Electrolysis — TRIPLE CONTENT

What is electrolysis? Why did electrolysis not exist in for example 1760?

Electrolysis is a chemical change that breaks down ionic compounds into elements which are molten or in aqueous solution by an electric current being passed through.

Electricity did not exist

Why do covalent compounds not conduct electricity?

Covalent compounds consist of molecules that are uncharged, they contain no mobile charged particles (either ions or delocalised electrons) to carry a current.

How do the ionic compounds need to be in order to carry out electrolysis? Explain.

• Molten or in aqueous solution
• because they can't conduct electricity when solid as the ions can't move freely because they are in a fixed positions in a lattice.

TERMS: electrolytes, electrode

Electrolyte: the liquid/solution that can conduct an electric current because of ionic compounds melted or dissolved in it (ions are free to move)
Electrode: a rod of metal through which an electric current flows into or out of an electrolyte. It is the location where oxidation-reduction happens. Fairly inert.

What substances are electrodes made of + examples? Why?

Inert substances such as graphite or platinum and can conduct electricity.
In hopes that they won't react with the electrolyte or products (esto es gracias a su extraction process)

What are the positive and negative electrodes called? And the ions?

PANIC (Positive Anode Negative Is Cathode)

Anion-negative
Cation-positive (cat-ion paw-sitive)

Electrolysis of molten compounds:
What are binary ionic compounds? Example? Why is the example melted for electrolysis?

These are compounds consisting of just 2 elements joined together.
Lead(II)bromide.
Because lead(II) bromide is insoluble (insoluble salt) so that there are free to move ions.

How can you predict the products of any binary molten compound? What happens during the process of electrolysis?

Identify ions present:
The positive ion (cation) will migrate towards the cathode and the negative ion (anion) towards the anode.
Therefore the cathode product will always be a metal and the product formed e eethe anode will always be a non-metal.

Explain why copper(II) is formed at the cathode during the electrolysis of its salts (2)

Copper = Cu2+

• Because the ions in copper would be positive
• therefore it would be attracted to the negative cathode (opposites attract).
• There it gains electrons to form copper (Cu) atoms.

Electrolysis of molten lead(II) bromide:
What is the method?

1) Add lead(II) bromide into a crucible (un basin que soporta altas temp) and heat it so that it will turn molten, allowing ions to be free to move and conduct electricity.

2) Add two graphite rods as the electrodes and using wires (so electrons can flow) connect this to a power pack or battery (to drive the flow of electrons)

3) Turn on the power pack/battery and allow electrolysis to take place

Electrolysis of molten lead(II) bromide:
Explain happens at the anode + observations (3) (+ what is it?)

• Negative bromide ions are attracted to the positive anode

• Where they undergo oxidation so they gain one electron to form a bromine atom. These pair up to form bromine molecules (diatomic)

• Brown bubbling at the anode. It is brown bromine gas being given off

Electrolysis of molten lead(II) bromide:
Explain what happens at the cathode + observations (3) (+ what is it?)

• Positive lead ions are attracted to the negative cathode.
• At the cathode, lead ions undergo reduction where they gain 2 electrons to form lead ATOMS.
• Layer of grey solid deposits on the bottom of the anode. It is grey lead metal.

What would be the half ionic equation for lead(II) bromide?

Bromine: 2Br- → Br2 + 2e- (oxidation)
Lead: Pb2+ + 2e- → Pb (reduction)

The lead bromide is heated until it melts. When the lead bromide melts, the lamp lights.
Two possible questions (1):

a) The teacher stops heating the mixture and allows it to solidify.
Suggest why the lamp stays alight.

b) Suggest why the lamp stops working after the teacher stops heating the lead(II) bromide.

a) - It completes the circuit as it is connected to the electrode (?)

b) - Ions stop moving

Basically, using the power of the batteries the electrons from the bromide ions are being passed to the positive anode and transported through the wires to the cathode. These electrons are given to lead ions.

So we are using ELECTRICITY.

Explain in terms of electrons what happens in both the anode and the cathode.

At the cathode each lead ion gains 2 electrons from 2 bromide ions.
At the anode each bromide ion loses 1 electron.

Explanation:

• As soon as you connect power source, it pumps any mobile electrons away from the left hand electrode to right hand one.
• This means extra electrons in right hand so it becomes negative (it has become the cathode) and left hand positive (becomes anode) and attracting its opposite charged ions
• When lead(II) Pb2+ gets to the cathode, it gains 2 electrons and forms neutral lead atoms and fall to the bottom of container as molten lead.
• When bromide gets to anode, bromide discharges 1 extra electron onto the anode and converts bromide ion into bromine atom

What does the power source do?

Pumps new electrons along the wire to replace the electrons that have been removed from cathode (by lead ions)

Electrolysis of aqueous solutions:
Aqueous solutions with water (H2O) as solvent. During electrolysis of aqueous solutions, what do the water molecules produce? What does this mean in relation to the electrolyte?

H+ and OH- ions
The electrolyte will contain ions from the compound PLUS ions from the water.

Water is a weak electrolyte. It ionises very slightly to give hydrogen ions and hydroxide ions

Electrolysis of aqueous solutions:
At the anode…

• Anions (non-metal) and OH- ions are attracted to the negative anode but only one will lose electrons.

1) If a halide is present (an ion from group 7/halogen ion) then it will form at the anode. The halide then loses one electron (undergoes oxidation) becoming a (diatomic) halogen molecule/element.

2) If no halide is present but water is, oxygen gas (bubbles) is formed at the anode. OH- (no preguntes adonde se va el hydrogen, simplemente se va) is attracted to the anode and undergoes oxidation, losing electrons nd forms (diatomic) oxygen (O2).

