*Please click video title headings to play video if video does not load properly.
Electrolysis
Electrolysis is the passing of an electric current through an electrolyte to decompose it in a chemical reaction.
Electrolytes are ionic compounds which are either in a molten or aqueous state. As such, there are free moving ions present that can conduct electricity.
Electrolysis thus provides evidence that ions are free to move in the molten or aqueous state of an ionic compound.
Electrolysis reactions are redox reactions:-
Reduction occurs at cathode (-ve electrode) <Cations gain electrons at cathode>
Oxidation occurs at anode (+ve electrode) <Anions lose electrons at anode>
This can be remembered as red cat an ox.
Rules for electrolysis:-
At an inert cathode,
Hydrogen ions (H+) and ions of low reactivity metals such as Cu²+ are preferentially discharged. The lower the reactivity of the metal cation is, the more it is preferentially discharged. Note that Cu²+ will be preferentially discharged than H+ since Cu²+ has a lower reactivity than H+.
Ions of very reactive metals like Na+ and K+ are not discharged as the metals sodium and potassium will immediately react with cold water. Instead, H+ ions from water are discharged to give hydrogen gas.
At an inert anode,
Chloride (Cl-), bromide (Br-) and iodide (I-) ions which are the halides (group 7 ions) are preferentially discharged in their concentrated solutions. In diluted solutions, hydroxide (OH-) ions from water are preferentially discharged to give water and oxygen gas.
Sulphate (SO4²-) and nitrate (NO3-) ions are never discharged. Instead, hydroxide (OH-) ions from water are preferentially discharged to give water and oxygen gas.
Electrolytes are ionic compounds which are either in a molten or aqueous state. As such, there are free moving ions present that can conduct electricity.
Electrolysis thus provides evidence that ions are free to move in the molten or aqueous state of an ionic compound.
Electrolysis reactions are redox reactions:-
Reduction occurs at cathode (-ve electrode) <Cations gain electrons at cathode>
Oxidation occurs at anode (+ve electrode) <Anions lose electrons at anode>
This can be remembered as red cat an ox.
Rules for electrolysis:-
At an inert cathode,
Hydrogen ions (H+) and ions of low reactivity metals such as Cu²+ are preferentially discharged. The lower the reactivity of the metal cation is, the more it is preferentially discharged. Note that Cu²+ will be preferentially discharged than H+ since Cu²+ has a lower reactivity than H+.
Ions of very reactive metals like Na+ and K+ are not discharged as the metals sodium and potassium will immediately react with cold water. Instead, H+ ions from water are discharged to give hydrogen gas.
At an inert anode,
Chloride (Cl-), bromide (Br-) and iodide (I-) ions which are the halides (group 7 ions) are preferentially discharged in their concentrated solutions. In diluted solutions, hydroxide (OH-) ions from water are preferentially discharged to give water and oxygen gas.
Sulphate (SO4²-) and nitrate (NO3-) ions are never discharged. Instead, hydroxide (OH-) ions from water are preferentially discharged to give water and oxygen gas.
Electrolysis of concentrated sodium chloride solution
Step 1: Identify the ions present. Na+, H+, Cl- and OH-.
Step 2: Apply rules of electrolysis.
At cathode, H+ ions are preferentially discharged than Na+ ions as H+ ions are of much lower reactivity than Na+ ions.
At anode, Cl- ions are preferentially discharged than OH- ions as the solution is concentrated with far more Cl- ions
than OH- ions.
Step 3: Write the balanced ionic equations of the redox reactions.
At cathode (-ve electrode), 2H+ (aq) + 2e- → H2 (g) <Reduction occurs = H+ ions gain electrons> Red Cat
At anode (+ve electrode), 2Cl- (aq) → Cl2 (g) + 2e- <Oxidation occurs = Cl- ions lose electrons> An Ox
Step 4: Identify the products formed at cathode and anode.
Hydrogen gas is formed at the cathode while chlorine gas is formed at the anode. The solution
gradually becomes more alkaline forming sodium hydroxide as Na+ and OH- ions are left behind and not
discharged.
Step 2: Apply rules of electrolysis.
At cathode, H+ ions are preferentially discharged than Na+ ions as H+ ions are of much lower reactivity than Na+ ions.
At anode, Cl- ions are preferentially discharged than OH- ions as the solution is concentrated with far more Cl- ions
than OH- ions.
Step 3: Write the balanced ionic equations of the redox reactions.
At cathode (-ve electrode), 2H+ (aq) + 2e- → H2 (g) <Reduction occurs = H+ ions gain electrons> Red Cat
At anode (+ve electrode), 2Cl- (aq) → Cl2 (g) + 2e- <Oxidation occurs = Cl- ions lose electrons> An Ox
Step 4: Identify the products formed at cathode and anode.
Hydrogen gas is formed at the cathode while chlorine gas is formed at the anode. The solution
gradually becomes more alkaline forming sodium hydroxide as Na+ and OH- ions are left behind and not
discharged.
