chapter i - electrolysis

Section 1: Principles of ChemistryChapter I: Electrolysis


Electrolysis: Electrolysis is the process of decomposing an ionic compound into its consistuent elements through the conduction of electricity through ionic compound, when molten or dissolved in water.

Electrode: A rod or plate which carries electricity in or out of an electrolyte during electrolysis.

Electrons: Negatively charged sub-atomic particles that surround the nucleus of an atom.

Ion: A positively or negatively charged particle. It is formed when an atom or group of atom loses or gains electrons.

Anion: A negatively charged ion which moves towards the anode during electrolysis.

Anode: A positively charged electrode in an electrolytic cell.

Cation: A positively charged ion which moves towards the cathode during electrolysis.

Cathode: A negatively charged electrode in an electrolytic cell.

Electroplating: The process of depositing a layer of metal on another substance using electrolysis is called electroplating.

Electrolyte: An electrolyte is a substance that undergoes electrolysis. Electrolytes all contain ions. The movement of ions is responsible for both the conduction of electricity and the chemical changes that take place.

Non-electrolytes : Substances that do not conduct electricity under any condition are called non-electrolytes. Examples are Sulfur, sugar, pure water and organic compounds such as ethanol.

Current:

An electric current is a flow of electrons. For a substance to conduct electricity it must have mobile charged particles (electrons or ions). For instance metals can conduct electricity (a current passes through them) because metals have delocalized electrons.

Khadiza Karim Chowdhury Rodella Edexcel IGCSE in Chemistry

1.47 understand an electric current as a flow of electrons or ions1.48 understand why covalent compounds do not conduct electricity1.49 understand why ionic compounds conduct electricity only when molten or in solution1.50 describe simple experiments to distinguish between electrolytes and non-electrolytes1.51 recall that electrolysis involves the formation of new substances when ionic compounds conduct electricity1.52 describe simple experiments for the electrolysis, using inert electrodes, of molten salts such as lead(II) bromide1.53 describe simple experiments for the electrolysis, using inert electrodes, of aqueous solutions of sodium chloride, copper(II) sulphate and dilute sulphuric acid and predict the products1.54 write ionic half-equations representing the reactions at the electrodes during electrolysis1.55 recall that one faraday represents one mole of electrons1.56 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions


Q. Why covalent compounds do not conduct electricity?

Ans: Most covalent compounds do not conduct electricity whether in solid, liquid or gaseous state. This is because they do not have free-moving ions or electrons to conduct electricity. There are exceptions. Carbon, in the form of graphite, conducts electricity. Hydrogen Chloride, sulfur dioxide and ammonia react with water to form solutions that conduct electricity.

Q. Why ionic compounds conduct electricity only when molten or in solution?

Ans: When an aqueous solution of an ionic compound is electrolyte, a metal or hydrogen at the cathode. At the anode, non-metal such as oxygen or a halogen is given off.

A solid ionic compound will not conduct electricity. But when it melts the lattice breaks up and the ions are free to move. Since they are charged, this means they can conduct electricity. The ions also move freely when the compound is dissolved in water. So the aqueous solution of an ionic compound conducts too.

Ionic compounds do not conduct electricity in solid state because the ions are not free to move about. However, when an ionic compound is melted or dissolved in water to form an aqueous solution, it can conduct electricity. This is because the ions are free to move in the molten state or in aqueous solution.

A liquid (molten or dissolved in water) which conducts electricity is called an electrolyte.



Experiment to distinguish between electrolytes and non-electrolytes:

We can tests substances if they are electrolytes or non-electrolytes (liquids which do not conduct electricity) by doing simple tests.

A circuit is set up as shown below. It consists of a battery or power supply, wires, carbon electrodes and an indicator of a current i.e. a bulb or an ammeter. Carbon is picked as a material for the electrodes because it is inert i.e. it does not react and does therefore not interfere in the electrolysis reaction. The electrode which is connected to the positive terminal of the battery is named the anode; the electrode connected to the negative one is the cathode.

The substance to be tested is either molten or dissolved in water. If the bulb lights or a current is measured by an ammeter, the substance tested is an electrolyte. In the diagram below the compound on the left is ionic whilst the one on the right is covalent.


molten or dissolved ionic compound

(electrolyte) Covalent compound


Electrolysis:

Electrolysis is the process of using electricity to break down or decompose a compound. The compound is usually molten or dissolved in water. Electrolysis is important for extracting useful pure elements from compounds.

Electrolysis takes place in an electrolytic cell, as shown in figure below. The electrolytic cell works like an electrical circuit and has three main components – a battery, electrodes and an electrolyte.

Figure of an electrolytic cell.

How does an electrolytic cell work?When the circuit is complete in an electrolytic cell, the battery acts as an ‘electron pump’. It draws

electrons away from the anode, which becomes positively charged. These electrons are supplied to the cathode, which becomes negatively charged.

Therefore, electrons enter the positive terminal of the battery and are ‘pumped out’ at the negative terminal.

What happens when electricity is passed through an electrolyte?When electricity is passed through an electrolyte, chemical reactions take place at the electrodes, and

the electrolyte is decomposed. Reactions taking place at the electrodes are called electrolytic reactions.

How can electrolysis account for the existence of mobile ions in molten or aqueous ionic compounds? Electrolytes only conduct electricity because they contain charged particles (i.e. Ions) that are mobile.

Since molten or aqueous ionic compounds can act as electrolytes, they must contain mobile ions.

In the solid state, ionic compounds do not conduct electricity and thus cannot act as electrolytes. This is evidence that the positive ions and negative ions in solid ionic structures are held is a fixed lattice. They are unable to move freely and thus unable to conduct electricity.


Electrolyte:

A molten ionic compound or an aqueous solution that conducts electricity.

Dissociates to form positive ions (cations) and negative ions (anions).

Ions present in the electrolyte allow electricity to flow through it.

Examples are dilute sulfuric acid, molten sodium chloride and copper (II) sulfate solution.

Electrodes:

Conduct electric current. Usually carbon (i.e.

graphite) rods, or metal plates.

Electrode that is connected to the positive terminal of the battery is called the anode.

Electrode that is connected to the negative terminal of the battery is called the cathode.


What are the differences between how elements (metals and graphite) and compounds (electrolytes) conduct electricity?

Metals, carbon in form of graphite and ionic compounds are all conductors of electricity. Both metals and graphite are known as electrical conductors whereas ionic compounds are called electrolytic conductors. Table below shows the two major differences between electrical and electrolytic conduction.

Electrical conduction by metals and graphite

Electrical conduction by electrolytes

Method of Conduction Electricity is conducted by the flow of electrons from one end of the conductor to the other end.

Electricity is conducted by the movement of positive ions and negative ions across the (molten or aqueous) electrolyte.

Effect of Conduction Metals and graphite remain unchanged chemically when an electric current flows through them.

The electrolytes are decomposed to form new substances when they conduct electricity.

An electrolyte is a different type of conductor than a solid metal. When it conducts, the current passing through the electrolyte causes a chemical reaction to take place i.e. new substances are formed. The ionic compound which makes up the electrolyte is decomposed. A reaction in which a compound is decomposed using an electrical current is called an electrolysis reaction.