3) An other negative ions will form if there is ni halide or water.

4(?) If dilute OH- oxidised
If concentrated: Cl- oxidised.
EXCEPT SO4 2-/NO3 - WHICH NEVER (creo) DISCHARGE

• In first 2 cases the other negative ion remains in solution.
  First case is the OH- ions and in second case other ion such as, e.g sulfur (S 2-)

Why don't sulfates (SO4 2-) and nitrates (NO3 -) discharge during electrolysis?

Both these ions are very stable so it is difficult to oxidise compared to other ions.

How does concentration affect electrolysis?

A solution of e.g sodium chloride, at the anode chloride ions will get oxidised because of the rule.

However if it gets very dilute, it is easier for OH- to get discharged. So O2 is produced instead of Cl2 gas.

Electrolysis of aqueous solutions:
At the cathode (what is attracted? What will be formed either___ or ___? What is the order which dictate which one will undergo reduction?)

• Cations and H+ ions are attracted to the cathode but only one will gain electrons.

• Either hydrogen gas (bubbles) or the metal will be produced (solid deposit forms at the cathode)

1) If the metal is less reactive than hydrogen it will form on the electrode. These metals are: copper (Cu), gold (Au), silver (Au) and platinum (Pt).

2) If the metal is more reactive than hydrogen, then hydrogen will form (bubbles are seen).

• This is because the more reactive ions will remain in solution.

Suggest why the solution contains some OH- or/and H+ ions.

Because these ions come from the water/are present in water.
It is in an aqueous solution of water.

State the half ionic equation for zinc chloride (ZnCl2), magnesium oxide (MgO) and lithium oxide (Li2O).
Include which electrode it is and if it's oxidation or reduction.

ZnCl2
At the cathode (reduction): Zn 2+ + 2e- → Zn
At the anode (oxidation): 2Cl- → Cl2 + 2e-

MgO
At the cathode (reduction): Mg 2+ + 2e- → Mg
At the anode (oxidation): 2O 2- → O2 + 4e-

Li2O
At the cathode (reduction): Li+ + e- → Li (lithium is NOT diatomic)
At the anode (oxidation): 2O 2- → O2 + 4e-

If OH- is discharged, what is its half equation? Where? FROM MEMORY

At the anode (oxidation): 4OH- → O2 + 2H2O + 4e-

Produces both oxygen AND water.

What are the products formed at each electrode for sodium chloride? + Half equations + observations

Ions present: H+, OH-, Na+, Cl-

At cathode: Hydrogen (bubbles, test for hydrogen)
2H+ (aq) + 2e- → H2 (g)

At anode: Chlorine (green bubbles, test for chlorine)
2Cl- (aq) → Cl2 (g) + 2e-

Sodium hydroxide solution is the product remaining in the electrolysis container.

Why will the amount of chlorine gas produced be lower than the expected?

Chlorine is more soluble in water (dissolves in it) or it may react with water.

What are the products formed at each electrode for dilute sulfuric acid?

Ions present: H+, OH-, H+, SO4 2-
Sulfuric acid = H2SO4

At cathode: Hydrogen (bubbles, test for hydrogen)
2H+ (aq) + 2e- → H2 (g)

At anode: Oxygen (bubbles, test for oxygen)
SULFATES NEVER DISCHARGE so
4OH- → O2 + 2H2O + 4e-

What can we notice about the quantity formed of hydrogen and oxygen?

If we were to do the electrolysis of sulfuric acid using inert electrodes, how what would be the amount of hydrogen and oxygen different from the theory? Why? How might we avoid this?

Twice at much hydrogen is produced as oxygen.

If we were to do this the amount of hydrogen would be more than twice oxygen's because oxygen is more soluble in water than hydrogen.
To avoid this, we would carry out the electrolysis experiment for a few minutes first in order to saturate the water with O2 and then start to tolled the gases.

What are the products formed at each electrode for copper(II) sulfate?

Ions present: H+, OH-, Cu 2+, SO4 2+

At cathode: Copper (pink-brown solid formed) NOT orange.
Cu 2+ (aq) + 2e- → Cu

At anode: Oxygen (bubbles, test for oxygen)
4OH- → O2 + 2H2O + 4e-

Explain why copper(II) sulfate solution becomes paler blue during the electrolysis (2).

• The blue colour is caused by copper IONS (Cu 2+)
• During electrolysis, copper ions Cu 2+ are being discharged/removed from the solution/concentration of copper ions decreases/ copper ions form copper.

During electrolysis of copper(II) sulfate, a lightbulb is connected to the circuit. Why do

The copper(II) sulfate solution completes the circuit

What are the tests for: oxygen, hydrogen, chlorine, ammonia (NH3) and carbon dioxide?

Oxygen: relight a glowing splint.

Hydrogen: a lit splint is dipped into sample of gas, a squeaky pop will be heard.

Chlorine: if damp litmus paper is dipped into a sample of gas, it will turn red and then bleach to a white colour.

Ammonia: use damp red litmus paper, if positive turns blue.

Carbon dioxide: bubble the gas through limewater, if positive limewater turns cloudy.

Method for electrolysis of e.g copper(II) sulfate.

• Pour copper(II) sulfate solution into a beaker.
• Place two graphite/platinum rods into the copper sulfate solution.
• Attach electrodes to the power supply (e.g battery) using wires. One at the positive and the other one at the negative terminal.
• Turn on power supply.
• Test any gas produced.
• Record observations of what happens at each electrode.