Electrolysis Flashcards
What is an electric current?
A flow of electrons
Why can’t covalent compounds conduct electricity?
Covalent compounds do not conduct electricity because there are no delocalised electrons that are free to move.
Why can ionic compounds conduct electricity only when molten or in solution?
When solid the ions are not free to move, when molten or in solution the ions are free to move.
Describe experiments to distinguish between electrolytes and non- electrolytes
- Set up an electric circuit with a bulb and a break in the wire attached to two electrodes, put both electrodes into a solution/molten substance.
- If the bulb lights up, 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 bulb does not light up, then there is no current flowing. The solution has not conducted electricity meaning it must be a non-electrolyte.
What happens during electrolysis?
Ionic compounds conduct electricity only when molten or in solution. During electrolysis, positively charged ions move to the cathode and form metals. The negatively charged ions move to the anode, to form non metals. This is because opposite charges attract. These ions become atoms because they undergo a reaction where either a gain or loss of electrons occurs.
Describe an experiment to investigate electrolysis, using lead(II) bromide
When the solid melts, ions become free to move. As soon as you connect the power source, it pumps and mobile electrons away form the left-hand electrode, towards the right. The excess of electrons on the right-hand electrode makes it negatively charged, the left hand electrode is positively charged because it is short of electrons.
The positive lead ions are attracted to the cathode, each lead ion picks up two electrons to form neutral lead atoms. These fall to the bottom of the container as lead melts. Lead ions are reduced.
Bromide ions are attracted to the anode, each bromide ion loses an electron to the electrode because there is a shortage. Each bromide ion turns into a bromine atom, these join in pairs to form bromine molecules. Bromide ions are oxidised.
Describe experiments to investigate electrolysis, using Sodium Chloride
Cathode: Solution contains Na+ and H+ ions which are both attracted to the cathode. The H+ gets discharged because the hydrogen ion accepts an electron. Each hydrogen atom formed combines with another to make a molecule. Each time a water molecule ionises, it produces a hydroxide ion which makes the solution strongly alkaline around the cathode.
Anode: Cl- and OH- are both attracted to the anode. There are far more chloride ions present in the solution so that are discharged to chlorine molecules.
Describe experiments to investigate electrolysis, using Sulfuric acid
Cathode: hydrogen ions are discharged to give hydrogen gas
Anode: sulfate ions and hydroxide ions are attracted. Sulfate ions are too stable to be discharged so you get oxygen form the discharge of hydroxide ions from the water.
Describe experiments to investigate electrolysis, using Copper (II) Sulfate
Cathode: copper ions and hydrogen ions will be attracted. Copper is below hydrogen in the reactivity series which means that copper ions are easier to discharge. The carbon electrode will get coated with brown copper.
Anode: Sulfate ions and hydroxide ions will be attracted. Sulfate ions are very stable so hydroxide ions are discharged to give oxygen.
Once all the copper ions are used up, you are left with hydrogen ions and sulfate ions. The solution turns into dilute sulfuric acid.
Write ionic half-equations representing the reactions at the electrodes during electrolysis
Lead (II) bromide, PbBr2:
At the cathode
Pb2+ (l) + 2e– –> Pb (l) (reduction)
At the anode
2Br– (l) –> Br2 (g) + 2e– (oxidation)
Sodium chloride solution, NaCl (aq)
At the cathode
2H+ (aq) + 2e– –> H2 (g) (reduction)
At the anode
2Cl– (aq) –> Cl2 (g) + 2e– (oxidation)
Copper sulfate solution, CuSO4 (aq
At the cathode
Cu2+ (aq) + 2e– –> Cu (s) (reduction)
At the anode
4OH– (aq) –> O2 (g) + 2H2O (l) + 4e– (oxidation)
What is a faraday?
One faraday represents one mole of electrons
Calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions
Step 1: Deduce the number of moles of electrons. If 1 F (faraday) represents 1 mole of e– then 10 F will represent 10 moles of e–.
Step 2: Deduce the ratio of electrons to chlorine. From the half equation you can see that 2 moles of e– are produce for every 1 mole of Cl2(g). The ratio is 2:1.
Step 3: Deduce the amount, in moles of chlorine. Therefore 10 moles of e– will produce 5 moles of Cl2(g).