We need to be able to predict the products of electrolysis reactions. To be able to do this we need to know what the ionic compound is but we also need to know if it is molten or dissolved in water.

Some important backgrounds:

The conduction of electricity by metals and carbon:

In metal or carbon, electricity is simply a flow of electrons. The movement of electrons doesn’t produce any chemical change in the metal or carbon.

Metals and carbon contain mobile electrons, and it is these that move. That’s equally true even for liquid metal such as mercury. In electrical circuits, we can think of a battery or a power pack as an ‘electron pump’, pushing the electrons through various bits of metal or carbon.

Passing electricity through compounds – electrolysis:

Hardly any solid compounds conduct electricity. On the other hand, a lot of compounds will conduct electricity when they are dissolved in water. All of these show signs of chemical reaction while they are conducting.



Explaining Electrolysis:

The process of electrolysis involves three things: the external circuit, the reaction within the electrolyte, and the reactions at the surface of the electrodes.

Fig – Movement of ions and electrons during electrolysis.

Electrical energy is converted to chemical energy as chemical reactions (redox reactions) take place at the electrodes.

What happens to the ions at the electrodes?As shown in the figure above, cations (positive ions) receive electrons from the cathode while anions (negative ions) give up electrons to the anode. The process of gaining or losing electrons at the electrodes is called discharge. When ions are discharged at electrodes, they form atoms or molecules.


In the external circuit: During electrolysis, electrons flow from the positive terminal to the negative terminal of the battery.

At the anode: During electrolysis,

Anions (negative ions) move to the anode.

Anions give up electrons at the anode.

Oxidation occurs at anode.

At the Cathode: During electrolysis,

Cations (positive ions) move to the cathode.

Cations receive electrons from the cathode.

Reduction occurs at the cathode.

Within the electrolyte: During electrolysis,

Cations move toward the cathode.

Anions move toward the anode. The flow of ions towards the

electrodes constitutes the flow of the electric current through the electrolyte.


Electrolysis of molten ionic compounds or salts

When a molten salt is electrolyzed, the positive metal ions are attracted to the negative electrode or cathode where they lose their charge as they gain electrons (reduction) and become metal atoms. The negative non metal ions are attracted to the anode where they lose their negative charge. They lose electrons (oxidation) and become non metal atoms again and usually form molecules.

This is shown by the example below which shows the electrolysis of molten lead bromide. The products are bromine and lead.

The actual changes going on at each electrode can be shown in ionic half equations:

The flow of charge is maintained in the following way:

by the flow of electrons in the wires and the electrodes and

By the movement of ions towards the oppositely charged electrode in the electrolyte.

When a molten salt is electrolyzed the metal is produced at the cathode whilst the non-metal is formed at the anode.

Some ionic compounds are binary compounds. A binary compound is a compound containing only two elements. A binary compound often contains a metal Cation and a non-metal anion. When such a molten binary compound undergoes electrolysis, a metal and a non-metal are formed.

Electrolysis of molten Lead (II) Bromide, PbBr2


Half equation at the cathode:

Pb2+(l) + 2e- Pb (s)

Half equation at the anode:

2Br-(l) Br2 (l) + 2e-


Nothing at all happens until the lead (II) bromide melts. Then:

The bulb lights up, showing that electrons are flowing through it. There is bubbling around the electrode (the anode) connected to the positive terminal of the

power source as brown bromine gas is given off. Nothing seems to be happening at the electrode (the cathode) connected to the negative

terminal of the power source, but afterwards metallic lead is found underneath. When you stop heating and the lead (II) bromide solidifies again, everything stops – there is no

more bubbling and the bulb goes out.

Explanation:

Lead (ii) bromide is an ionic compound. The solid consists of a giant structure of lead (II) ions and bromide ions packed regularly in a crystal lattice. It doesn’t have any mobile electrons, and the ions are locked tightly in the lattice and aren’t free to move. The solid lead (II) bromide doesn’t conduct electricity.

As soon as the solid melts, the ions become free to move around, and it is this movement that enables the electrons to flow in the external circuit. This is how it works…

As soon as you connect the power source, it pumps any mobile electrons away from the left-hand electrode towards the right-hand one. At the moment the lead (II) bromide is still solid.

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. There is a limit to how many extra electrons the ‘pump’ can squeeze into the negative electrode because of the repulsion by the electrons already there.

Things change when the lead (II) bromide melts, and the ions become free to move.



The positive lead (ii) ions are attracted to the cathode. When they get there, each lead (II) ion picks up two electrons from the electrode and forms neutral lead atoms. These fall to the bottom of the container as molten lead.

Pb2+ + 2e- Pb (l)This leaves spaces in the electrode that more electrons can move into. The power source pumps new electrons along the wire to fill those spaces.

Bromide ions are attracted to the positive anode. When they get there, the extra electron which makes the bromide atom negatively charged moves onto the electrode because this electrode is short of electrons.

The loss of extra electron turns each bromide on into a bromine atom. These join in pairs to make bromine molecules. Overall:

2Br- Br2 + 2e-

The new electrons on the electrode are pumped away by the power source to help fill the spaces being created at the cathode. Because electrons are flowing in the external circuit the bulb lights up.

Electrons can flow in the external circuit because of the chemical changes to the ions arriving at the electrodes. We say that ions are discharged at the electrodes. Discharging an ion simply means that it loses its charge – either giving up electron(s) to the electrode or receiving electron(s) from it.

Electrolysis and redox:

If you look at the electrode equations above, you will see that the lead (II) ions gain electrons at the cathode. Gain of electrons is reduction. The lead(II) ions are reduced to lead atoms.

The bromide ions lose electrons at the anode. Loss of electrons is oxidation. Bromide ions oxidized to bromine molecules.

The electrolysis of other molten substances:

In each case, the positive ions are attracted to the negative cathode, where they are discharged by gaining electrons. Positive ins are known as cations because they are attracted to the cathode. The



negative ions move to the anode, where they are discharged by giving electrons to the electrode. Negative ions are known as anions.

Not all ionic compounds can be electrolysed molten. Some break up into simpler substances before their melting point. For example, copper(II) carbonate breaks into copper(II) oxide and carbon dioxide, even on gentle heating. It is impossible to melt it.

Other ionic compounds have such high melting points that it isn’t possible to melt them in lab, although it can be done industrially. For example, it is difficult to keep sodium chloride molten in lab because its melting point is 801 oC. If you could keep it molten you would get sodium at the cathode and chlorine at the anode. Sodium is manufactured by electrolysing molten sodium chloride.

Electrolysis of molten sodium chlorideElectrolyte: Molten sodium chlorideElectrode: Carbon anode and cathode


At the cathode: Na+ (l) + e- Na (l) (molten sodium chloride is produced)

At the anode: 2Cl- (l) Cl2(g) + 2e- (Chlorine gas is produced)


Ions in the electrolyte: sodium ions, Na+ and chloride ions, Cl-.

Anode reactions:1. Chloride ions migrate to the anode and are discharged.2. Each chloride ion loses 1 electron and a chlorine atom is formed through oxidation.3. Finally, greenish-yellow chlorine gas is liberated.

