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Electrolysis - decomposition of compound using electricity
Electrolyte - an ionic compound which conducts electric current in molten or aqueous solution, being decomposed in the process.
Electrode - a rod or plate where electricity enters or leaves electrolyte during electrolysis.
Reactions occur at electrodes.
Discharge - the removal of electrons from negative ions to form atoms or the gain of electrons of positive ions to become atoms.
Anode
Electrolysis of molten PbBr[sub]2[/sub]
To make molten lead(II) bromide, PbBr[sub]2[/sub], we strongly heat the solid until it melts. To electrolyse it, pass current through the molten PbBr[sub]2[/sub].
What happens:
Ions present: Pb[sup]2+ [/sup]and Br[sup]-[/sup]
Reaction at Anode
Br- loses electrons at anode to become Br atoms. Br atoms created form bond together to make Br2 gas.
2Br[sup]-[/sup](aq) --> Br[sub]2[/sub](g)+ 2e[sup]-[/sup]
Reaction at Cathode
Pb[sup]2+[/sup] gains electrons at cathode to become Pb atoms becoming liquid lead (II).
Pb[sup]2+[/sup](aq) + 2e[sup]-[/sup] --> Pb(l)
Overall equation
PbBr[sub]2[/sub](l) --> Pb(l) + Br[sub]2[/sub](g)Electrolysis of Aqueous Solution
At cathode
Electrolysis of Concentrated NaCl
What happens:
Ions Present: Na[sup]+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and Cl[sup]-[/sup]
Reaction at Anode
2HCl(l) --> H[sub]2[/sub](l) + Cl[sub]2[/sub](g)
Note: any cation and anion left undischarged in solution forms new bonds between them.
E.g. in above, leftovers Na[sup]+[/sup] and OH[sup]-[/sup] combine to form NaOH.
B. Very Dilute Solutions
Electrolysis of Dilute H2SO4
What happens:
Ions Present: H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
What happens:
Ions Present: Cu[sup]2+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
Ions Present: Cu[sup]2+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
Electrolyte - an ionic compound which conducts electric current in molten or aqueous solution, being decomposed in the process.
Electrode - a rod or plate where electricity enters or leaves electrolyte during electrolysis.
Reactions occur at electrodes.
Discharge - the removal of electrons from negative ions to form atoms or the gain of electrons of positive ions to become atoms.
Anode
- positive electrode connected to positive terminal of d.c. source.
- Oxidation occurs here.
- Anode loses negative charge as electrons flow towards the battery, leaving anode positively charged.
- This causes anion to discharge its electrons here to replace lost electrons and also, negative charge are attracted to positive charge.
- negative electrode connected to negative terminal of d.c. source.
- Reduction occurs here.
- Cathode gains negative charge as electrons flow from the battery towards the cathode, making cathode negatively charged.
- This causes cation to be attracted and gains electrons to be an atom.
- negative ion
- attracted to anode.
- positive ion
- attracted to cathode.
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- Molten/aqueous ionic compounds conduct electricity because ions free to move.
- In solid state, these ions are held in fixed position within the crystal lattice.
- Hence solid ionic compounds do not conduct electricity.
- When molten binary compound is electrolysed, metal is formed on cathode while non-metal is formed on anode.
Electrolysis of molten PbBr[sub]2[/sub]
To make molten lead(II) bromide, PbBr[sub]2[/sub], we strongly heat the solid until it melts. To electrolyse it, pass current through the molten PbBr[sub]2[/sub].
What happens:
Ions present: Pb[sup]2+ [/sup]and Br[sup]-[/sup]
Reaction at Anode
Br- loses electrons at anode to become Br atoms. Br atoms created form bond together to make Br2 gas.
2Br[sup]-[/sup](aq) --> Br[sub]2[/sub](g)+ 2e[sup]-[/sup]
Reaction at Cathode
Pb[sup]2+[/sup] gains electrons at cathode to become Pb atoms becoming liquid lead (II).