Electrolysis of dilute sodium chloride solution
Note that if dilute sodium chloride solution is used, hydrogen (H+) ions are still preferentially discharged at the cathode (-ve electrode) forming hydrogen gas (see above electrolysis of concentrated sodium choride solution for the same ionic equation).
However, at the anode (+ve electrode), hydroxide ions (OH-) are preferentially discharged since there are far more OH- ions present than Cl- ions in the dilute solution. Water and oxygen gas are formed at the anode.
4OH- (aq) → 2H2O (l) + O2 (g) + 4e-
After a period of time, a more concentrated sodium chloride solution is formed since Na+ and Cl- ions are not discharged at the electrodes and are left behind.
However, at the anode (+ve electrode), hydroxide ions (OH-) are preferentially discharged since there are far more OH- ions present than Cl- ions in the dilute solution. Water and oxygen gas are formed at the anode.
4OH- (aq) → 2H2O (l) + O2 (g) + 4e-
After a period of time, a more concentrated sodium chloride solution is formed since Na+ and Cl- ions are not discharged at the electrodes and are left behind.
Electrolysis of copper (II) sulphate solution using carbon electrodes
Ions present in the solution are Cu²+ (aq), SO4²- (aq), H+ (aq) and OH- (aq).
At cathode, copper (II) ions are preferentially discharged than H+ ions to give copper metal since copper (II) ions are of lower reactivity than hydrogen ions. Cu²+ (aq) + 2e- → Cu (s) <reduction occurs = Cu²+ ions gain electrons>
At anode, hydroxide ions are discharged to give water and bubbles of oxygen gas. Sulphate ions are never discharged.
4OH- (aq) → 2H2O (l) + O2 (g) + 4e- <oxidation occurs = OH- ions lose electrons>
Solution gradually turn from blue to become colourless as copper (II) ions are discharged. At same time, H+ and SO42- ions are left behind and the solution gradually becomes sulphuric acid which is acidic.
At cathode, copper (II) ions are preferentially discharged than H+ ions to give copper metal since copper (II) ions are of lower reactivity than hydrogen ions. Cu²+ (aq) + 2e- → Cu (s) <reduction occurs = Cu²+ ions gain electrons>
At anode, hydroxide ions are discharged to give water and bubbles of oxygen gas. Sulphate ions are never discharged.
4OH- (aq) → 2H2O (l) + O2 (g) + 4e- <oxidation occurs = OH- ions lose electrons>
Solution gradually turn from blue to become colourless as copper (II) ions are discharged. At same time, H+ and SO42- ions are left behind and the solution gradually becomes sulphuric acid which is acidic.
Electrolysis of copper (II) sulphate solution using copper electrodes
Ions present in the solution are Cu²+ (aq), SO4²- (aq), H+ (aq) and OH- (aq).
At cathode, copper (II) ions are preferentially discharged than H+ ions to give copper metal since copper (II) ions are of lower reactivity than hydrogen ions. Cu²+ (aq) + 2e- → Cu (s) <reduction occurs = Cu²+ ions gain electrons>
At anode, neither hydroxide ions nor sulphate ions are discharged (*contrast this with electrolysis of copper (II) sulphate solution using carbon electrodes).This is because the type of electrodes used affect what happens here which is copper instead of carbon used as electrodes. Instead, copper atoms from the anode lose electrons to become copper (II) ions. The mass of the anode decreases as copper dissolves.
Cu (s) → Cu²+ (aq) + 2e- <oxidation occurs = Copper atoms lose electrons>
Copper (II) sulphate solution remains blue as Cu²+ ions that are discharged at the cathode are constantly replaced by more Cu²+ ions formed from the copper atoms at anode.
This electrolysis is used to refine (make pure) copper. Impure copper is used as anode while pure copper is used as cathode. Over time, pure copper is transferred from anode to cathode. Anode decreases in mass while cathode increases in mass. The impurities at anode will fall off to bottom of electrolysis bath.
At cathode, copper (II) ions are preferentially discharged than H+ ions to give copper metal since copper (II) ions are of lower reactivity than hydrogen ions. Cu²+ (aq) + 2e- → Cu (s) <reduction occurs = Cu²+ ions gain electrons>
At anode, neither hydroxide ions nor sulphate ions are discharged (*contrast this with electrolysis of copper (II) sulphate solution using carbon electrodes).This is because the type of electrodes used affect what happens here which is copper instead of carbon used as electrodes. Instead, copper atoms from the anode lose electrons to become copper (II) ions. The mass of the anode decreases as copper dissolves.
Cu (s) → Cu²+ (aq) + 2e- <oxidation occurs = Copper atoms lose electrons>
Copper (II) sulphate solution remains blue as Cu²+ ions that are discharged at the cathode are constantly replaced by more Cu²+ ions formed from the copper atoms at anode.
This electrolysis is used to refine (make pure) copper. Impure copper is used as anode while pure copper is used as cathode. Over time, pure copper is transferred from anode to cathode. Anode decreases in mass while cathode increases in mass. The impurities at anode will fall off to bottom of electrolysis bath.