2Cl-(l) Cl2 (g) + 2e-

Cathode reactions:1. Sodium ions will migrate to the cathode and are discharged,2. Each sodium ion gains 1 electron and forms sodium atoms through reduction reaction.

Na+ (l) + e- Na(l)3. Shiny, silvery deposits of sodium metal are formed in the crucible.

Overall reaction:2 moles of sodium and 1 mole of chlorine gas are obtained from 2 moles of sodium chloride.

2NaCl (l) 2Na(l) + Cl2(g)

Electrolysis of molten magnesium oxideElectrolyte: Molten magnesium chloride.Electrode: Carbon anode and cathode

Ions in the electrolyte: magnesium ions, Mg2+ and oxide ions, O2-

Anode reactions:1. Oxide ions migrate to the anode and are discharged.2. Each oxide ion loses 2 electrons and a oxygen atom is formed through oxidation.3. Finally, colourless, odourless oxygen gas is liberated.

2O2-(l) O2(g) + 4e-

Cathode reactions:

1. Magnesium ions will migrate to the cathode and are discharged,.2. Each magnesium ion gains 2 electrons and forms magnesium atoms through reduction reaction.3. Shiny, silvery deposits of magnesium metal are formed in the crucible.

Mg2+(l) + 2e- Mg(l)

Overall reaction:2 moles of magnesium and 1 mole of oxygen gas are obtained from 2 moles of magnesium oxide.

2MgO(l) 2Mg(l) + O2(g)

Electrolysis of Aqueous Electrolytes:

Aqueous electrolytes are mixtures of two electrolytes – the compound and water.



Water ionises, to a small extent, to give H+ ions and OH- ions as follows:

H2O (l) H+ (aq) + OH- (aq)

As more than 1 type of anions and cations are present, preferential discharge is carried out based on the following criteria:

Position of ion in the electrochemical series Concentration of ion Nature of electrolyte.

Fig. Position of ion in the electrochemical series:

CATIONS ANIONSK+ (aq)Na+ (aq)Ca2+ (aq)Mg2+ (aq) SO4

2- (aq)Zn2+ (aq) NO3

- (aq) Fe2+ (aq) Cl- (aq)Pb2+ (aq) Br- (aq)H+ (aq) I- (aq)Cu2+ (aq) OH- (aq)Ag+ (aq)Note: sulfate ions and nitrate ions will never be discharged during electrolysis.

The electrolysis of aqueous solutions

The electrolysis of sodium chloride solution:

If we electrolyse sodium chloride solution will not get the same products as if we electrolysed it molten. Although chlorine is formed at the anode, hydrogen is produced at the cathode rather than sodium. The hydrogen at the cathode is coming from water.

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

H2O (l) H+ (aq) + OH- (aq)

Whenever we have water present, we have to consider theses ions as well as the ions in the compound we are electrolysing.

A simple apparatus for electrolysing solutions:



At the cathode:The solution contains Na+ (aq) and H+ (aq), and these are both attracted to the negative cathode. The H+ (aq) gets discharged because it is much easier to persuade a hydrogen ion to accept an electron than it is a sodium ion. Each hydrogen atom formed combines with another one to make a hydrogen molecule.

2H+ (aq) + 2e- H2 (g)

The hydrogen ions come from water molecules splitting up. Each time a water molecule ionises, it also produces a hydroxide in. There is a build up of these in the solution around the cathode.

2H2O (l) + 2e- H2 (g) + 2OH- (aq)

These hydroxide ions make the solution strongly alkaline in the region around the cathode. Because of the presence of the sodium ions attracted to the cathode, you can think of the electrolysis as also forming sodium hydroxide solution.

At the Anode:Cl- (aq) and OH- (aq) are both attracted by the positive anode. The hydroxide ion is slightly easier to discharge than the chloride ion is, but there isn’t that much difference. There are far, far more chloride ions present in the solution, and so it is mainly these are discharged.

Electrolysis of dilute Sodium Chloride:

An aqueous solution of sodium chloride contains four different types of ions. They are:


H2(g)

H2O H+ (aq) + OH-

(After two hydrogen atoms have paired up)

These are discharged as hydrogen atoms.

There is a build-up of OH- ions in the solution around the cathode.

The water keeps on ionising.

2Cl-(aq) Cl2 +2e-


Ions from sodium chloride – Na+ (aq) and Cl- (aq) Ions from water – H+ (aq) and OH- (aq)

When dilute sodium chloride solution is electrolyzed using inter electrodes, the Na+ and H+ ions are attracted to the cathode. The Cl- and OH- ions are attracted to the anode.

The overall reaction shows that the electrolysis of dilute sodium chloride is equivalent to the electrolysis of water.

Electrolysis of sodium chloride solution


ElectrolysisMolten sodium chloride:Cathode : Na+ ions are discharged. Anode: Cl- dscharged.At the Cathode: The Na+ and H+ ions are attracted to the platinum cathode. H+ ions gain electrons from the cathode to form hydrogen gas.

2H+ (aq) + 2e- H2 (g)Na+ ions remain in the solution.

At the anode: Cl- and OH- ions are attracted to the platinum anode. OH- ions give up electrons to form water and hydrogen gas.

4OH- (aq) 2H2O (l) + O2 (g) + 4e-

Cl- ions remain in the solution.

Summary: The overall reaction is:

2H2O (l) → (electrolysis) 2H2 (g) + O2 (g)

Since water is being removed (by decomposition into hydrogen and oxygen), the concentration of sodium chloride solution increases gradually.


When a solution is electrolyzed predicting the products becomes more complicated as also the water (the solvent) also becomes involved. Water although a covalent compound ionizes a little bit producing hydrogen and hydroxide ions which are also attracted to the electrodes. In most cases a gas will be produced which needs to be collected and tested.

Starting chemicals

concentrated sodium chloride solution, NaCl (aq) ions present:

at cathode: Na+ H+ at the anode: Cl- OH- only 1 ion can react or be discharged at each electrode, the other remains in

solution conditions Electrolytic cell with carbon electrodes

observations anode bubbles, yellow green gas, nearly 1:1 ratio with gas at cathode test: gas bleaches damp litmus papercathode bubbles, colourless gas test: ‘pop’ sound with lit splint, more than 1:1 ratio red litmus turns blue

products anode chlorine is discharged and not the hydroxide ions ionic half equation: 2Cl- (aq) Cl2 (g) + 2e-

oxidation hydroxide ions remain in solutioncathode hydrogen is discharged:

o ionic half equation: 2H+(aq) + 2e- H2 (g)o reductiono Na+ remain in solution

sodium hydroxide: Na+ and OH- are left in the solution as sodium hydroxide which can be detected using an indicator

overall equation: at anode: 2Cl- (aq) Cl2 (g) + 2e-

at cathode: 2H+ (aq) + 2e- H2(g)

: 2H+(aq) +2Cl- (aq) Cl2 (g)+ H2 (g)

Trend: for any concentrated solution of a salt made from a group 1 metal and a group 7 non-metal there will always be three products: (1) hydrogen, (2) the halogen, and (3) the hydroxide of the group 1 metal.