Pb[sup]2+[/sup](aq) + 2e[sup]-[/sup] --> Pb(l)
Overall equation
PbBr[sub]2[/sub](l) --> Pb(l) + Br[sub]2[/sub](g)Electrolysis of Aqueous Solution
- Aqueous solutions contain additional H[sup]+[/sup] and OH[sup]-[/sup] ions of water, totaling 4 ions in the solution :
- 2 from electrolyte, 2 from water.
- Only 2 of these are discharged.
- 2 from electrolyte, 2 from water.
- Electrolysis of aqueous solutions use the theory of selective discharge.
At cathode
- In CONCENTRATED solutions of nickel/lead compound, nickel/lead will be discharged instead of hydrogen ions of water which is less reactive than nickel/lead.
- In VERY DILUTE solutions, hydrogen, copper and silver ions are preferable to be discharged, according to its ease to be discharged.
- Reactive ions (potassium, sodium, calcium, magnesium, aluminium) will NEVER BE DISCHARGED in either concentrated or dilute condition. Instead, hydrogen ions from water will be discharged at cathode.
- In CONCENTRATED solutions, iodine/chlorine/bromine ions are preferable to be discharged, although it’s harder to discharged compared to hydroxide ions.
- In VERY DILUTE solutions containing iodide/chloride/bromide ions, hydroxide ions of water will be discharged instead of iodide/chloride/bromide, according to ease of discharge.
- Sulphate and nitrate are NEVER DISCHARGED in concentrated/dilute solutions.
Electrolysis of Concentrated NaCl
What happens:
Ions Present: Na[sup]+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and Cl[sup]-[/sup]
Reaction at Anode
- Cl[sup]-[/sup] loses electrons at anode to become Cl atoms, although OH[sup]-[/sup] is easier to discharge.
- Cl atoms created form covalent bond together to make Cl[sub]2 [/sub]gas.
- 2Cl[sup]- [/sup](aq) --> Cl[sub]2 [/sub](g) + 2e[sup]-[/sup]
- H[sup]+[/sup] gains electrons at cathode to become H atoms becoming hydrogen gas
- 2H[sup]+ [/sup](aq) + 2e[sup]-[/sup] --> H[sub]2[/sub] (l)
2HCl(l) --> H[sub]2[/sub](l) + Cl[sub]2[/sub](g)
Note: any cation and anion left undischarged in solution forms new bonds between them.
E.g. in above, leftovers Na[sup]+[/sup] and OH[sup]-[/sup] combine to form NaOH.
B. Very Dilute Solutions
Electrolysis of Dilute H2SO4
What happens:
Ions Present: H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
- OH- loses electrons at anode to become O[sub]2[/sub] and H[sub]2[/sub]O.
- 4OH[sup]-[/sup] (aq) --> O[sub]2[/sub] (g) + 2H[sub]2[/sub]O (l) + 4e[sup]-[/sup]
- H[sup]+[/sup] gains electrons at cathode to become H atoms becoming hydrogen gas.
- 2H[sup]+[/sup](aq) + 2e[sup]-[/sup] --> H[sub]2[/sub] (g)
- Both equations must be balanced first.
- The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
- (2H[sup]+[/sup](aq) + 2e[sup]-[/sup] --> H[sub]2[/sub] (g)) x 2 = 4H[sup] +[/sup](aq) + 4e[sup]- [/sup]--> 2H[sub]2 [/sub](g)
- (2H[sup]+[/sup](aq) + 2e[sup]-[/sup] --> H[sub]2[/sub] (g)) x 2 = 4H[sup] +[/sup](aq) + 4e[sup]- [/sup]--> 2H[sub]2 [/sub](g)
- Now we can combine the equations, forming:
- 4H[sup]+[/sup] (aq) + 4OH[sup]+ [/sup](aq) --> 2H[sub]2 [/sub](g) + O[sub]2 [/sub](g) + 2H[sub]2[/sub]O (l)
- 4H[sup]+[/sup] (aq) + 4OH[sup]+ [/sup](aq) --> 2H[sub]2 [/sub](g) + O[sub]2 [/sub](g) + 2H[sub]2[/sub]O (l)
- 4H[sup]+[/sup] and 4OH[sup]+[/sup] ions, however, combine to form 4H[sub]2[/sub]O molecules.
