1.47 explain the electrical conductivity and malleability of a metal in terms of its structure and bonding.
Metals have delocalised electrons, electrons carry electricity; so because there are free electrons charge can pass easily through a metal.
The structure of a metal is with rows of atoms on top of one another, in pure metals as all the atoms will be the same size, the layers can slide easily over one another making them easy to bend.
1.51 describe experiments to distinguish between electrolytes and nonelectrolytes
Set up an electric circuit with an LED and a break in the wire, put both ends of wire into a solution/molten substance. If the LED lights up then there is a current flowing, this will only be able to happen if the solution is conducting: so it must be an electrolyte. Conversely if the LED does not light up then there is no current flowing, and so the solution has not conducted electricity meaning it must be a nonelectrolyte.
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
Place inert electrodes (ones that wont react) into an aqueous solution.
At the positive electrode the negatively charged ion from the compound will form an atom. At the negative electrode the atom of the positive ion will form.
sodium chloride: Hydrogen at the negative; chlorine at the positive
copper(II) sulfate: copper at the negative; oxygen at the positive
dilute sulfuric acid: Hydrogen at the negative; oxygen at the positive
If the metal in the solution is more reactive then hydrogen, the hydrogen from the water will be a product, as the metal will bond with the oxygen.
Test the products using known methods: eg damp blue litmus paper turned red by chlorine.
1.55 write ionic half-equations representing the reactions at the electrodes during electrolysis
At the positive electrode, electrons will be lost: to show this we write the lost electrons as products:
2Br- > Br2 + 2e-
Make sure the charges are equal on both sides: 1- > 1-.
At the negative electrode, electrons will be gained so we write them as reactants:
2H+ + 2e- > H2
And to make sure the charges are the same on both sides: 0 > 0.
1.56 recall that one faraday represents one mole of electrons
One Faraday is 96500 coulombs. That is the amount of coulombs in one mole of electrons.
1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions
One faraday is 96500 coulombs. It is also one mole of electrons.
If current of 0.2 Apms is passed through copper(ll) sulphate for tow hours, how much copper do you get?
- Write out the half equation
Cu2+ + 2e > Cu
- Work out coulombs of electrons flowing
Coulombs= current x time
time is 2x60x60 (times 60 makes minutes, times 60 again makes it seconds) Q= 0.2 x 7200= 1440 coulombs
- Convert C into moles…