The electrolysis of copper (II) sulfate solution using carbon electrodes:



The copper (II) ions and the hydrogen ions (from the water) will be attracted to the cathode. Copper is below hydrogen in the reactivity series, which means that its ion is easier to discharge. The carbon electrode will get coated with brown copper.

Cu2+ (aq) + 2e- Cu(s)

Sulfate ions and hydroxide ions (from the water) will be attracted to the anode. Sulfate ions are very stable and aren’t discharged. Instead, you get oxygen from the discharge of hydroxide ions from the water.

4OH-(a) 2H2O (l) + O2 (g) + 4e-

If we continue the electrolysis for a long time, the copper(II) ions will eventually all be used up, and so the colour of the solution will fade from blue to colourless.

What is left in the solution? Hydrogen ions from the water aren’t being discharged and neither are the sulfate ions. The solution turns into dilute sulphuric acid. The electrolysis will then continue for dilute sulfuric acid.

2H2O(l) O2(g) + 4H+(aq) +4e-

Electrolysis of Aqueous Copper (II) Sulfate Using Inert electrodes:



Copper (II) Sulfate solution can be electrolyzed using inert platinum electrodes.

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

Apparatus and chemicals

beaker (250 cm3) graphite electrodes (about 5 mm diameter 2 copper strips, retort stand and clamp to hold electrodes small pieces of emery paper

DC power supply light bulb (small, 6 volt, 5 watt) leads and crocodile clips aqueous copper(II) sulfate, 0.5 mol dm-3

balance, ± 0.01g

Electrolysis of copper (II) sulphate solution


Rule 1: An aqueous solution of copper (II) sulfate contains four types of ions:Ions from copper (II) sulfate – CuSO4 Cu2+ + So4

2- Ions from water – H2O H+ (aq) + OH- (aq)

Rule 2: At the anode.

OH- ions and SO42- ions are attracted

to the anode. OH- ions give up electrons more readily than SO4

2+ ions. Consequently, OH- ions are preferentially discharged to give oxygen gas.

4OH-(a) 2H2O (l) + O2 (g) + 4e-

The SO42- ions remain in the solution

Rule 3: At the cathode

H= ions and Cu2+ ions are attracted to the cathode. Copper is lower than hydrogen is reactivity series. Cu2+ ions accept electrons more readily than H=

ions. As a result, Cu2+ ions are preferentially discharged as copper metal (atoms).

Cu2+(aq) + 2e- Cu(s)

The H+ ions remain in the solution.

Rule 4: Summary

When aqueous copper (II) sulfate is electrolyzed using platinum electrodes, copper metal is deposited at the cathode and oxygen gas is given off at the anode. The overall reaction is:

2CuSO4 + 2H2O (electrolysis) -------> 2Cu(s) + O2 (g) + 2H2SO4 (aq)


starting chemicals

copper sulphate solution, CuSO4 (aq) ions present:

at cathode: Cu2+ H+ at the anode: SO4

2- OH-

conditions Electrolytic cell with carbon electrodes

products anode oxygen is produced ionic half equation: 4OH- (aq) O2 (g) + 2H2O (l) + 4e- (oxidation) sulphate, SO4

2- remains in solution

cathode copper

o is discharged and not the hydrogen ions as they remain in the solution: o ionic half equation: Cu2+ (aq) + 2e- Cu (s) o reduction

hydrogen remains in solution

Overall equation at anode: 4OH- (aq) O2 (g) + 2H2O (l) + 4e-

at cathode: 2Cu2+ (aq) + 4e- 2Cu (s)

4OH- (aq) + 2Cu2+ (aq) O2 (g) + 2H2O (l)+ 2Cu (s)

The electrolysis of dilute sulfuric acid using carbon electrodes:


Observations

anode bubbles, colourless gas test: relights glowing splint

cathodered deposit on electrodesolution

Whole solution becomes colourless


In this case, the only positive ions arriving at the cathode are hydrogen ions (from the acid and the water). These are discharged to give hydrogen gas.

2H-(aq) + 2e- H2(g)

At the anode – sulfate ions and hydroxide ions (from water) arrive. The sulfate ions are too stable to be discharged, and so we get oxygen from discharge of hydroxide ions from the water.

4OH-(a) 2H2O (l) + O2 (g) + 4e-

Twice as much as hydrogen is produce as oxygen.

Look at the equations above. For very four electrons that flow around the circuit. We get one molecule of oxygen. But four electrons will produce two molecules of hydrogen.

We get twice the number of molecules of hydrogen as of oxygen. Twice the numbers of molecules occupy twice the volume.

Electrolysis of dilute sulphuric acid



The two previous solutions are considered concentrated solutions. The electrolyte in this example is diluted and this also has an effect on the outcome as none of the ions of the electrolyte will be discharged instead they remain in the solution. The ions which are discharged are the hydrogen and hydroxide ions which form hydrogen gas and oxygen gas. In essence when a dilute solution is electrolyzed water is decomposed. The identity and nature of the electrolyte does not really matter.

Starting chemicals

dilute sulphuric acid, H2SO4 (aq) ions present:

at cathode: H+ at the anode: SO4

2- OH-

conditions Electrolytic cell with carbon electrodes

observations anode bubbles, colourless gas, volume approximately half of the gas produced at the cathode test: relights glowing splint

cathode bubbles, colourless gas volume: twice as much as gas at the anode test: ‘pop’ sound with lit splint,

products anode oxygen is produced ionic half equation: 4OH- (aq) O2 (g) + 2H2O (l) + 4e-

oxidation

cathode hydrogen is discharged: ionic half equation: 2H+(aq) + 2e- H2 (g) reduction

overall equation at anode: 4OH- (aq) O2 (g) + 2H2O (l) + 4e-

at cathode: 4H+ (aq) + 4e- 2H2 (g)

4OH- (aq) + 4H+ (aq) O2 (g) + 2H2O (l)+ 2H2 (g)

1. The decomposition of a compound by electricity is called electrolysis.2. Electrolysis takes place in an electrolytic cell. An electrolytic cell consists of battery,

electrodes and an electrolyte.



3. A cathode is the negatively charged electrode and the anode is the positively charged electrode.

4. An electrolyte is a compound that conducts electricity in the molten state or in aqueous solution.

5. An electrolyte conducts electricity because it dissociates into positive ions called cations and negative ions called anions.

6. During electrolysis, cations move towards the cathode while anions move towards the anode.

7. Redox reactions take place at the electrodes during electrolysis. Oxidation occurs at the anode and reduction occurs at the cathode.

8. Electrolytes are made up of: Molten or aqueous solutions of ionic compounds. Aqueous solutions of acids and alkalis.

9. Chemicals can be classified as: Non-electrolytes Weak electrolytes, or Strong electrolytes.

Non-electrolytes Weak Electrolytes Strong ElectrolytesOrganic liquids or solvents, like ethanol, tetrachloromethane, pure water, sugar and molten sulfur.

Include weak acids and alkalis like limewater, ethanoic acid, sulfurous acid and carbonic acid.

Strong acids, alkalis and salt solutions like sulfuric acid, aqueous sodium hydroxide, copper(II) sulfate and sodium chloride.

