts electrons easier that hydrogen so copper is produced at the cathode.
Hydroxide loses electrons easier so at the anode oxygen and water are produced.

- H2SO4 contains SO4^2-, H+ and OH-
Hydrogen accepts electrons so at the cathode hydrogen is produced.
Hydroxide loses electrons easier than sulfate and so at the anode oxygen and water are produced.

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

INVESTIGATING ELECTROLYSIS:

- Inert means that they do not react with anything.
- An example of this is with lead(II) bromide. With this you can write half-equations to show the transitions of the electrons at each electrode.

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

ELECTROLYSIS:

- Is mainly used to create new substances. In electrolysis the positive ions move to a cathode (since opposites attract) and vice versa with the negative ions. This means that the ions lose their charge and become atoms again to form new substances.

1.51 describe experiments to distinguish between electrolytes and nonelectrolytes

DISTINGUISH DIFFERENCES:

- Create a circuit that has a break in. In the circuit you should have a component that indicates a full circuit. For example a motor or LED. Place the two brakes into the dissolved or molten substance and if it is an electrolyte the component will indicate. This is because the substance conducts.