Electrolysis
NOTE - PAPER 2 CHAPTER
Conduction of electricity is very important for compounds to undergo electrolysis and the charged particles present in a compound move allowing it to conduct electricity.
These compounds would mainly be -
Ionic Compounds –
These don’t conduct electricity when solid but do conduct electricity in the molten or aqueous state (which is when they are dissolved in water) as the IONS are free to move.
The ionic compounds consist of cations which are positive and anions which are negative.
Covalent Compounds –
These don’t conduct electricity regardless of the state they are in as they consist of individual molecules and hence don’t have any overall electrical charge so can't move.
Furthermore, they have covalent bonds which are very strong and so hold the electrons in the atoms firmly in place.
Electrolysis is a chemical change caused by passing a current through a compound that’s molten or in a solution.
An electrolyte is a liquid or solution that undergoes the process of electrolysis.
Electrodes are electric conductors that are used to make contact and carry the electric current into the non-metallic solids, liquids or gases. The electrodes are made of carbon and platinum and are inert so they don’t react
Electrons flow from positive to negative
Positive ions go towards the cathode while the negative ions go to the anode.
PANIC: Positive (is) Anode, Negative is Cathode
Electrolysis of Ionic Compounds
Molten compounds undergo electrolysis and always produce their respective elements.
Lead (II) Bromide is an ionic compound and consists of lead (II) ions and Bromine ions packed together. When heated they melt and the ions become free to move.
The lead ions are attracted to the cathode and pick up two ions and become lead atoms and fall to the bottom forming molten lead.
The bromine ions are attracted to the anode and lose an electron forming a bromine atom.
Half-ionic equations are used to show what happens at each electrode.
At the cathode: Pb2+Pb2+ + 2e−2e− ⟹ ⟹ Pb
At the anode: 2Br−2Br− ⟹ ⟹ Br2Br2 + 2e−2e− (don’t forget that Bromine is di-atomic)
The discharge of ions means that ions are losing their charge.
Electrolysis of Aqueous Solutions
Aqueous solutions will always have water present and some split up into hydrogen and hydroxide ions, H+ and OH- and participate in electrolysis reactions.
At the positive electrode, the OH- ions or the non-metal ions are discharged and lose electrons or gain oxygen (oxidised)
At the negative electrode, the H+ ions or the metallic ions are discharged but only one would gain electrons or lose oxygen (reduction)
Aqueous Sodium Chloride solution contains Na+ and H+ ions which are attracted to the cathode.
Hydrogen is less reactive than sodium so we can more easily add an electron to hydrogen ions to form a hydrogen molecule.
Hydroxide ions present in the solution due to the water splitting make the solution alkaline \n 2H+2H+(aq) + 2e−2e− ⟹ ⟹ H2H2 (g)
At the anode, Cl- and OH- ions are present but as there are many more chloride ions these are oxidised and form Cl2Cl2
2Cl−2Cl−(aq) - 2e−2e− ⟹ ⟹ Cl22(g)
The remaining solution now consists of Na+Na+ and OH−OH− ions. So we are left with Sodium Hydroxide (NaOH)
We get less chlorine than we expect as chlorine is more soluble in water, reducing its yield
During the electrolysis of Dilute Sulfuric acid, twice as much hydrogen is produced
2H+2H+ (aq) + 2e−2e− ⟹ ⟹ H2(g)
4OH−4OH− (aq) ⟹ ⟹ 2H2O2H2O (l) + O2O2 + 4e−4e−
If you look at the equations above you would see that 2 electrons produce 1 mol of hydrogen meaning 4 electrons would produce 2 mols of hydrogen.
In the second, 4 electrons produce 1 mol of oxygen
This means that twice as much hydrogen is produced compared with oxygen
Oxidation and Reduction
Reduction is the gain of electrons or the loss of oxygen.
Oxidation is the loss of electrons or the gain of oxygen.
OILRIG: Oxidation is loss (of electrons), Reduction is gain (of electrons)
Reduction always occurs at the cathode while oxidation occurs at the anode.
Not all ionic compounds can be electrolysed as they break up into similar chemical compounds before their boiling point making it impossible to melt.
CCRG: Cathode, Cation, Reduction (is) Gain of electrons, Reducing Agent
AAOL: Anode, Anion, Oxidation (is) Loss of electrons, Oxidising Agent
Reactivity Series
If the metal is above hydrogen in the reactivity series, you would get Hydrogen produced at the cathode eg: Potassium would produce Hydrogen at the cathode
If the metal is below hydrogen, you would get the metal at the cathode, eg: Gold would produce gold itself at the cathode.
If you have halides, you get that at the anode but any other negative ions would produce oxygen at the anode.
IMPORTANT IONS AND CHARGES -
CationsAnions | |
K+K+ : Potassium ion | SO42−SO42− : Sulfate ions |
Na+Na+ : Sodium ion | No3−No3− : Nitrate ions |
Ca2+Ca2+ :Calcium ion | Cl−,Br−,I−Cl−,Br−,I− : Halide ions |
Al3+Al3+ : Aluminum ions | OH−OH− : Hydroxide ions |
Zn2+Zn2+ : Zinc ions | |
Fe2+Fe2+ : Iron (II) ions | |
Pb2+Pb2+ : Lead ions | |
H+H+ : Hydrogen ions | |
Cu2+Cu2+ : Copper ions | |
Ag+Ag+ : Silver ions |
Experiments
Electrolysis of aqueous NaCl (Sodium Chloride) solution
Take a glass tube, close it with a rubber bung, place the electrodes through it and connect it to the battery.
Pour the concentrated NaCl into the glass tube
Invert a test tube and place it over the electrodes and ensure that the electrodes aren’t completely covered, or the ions won't be able to flow
Connect the electrodes to the battery
Conduct the experiment in a fume cupboard to ensure that the poisonous fumes produced from chlorine gas are ventilated away
Hydrogen forms at the cathode and Chlorine gas forms at the anode
Quantitative electrolysis
Take a glass tube, close it with a rubber bung, place the electrodes through it, connect it to the battery and add a variable resistor and ammeter to the circuit.
Place an inverted test tube over the electrodes to collect the Chlorine and Hydrogen gas
Pour 50cm350cm3 of concentrated NaCl into the glass tube
Place a glass burette with NaCl solution over the cathode
Take the initial reading on the glass burette
Set the current to 0.2A using the variable resistor and connect the battery
Start the timer
Repeat for different currents (0.4A, 0.6A, 0.8A and 1A)
Repeat the experiment to get accurate and reliable results
Collect the data in a table and plot a graph