10. At the anode, the anions are discharged through loss of electrons. This means the anions undergoes oxidation.

11. At the cathode, the cations are discharged through the gain of electrons. Cations under goes reduction.

12. A simple binary compound contains only two elements – a metal and a non-metal.13. The metallic ions (Mn+) will discharge at the cathode to form a metal atom and the anion (Nn-)

will discharge at the anode to for, a non-metallic atom. Mn+ + ne- MNn- N + ne-

The electrolysis of other solutions using electrodes:How to work out what will happen:



If the metal is above hydrogen in the reactivity series, we get hydrogen produced instead of the metal.

If the metal is below hydrogen in the reactivity series, we get the metal produced. If we have reasonably concentrated solutions of halides (chlorides, bromides or iodides), we get

the halogen. With other common negative ions, we get oxygen.

This leaves the problem of what we get if we have a moderately reactive metal such as zinc, for example. Reasonably concentrated solutions will give us the metal. Very dilute solutions will give you mainly hydrogen, in between, we will get both.

In general, when an aqueous solution of an ionic compound is electrolysed, a metal or hydrogen is produced at the cathode. At the anode, a non-metal, oxygen or a halogen is given off.

CATHODE ANODEProduct Equation Product Equation

KI (aq) Hydrogen 2H+(aq) + 2e- H2 (g) iodine 2I- (aq) I2 aq) + 2e-

MgBr2 (aq) Hydrogen 2H+(aq) + 2e- H2 (g) bromine 2Br- (aq) Br2 aq) + 2e-

H2SO4 (aq) Hydrogen 2H+(aq) + 2e- H2 (g) oxygen 4OH- (aq) O2 (g) + 2H2O (l) + 4e-

CuSO4 (aq) Copper Cu2+(aq) + 2e- Cu(s) oxygen 4OH- (aq) O2 (g) + 2H2O (l) + 4e-

What would happen with non-electrolytes?

For electrolysis to work, there have to be ions present. The current in the external circuit with bulb and power source) can flow only if there are ions which can move and be discharged.

If we try to electrolyses can ionic compound (either molten or in solution), there wouldn’t be any current flow, because there aren’t any ions. Nothing else would happen either. Sugar, for example, is a non-electrolyte – it doesn’t undergo electrolysis. It won’t conduct electricity and won’t be decomposed by it– Either solution or molten. If something undergoes electrolysis, it must contain ions. If it doesn’t undergo electrolysis, it doesn’t contain ions.

There are cases, of course, where it s impossible to test the substance in this way. Copper can be electrolyses molten because it decomposes before melting. It isn’t soluble in water either. So we can’t test it, despite the fact it does contain ions.

Why is carbon rods used in electrolysis?



The halogens produced during electrolysis are very reactive. Inert electrodes such as carbon electrodes are used to prevent reactions from occurring between the halogen and the electrode.

Inert electrodes are electrodes that do not take part in any chemical reactions during electrolysis. They simply provide the surface for electron transfer to occur during electrolytic reactions. Inert electrodes are used to prevent reactions from occurring between the electrodes and the products of electrolysis. Carbon and platinum electrodes are considered to be inert electrodes because they are rarely involved n electrolytic reactions.

Reactivity series and selective discharge of ions

In electrolysis, when more than one type of cation and anion is present in a solution, only one cation and one anion are preferentially discharged. This is known as selective discharge of ions.

How do you predict which ions are discharged in the electrolysis of a compound in aqueous solution?

If inert electrodes are used during electrolysis, the ions discharged and hence are the products formed depend on three factors:

1. The position of the metal (producing the cation) in the reactivity series.2. The relative ease of discharge of an anion.3. The concentration of the anion in the electrolyte.

CATIONS ANIONSK+ (aq)

Na+ (aq)Ca2+ (aq)Mg2+ (aq) SO4

2- (aq)Zn2+ (aq) NO3

- (aq) Fe2+ (aq) Cl- (aq)Pb2+ (aq) Br- (aq)H+ (aq) I- (aq)Cu2+ (aq) OH- (aq)Ag+ (aq)Note: sulfate ions and nitrate ions will never be discharged during electrolysis.

Selective discharge of cations during electrolysis:

The cations of an element lower in the reactivity series are discharged at the cathode in preference to other cations in the solution. This is because cations of a less reactive element accept electrons more


Difficulty of discharge decreases down the series.

Ease of discharge increases.


readily. For example, if a solution containing Sodium ions and Hydrogen ions are electrolysed. Hydrogen ions are discharge in preference to sodium ions.

Selective discharge of anions during electrolysis:

Sulfate (SO42-) and nitrate (NO3

--) ions remain in the solution and are not discharged during electrolysis. If a solution containing SO4

2- , NO3- and OH- ions is electrolysed, the OH- ions will be discharged in

preference to SO42- and NO3

-. The OH- ions give up electrons most readily during electrolysis to form and water and oxygen.

4OH- (aq) O2 (g) + 2H2O (l) + 4e-

Effect of concentration on the selective discharge of anions:

An increase in the concentration of an anion tends to promote its discharge. For example, in electrolysis of concentrated sodium chloride solution, two types of ions are attracted to the anode: Cl - and OH- . According to their own relative ease of discharge, OH- should be discharged preferentially. However, in concentrated sodium chloride solution, Cl- ions are far more numerous than OH- ions and so are discharged at the anode instead.

2Cl- (l) Cl2(g) + 2e-

What are the general rules for predicting selective discharge?

The following rules can be applied when predicting the products of electrolysis of any aqueous solution using (inert electrode):

Rule 1 Identify the cations and anions in the electrolyte. Remember that an aqueous solution also contains H+ and OH- ions from the dissociation of water molecules.

Rule 2 At the anode, the product is always oxygen unless the electrolyte contains high concentration of the anions, Cl-, Br- and I- ions.

Rule 3 At the cathode, reactive metals such as sodium potassium are never produced during electrolysis of the aqueous solution. If the metal is above hydrogen in the reactivity series, we get hydrogen produced instead of the metal. If the metal is below hydrogen in the reactivity series, we get the metal produced.

Rule 4 Identify the cations and anions that remain the solution after electrolysing. They form the product remaining in solution after H+ and OH- ions have been discharged. Summarise the reaction.

For example, in the electrolysis of dilute sodium chloride solution, Na+ (aq) and Cl- (aq) ions remain in the solution after H+ (aq) and OH- (aq) ions have been discharged. Hence, the solution of sodium chloride becomes more concentrated after electrolysing.



Electrolysis of aqueous sodium hydroxide: Electrolyte: Aqueous sodium hydroxideNaOH Na+ + OH-

H2O H+ (aq) + OH- (aq)

Electrodes: carbon anode and cathodeIons in electrolyte: Na+, OH- & H+

Anode reactions: OH- migrate to the anode and are discharged. Each OH- ions loses 1 electron and is oxidised to oxygen gas and water.

4OH- (aq) O2 (g) + 2H2O (l) + 4e-

Cathode reactions: Na+ and H- ions migrate to the cathode and H+ ions are preferentially discharged. Each H+ ion gains electron and a neutral H atom is formed. Bubbles of hydrogen gas are eventually liberated.