- Hence: 4H[sub]2[/sub]O (l) --> 2H[sub]2[/sub] (g) + O[sub]2 [/sub](g)+ 2H[sub]2[/sub]O (l)
- Hence: 4H[sub]2[/sub]O (l) --> 2H[sub]2[/sub] (g) + O[sub]2 [/sub](g)+ 2H[sub]2[/sub]O (l)
- H[sub]2[/sub]O molecules are formed on both sides.
- Therefore, they cancel the coefficients: 2H[sub]2[/sub]O (l) --> 2H[sub]2 [/sub](g) + O[sub]2 [/sub](g)
- Therefore, they cancel the coefficients: 2H[sub]2[/sub]O (l) --> 2H[sub]2 [/sub](g) + O[sub]2 [/sub](g)
- Since only water is electrolysed, the sulfuric acid now only becomes concentrated.
- Inert Electrodes are electrodes which do not react with electrolyte or products during electrolysis.
- Eg. platinum and graphite.
- Eg. platinum and graphite.
- Active Electrodes are electrodes which react with products of electrolysis, affecting the course of electrolysis.
- Eg. copper.
- Eg. copper.
What happens:
Ions Present: Cu[sup]2+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
- OH[sup]-[/sup] loses electrons at anode to become O[sub]2[/sub] and H[sub]2[/sub]O.
- 4OH[sup]-[/sup] (aq) --> O[sub]2 [/sub](g) + 2H[sub]2[/sub]O (l) +4e[sup]-[/sup]
- Cu[sup]2+[/sup] gains electrons at cathode to become Cu atoms becoming liquid copper.
- Hydrogen ions are not discharged because copper is easier to discharge.
- Cu[sup]2+[/sup] (aq) + 2e[sup]-[/sup] --> Cu (s)
- Both equations must be balanced first.
- The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
- (Cu[sup]2+ [/sup](aq) + 2e[sup]-[/sup] --> Cu (s)) x 2 = 2Cu[sup]2+ [/sup](aq) + 4e[sup]-[/sup] --> 2Cu (s)
- (Cu[sup]2+ [/sup](aq) + 2e[sup]-[/sup] --> Cu (s)) x 2 = 2Cu[sup]2+ [/sup](aq) + 4e[sup]-[/sup] --> 2Cu (s)
- Now we can combine the equations, forming:
- 2Cu(OH)[sub]2 [/sub](aq) --> 2Cu (s) + O[sub]2 [/sub](g) + 2H[sub]2[/sub]O (l)
- 2Cu(OH)[sub]2 [/sub](aq) --> 2Cu (s) + O[sub]2 [/sub](g) + 2H[sub]2[/sub]O (l)
- Since copper ions in solution are used up, the blue colour fades.
- Hydrogen and sulphate ions left forms sulphuric acid.
Ions Present: Cu[sup]2+[/sup], H[sup]+[/sup], OH[sup]-[/sup] and SO[sub]4[/sub][sup]2-[/sup]
Reaction at Anode
- Both SO[sub]4[/sub][sup]2- [/sup]and OH[sup]-[/sup] gets attracted here but not discharged. Instead, the copper anode discharged by losing electrons to form Cu[sup]2+[/sup]. So, the electrode size decreases.
- Cu (s) --> Cu[sup]2+ [/sup](aq) + 2e[sup]-[/sup]
- Cu[sup]2+[/sup] produced from anode gains electrons at cathode to become Cu atoms becoming copper. Hence, the copper is deposited here and the electrode grows.
- Cu[sup]2+ [/sup](aq) + 2e[sup]-[/sup] --> Cu (s)
- There is no change in solution contents as for every lost of Cu[sup]2+[/sup] ions at cathode is replaced by Cu[sup]2+[/sup]ions released by dissolving anode.
- Only the cathode increases size by gaining copper and anode decreases size by losing copper.
- We can use this method to create pure copper on cathode by using pure copper on cathode and impure copper on anode.
- Impurities of anode falls under it.