2H+(aq) + 2e- H2 (g)

Overall reaction:Water is decomposed into hydrogen and oxygen.2H2O (l) O2 (g) + 4H+(aq) +4e-

Electrolysis of concentrated hydrochloric acid:

Electrolyte: concentrated hydrochloric acidHCl H+ + Cl-

H2O H+ (aq) + OH- (aq)

Anode reactions: OH- & Cl- migrate to the anode and Cl- ions are discharged as the concentration of is much higher. Each Cl- ion loses electron and a chlorine atom is formed. Bubbles of greenish yellow chlorine gas are eventually liberated.

2Cl- (l) Cl2(g) + 2e- Cathode reactions:

H- ions migrate to the cathode and H+ ions are preferentially discharged. Each H+ ion gains electron and a neutral H atom is formed. Bubbles of hydrogen gas are eventually liberated.

2H+ (aq) + 2e- H2 (g)

Overall reaction:Hydrogen chloride is decomposed into Hydrogen and chloride.2HCl (aq) H2 + O2

1 volume of hydrogen gas and 1 volume of chlorine gas are collected.



Electrolysis of concentrated sodium chloride:

Electrolyte: concentrated sodium chloride Ions from sodium chloride – Na+ (aq) and Cl- (aq) Ions from water – H+ (aq) and OH- (aq)

Electrodes: carbon anode and cathodeAnode reactions:

OH- & Cl- migrate to the anode and Cl- ions are discharged as the concentration of is much higher. Each Cl- ion loses electron and a chlorine atom is formed. Bubbles of greenish yellow chlorine gas are eventually liberated.

2Cl- (l) Cl2(g) + 2e-

Cathode reactions: H- ions migrate to the cathode and H+ ions are preferentially discharged. Each H+ ion gains electron and a neutral H atom is formed. Bubbles of hydrogen gas are eventually liberated.

2H+ (aq) + 2e- H2 (g)

Overall reactions:2NaCl (aq) + 2H2O(l) 2NaOH(aq) + H2(g)+ Cl2(g)1 volume of hydrogen and 1 volume of chlorine gas is collected.

Electrolysis of Water:

Pure water is poor conductor of electricity because it consists almost entirely of molecules and has very few ions in it. However, if a small amount of an ionic compound or dilute sulfuric acid is added to water, the solution now becomes a good conductor of electricity. The product of electrolysis of water is always 2 volumes of hydrogen at the cathode and 1 volume of oxygen at the anode.

Electrolysis is used in industry for:

Extraction of reactive metals, Manufacture of chemicals, and Electroplating of metals. Electrolytic Purification.

Electrolytic Purification:

In this age of information technology, there is a high demand for very pure metals in printed circuits and for other specialised uses. Electrolysis is used to purify metals, such as can electrolytic purification of copper.

The electrolytic purification of copper involves the electrolysis of copper (II) sulfate, but using copper electrodes.

Electrolysis of aqueous copper (II) sulfate using copper electrodes:



When copper (II) sulfate is electrolysed using inert electrodes, copper is formed at the cathode and oxygen at the anode. However, if copper electrodes are used, the electrodes may undergo oxidation in preference to other possible reactions. In this case, the electrodes are called reactive electrodes.

What happens when aqueous copper (II) sulfate is electrolysed using reactive electrodes?

When aqueous copper (II) sulfate is electrolysed using copper electrodes:

a) No gas is evolved at the anodeb) The copper anode slowly becomes smallerc) Copper is deposited on the cathode, andd) The blue colour of the aqueous solution remains unchanged.

How do you account for these observations?

Electrolysis of copper (II) sulfate solution using copper electrodes:

Electrolyte: copper (II) sulfate solution Ions from copper (II) sulfate – CuSO4

Cu2+ + SO42-

Ions from water – H2O H+ (aq) + OH- (aq)

Electrodes: Copper anode and cathode

Anode reactions: OH- and SO4

2- - migrate to the anode and neither are discharged. The copper electrodes lose electrons more readily than the anions.

Each copper atom loses 2 electrons to form Cu2+ ion. Eventually the copper anode dissolves.Cu(s) Cu2+ + 2e-

Cathode reactions: Cu2+ and H+ ions migrate to the cathode and Cu2+ ions are preferentially discharged as they are

lower than H+ in the reactivity series. Each Cu2+ ions gains 2 electrons and pink solid of copper metal is deposited on the cathode.

Cu2+ + 2e- --> Cu(s)

Overall reactions:Concentration of copper(II) sulfate remains unchanged as the copper from the anode is transferred to the cathode, copper(II) sulfate solution remains blue in colour.

What is the net result of electrolysis of aqueous copper (II) sulfate using copper electrodes?



The net result is copper is transferred from the anode to cathode. The copper cathode slowly increases in mass while the mass of the anode decreases. The colour and concentration of the copper (II) sulfate solution remains unchanged. This is because the copper ions that get discharged come mainly from the copper anode. There is no effective loss of copper ions from the copper (II) sulfate solution.

Purification of copper:to refine copper, impure copper is used as the anode of the electrolytic cell. The cathode is a thin sheet of pure copper and the electrolyte is aqueous copper (II) sulfate. During electrolysis the impure copper anode dissolves and impurities such as silver and platinum fall to the bottom of the cell. In a finely divided state, they from an anode slime. A layer of pure copper is deposited on the cathode.

Electroplating:

The process of depositing a layer of metal on another substance using electrolysis is called electroplating.

Uses of electroplating:

Electroplating is used to coat a metal object with another metal to get a good decorative finish and to prevent rusting. Food cans are electroplated with tin for these reasons. Silver-plating or gold-plating is used to coat a relatively cheap metal to make it look expensive. Chrome-plating is also used to beautify a metal object and to protect it from corroding. Electroplating is mainly used to:

Enhance a metal’s appearance Prevent corrosion of the metal

A typical set up for electrolysis is shown in the figure below. The metal that is said to be plated s used as the cathode and the metal used to plate the cathode comes from the anode. The electrolyte is the aqueous salt of the metal on the anode.

How objects are electroplated?

The object to be plated is made the cathode of an electrolytic cell and the anode is the source of the plating metal. The electrolyte is an aqueous solution of a salt of the plating metal. The net result is the transfer of the plating metal from the anode to the cathode.

Copper plating:



Apparatus similar to the one shown in the figure below is used to coat a metal object with a thin layer of copper. The anode is pure copper, often called the plating copper. The metal object to be copper-plated is made the cathode and the electrolyte is the aqueous copper (II) sulfate.

At the anode: Cu(s) Cu2+ + 2e-

At the Cathode: Cu2+(aq) + 2e- Cu(s)

Electroplating aluminium glasses with copper

Electrolyte: copper (II) sulfate solution Ions from copper (II) sulfate – CuSO4

Cu2+ + SO42-

Ions from water – H2O H+ (aq) + OH- (aq)

Electrodes: Copper anode and aluminium cathode.

Anode reactions: OH- and SO4

2- - migrate to the anode and neither are discharged. The copper electrodes lose electrons more readily than the anions.

Each copper atom loses 2 electrons to form Cu2+ ion. Eventually the copper anode dissolves.Cu(s) Cu2+ + 2e-

Cathode reactions: Cu2+ and H+ ions migrate to the cathode and Cu2+ ions are preferentially discharged as they are

lower than H+ in the reactivity series. Each Cu2+ ions gains 2 electrons and pink solid of copper metal is deposited on the aluminium

glasses.Cu2+ + 2e- --> Cu(s)

Simple cells



Q. What is a simple cell?

Ans: A simple cell is a device which converts chemical energy to electrical energy. It is also known as an electric cell. It is made by placing two different metals in contact with an electrolyte. The metals act as the electrodes for the simple cell. Each simple cell consists of:

2 Electrodes of different metals. An electrolyte solution of an acid or a salt.

As the more reactive metal tends to undergo oxidation more easily, it naturally becomes the anode.

The less reactive metal becomes the cathode, as it has a weaker tendency to lose electrons

For example, when a zinc plate and a copper plate are placed in dilute sulfuric acid and connected by wires, a chemical reaction occurs to produce an electric current that flows through the wires. A potential difference is set up between the metal plates. This potential difference or voltage can be registered by a voltmeter.

How is electric current produced?



In the zinc-copper cell, the zinc plate dissolves in the acid and bubbles of hydrogen gas form at the copper electrode. The overall cell reaction for this simple cell is:

Zn(s) + H2SO4 ZnSO4 (aq) + H2(g)

Oxidation and reduction occur together in order to cause the flow of electrons in a simple cell. Thus, electrical energy is produced by redox reactions in the simple cell.

How do you determine if a particular electrode is positive r negative in a simple cell?

More reactive metals tend to give up electrons and form ions more readily than less reactive metals. Thus, in a simple cell, the flow of electrons is always from the more reactive metals to the less reactive metal. The more reactive metal becomes the anode and the cell reactive becomes the cathode.

What happens in a zinc-copper cell when the electrolyte is copper (II) sulphate?


At the zinc electrode:Oxidation occurs. Zinc atoms give up electrons and go into solution as zinc ions.Zn(s) Zn2+ + 2e- (oxidation)

The electrons leave the zinc plate, pass through the voltmeter and enter the copper plate.

At the copper electrode: The electrons are taken up by the positively charged hydrogen ions to form hydrogen gas. This is a reduction reaction.

2H+(aq) + 2e- H2(g) (reduction)

The electrode from which electrons flow out of is the negative electrode. The electrode into which the electrons flow is a positive electrode. In the simple cell, zinc acts as the negative and electrode and copper as the positive electrode. The electrolyte is a dilute hydrochloric acid. The electrons which flow in the external circuit constitute the electric current.

Overall reaction: Zn(s) + 2H+(aq) H2(g) + Zn2+


The figure below shows as zinc-copper cell that uses a 1.0 mol/dm3 copper (II) sulfate solution which acts as the electrolyte.

The ionic equation for the overall reaction is obtained by adding the half ionic equations for the reactions at the electrodes.

Overall ionic equation: Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)

Salt bridge simple cell:

A kind of simple cell that uses two different metals and two different electrolyte solutions, connected by a salt bridge is shown in the figure below.

Here the electrolytes are salt solutions of the electrode metals. Similarly, zinc becomes the anode and copper, the cathode. At the zinc electrode: Oxidation occurs. Zinc atoms give up electrons and go into solution as

zinc ions with the electrons flowing through the external set up to the copper cathode.Zn(s) Zn2+ + 2e- (oxidation)

At the cathode, each copper(II) ions from the copper(II) sulfate solution gains 2 electrons and is deposited as the copper on cathode.

The overall reaction:Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)

Simple cells and reactivity series:


At the zinc electrode:

Zinc dissolves to form zinc ions.

Zn(s) Zn2+ + 2e- (oxidation)

At the copper electrode:

Copper (II) ions from the solution receive electrons to form copper.

Cu2+ + 2e- --> Cu(s)


The reactions of metals with water and acids are used to place metals in order of their reactivity.

Simple cells can also be used to determine the relative positions of metals in the reactivity series. This is because the amount of electrical energy produced in a simple cell is determined by how far apart the metals used (as electrodes) are in the reactivity series.

By using different metals in the simple cell, different voltages are produced.

METAL ELECTRODES VOLTAGE (V)Magnesium/ Copper 2.7Aluminium/ Copper 2.0

Zinc/ Copper 1.1Iron/Copper 0.8Lead/Copper 0.5

Copper/ Copper 0.0

The voltage of magnesium-copper cell is 2.7 V. If the magnesium is replaced by zinc, a less reactive metal, the voltage decreases to 1.1 V. This shows that the further apart the two metals are in the reactivity series, the greater the voltage produced. No current will flow if both electrodes are made up of the same metal.

Faraday’s law of electrolysis:



The mass of the substance liberated during electrolysis is directly proportional to the quantity of electricity produced.

The quantity of electricity required to deposit one mole of any substance is a whole number of mole of electrons.

Electrolysis Calculations:

The amount of product of an electrolysis reaction can be calculated and depends on 3 factors.

The amount of charge passed which is expressed in a number of faraday.

1 faraday = the charge of 1 mole of electrons = 96 500 coulombs (C)

The amount of charge which passes in a circuit depends on the current (amperes or A) of the circuit and the amount of time (in seconds) the current is switch on.

The number of faraday can be calculated by :

o multiplying the amperes by seconds which gives coulombs (C)

o dividing the number of coulombs by 96 500 C

charge of the ions: the greater the charge of the ion, the greater the number of electrons needed or the greater the number of faraday

Na+ + e- Na this means that to produce 1 mole of sodium atoms (i.e. 23 grams) one mole of electrons or 1 faraday is needed.

Ca2+ + 2e- Ca this means that to obtain 1 mole of calcium atoms two moles of electrons or 2 faraday are needed

2Cl- (aq) Cl2 (g) + 2e- this means that to obtain 1 mole of chlorine molecules two moles of electrons or 2 faraday are needed

4OH- (aq) O2 (g) + 2H2O (l) + 4e- this means that to obtain 1 mole of oxygen molecules 4 moles of electrons or 2 faraday are needed

Worked example



A solution of copper (II) sulphate is electrolysed. How much copper will be deposited by a current of 2 A flowing for 20 minutes?

Step 1: calculate the amount of coulombs (=charge)

charge = 2 A x 1200 seconds = 2400 C

Step 2: calculate the number of faraday:

faraday = 2400/96 500 = 0.025 faraday or number of moles of electrons

Step 3: use ratio of redox equation Cu2+ + 2e- Cu

2 moles of electrons give 1 mole of copper

0.025 moles of copper give 0.0125 moles of copper

0.0125 moles of copper = 64 x 0.0125 = 0.8g

How to interpret electrode equations?



Moles of electrons:

Magnesium is manufactured by electrolysing molten magnesium chloride. Magnesium is produced at the cathode (the negative electrode) and chlorine at the anode (the positive electrode).The electrode equations:Mg2++ 2e- Mg(l)

2Cl-(l) Cl2 + 2e-

In terms of moles:

1 mole of Mg2+ ions gains 2 moles of electrons and produces 1 mole of magnesium, Mg 2 moles of Cl- ions form 1 moles of chlorine, Cl2 and releases 2 moles of electrons.

When we are doing calculations, we just read e- as ‘1 mole of electrons’

Quantities of electricity

The coulomb is the measure of quantity of electricity. 1 coulomb is the quantity of electricity which passes f 1 ampere (amp) flows for 1 second.

Number of coulombs = current in amps x time in seconds

So, if 2 amps flow for 20 minutes, we can calculate the quantity of electricity (1 st convert the time into seconds) as:

Quantity of electricity = 2 x 20 x 60 = 2400 coulombs

The faraday constant:

A flow of electricity is a flow of electrons. 1 faraday is the quantity of electricity which represents 1 mole of electrons passing a particular point in the circuit – in other approximately 6 x 1023 electrons.

1 faraday = 96, 500 coulombs.

Interpreting electrode equations:

In electrolysis calculations, we are usually interested in the quantity of electricity and the mass or volume of the product. For example:

Na+ (l) + e- Na (l)

1 mole of sodium, Na, is produced by the flow of 1 mole of electrons (= 1 faraday)

Cu2+ + 2e- --> Cu(s)1 mole of copper, Cu, is produced by the flow of 2 moles of electrons (= 2 faradays)



It takes twice as much as electricity to produce a mole copper as it does a mole of sodium. That’s because the Cu2+ ion carries twice the charge, and needs twice as many electrons to neutralise it.2Cl- (l) Cl2 (g) + 2e-

1 mole of chlorine is produced when 2 moles of electrons (= 2 faradays) flow around the circuit.

Some sample calculations:

Electrolysing copper(II) sulfate solution:

What mass of copper is deposited on the cathode during electrolysis of copper(II) sulfate solution if 0.15 amps flow for 10 minutes?

The electrode equation is:

Cu2+ + 2e- --> Cu(s)

(RAM: Cu= 64, 1 faraday = 96000 coulombs)

Start by working out the number of coulombs = amps x time in seconds

= 0.15 x10 x 60

= 90 coulombs

Now work from the equation:

Cu2+ + 2e- --> Cu(s)

2 moles of electrons give 1 mole of copper, Cu

2 x 96000 coulombs to give 64g of copper

192,000 coulombs give 64g of copper

90 coulombs give 90

192000×64 = 0.03 g



Calculations involving gases

During electrolysis of dilute sulfuric acid using platinum electrodes, hydrogen is released at the cathode and oxygen at the anode. Calculate the volumes of hydrogen and oxygen produced (measured at room temperature and pressure) if 1.0 amp flows for 20 minutes.

The electrode equations are:

2H+ (aq) + 2e- H2 (g)

4OH- (aq) O2 (g) + 2H2O (l) + 4e-

(The molar volume of a gas = 24,000 cm3 at rtp; 1 faraday = 96000 coulombs)

Number of coulombs = amps x time in seconds

= 1.0 x 20 x 60

= 1200

Calculating the volume of hydrogen:

2H+ (aq) + 2e- H2 (g)

2 moles of electrons give 1 mole of hydrogen, H2

2 x 96000 coulombs give 24,000 cm3 of hydrogen at rtp.

192, 000 coulombs give 24,000 cm3 of hydrogen at rtp.

1200 coulombs give 1200192000

×24,000=150 cm3

Calculating the volume of oxygen:

4OH- (aq) O2 (g) + 2H2O (l) + 4e-

A flow of 4 moles of electrons produces 1 mole of oxygen, O2

4 x 96000 coulombs produces 24000 dm3 of oxygen

384,000 coulombs produces 24,000 dm3 of oxygen

1200 coulomb produces 1200384,000

x24000=75 cm3

Therefore, 150 cm3 of hydrogen and 75 cm3 of oxygen are produced.



A reversed calculation:

How long will it take to deposit 0.500g of silver on the cathode during electrolysis of silver (I) nitrate solution using a current of 0.250 amps? The cathode equation is:

Ag+ (aq) + e- Ag(s)

(RAM: Ag = 108; 1 faraday = 96000 coulombs)

1 mole of electrons gives 1 mole of silver, Ag.

96000 coulombs give 108g of silver.

To produce 1g of silver we’d need 96000108 = 888.89 coulombs

To produce 0.500g of silver we’d need 0.500108

×96,000=444.4 coulombs

Number of coulombs = amps x time in seconds

444.4 = 0.250 x t

T = 444.40.250

=1780 seconds

The time needed to deposit 0.500g of silver is 1780 seconds/ 29.67 minutes.

Electrolysing more than one solution:

Suppose we have two solutions connected together in series, so that the same quantity of electricity flows through both.

At the end of the electrolysis, it was found that 2.07g of lead had been deposited on the cathode in the left-hand beaker.



a) Calculate the quantity of electricity that passes during the experiment.b) Calculate the mass of silver that was deposited on the cathode in the right hand beaker.

(RAMs: Pb= 207; Ag= 108; 1 faraday =96000 coulombs)

(a) Pb2++ 2e- Pb(s)

2 moles of electrons give 1 mole of lead, pb

2 x 96000 coulombs give 207 g of lead

192, 000 coulombs give 207 g of lead

1920 coulombs give 2.07 gram of lead.

The quantity of electricity that passed = 1920 coulombs.

(b) If 1920 coulombs passed through the beaker containing the lead(II) nitrate solution, then exactly the same amount passes through the rest of the circuit.

Ag+ (aq) + e- Ag(s)

1 mole of electrons gives 1 mole of silver, Ag

96000 coulombs give 108g of silver.

1920 coulombs give 192096000

×108=2.16

The mass of silver deposited is 2.16g.

Alternative way of solving part (b):

If we weren’t asked to find the quantity of electricity in part (a), we could do part (b) much more easily without knowing anything at all about the faraday’s constant or even about coulombs.

Equations:

Pb2++ 2e- Pb(s)

Ag+ (aq) + e- Ag(s)

If 2 moles of electrons flow, we will get 1 mole of lead and 2 moles of silver. However many electrons flow, we will always get twice as much as moles of silver as lead.

In this calculation, 2.07g of lead were formed, which is 0.01 mol

We will therefore get 0.02 mol of silver = 0.02 x 108 = 2.16g



A similar example involving just one solution

During the electrolysis of concentrated copper(II) chloride solution, 3.2g of copper was deposited at the cathode. What volume of chlorine (measured as rtp) would be formed at the anode? (RAM: Cu = 64 ; molar volume = 24,000 cm3 at rtp)

The electrode equations are:

Cu2+ + 2e- --> Cu(s)

2Cl- (l) Cl2(g) + 2e-

For every 2 moles of electrons that flow, we will get 1 mole of Copper, Cu and 1 mole chlorine, Cl2. We are bond to get the same number of moles of each.

In this case, 3.2g of copper is 3.264 mol = 0.05 mol.

So we will also get 0.05 mol of chlorine which is 0.05 x 24, 000 cm3 at rtp.

The volume of chlorine produced is 1200 cm3.


chapter i - electrolysis

Documents

dilute sulphuric acid

dilute sulfuric acid

dilute sodium chloride

4ecathode reactions

2ecathode reactions

ionic half equation

negatively charged electrode

cu2 2e