CHAPTER 16Oxidation and Reduction

Redox Reactions• Redox Reactions• Oxidation-Reduction reactions are one of the largest

groups of chemical reactions.• Oxidation / Reduction reaction• Redox chemistry is an important aspect of everyday life• Our bodies work by redox reactions – the food we eat is

oxidised to enable us to obtain the energy we need to live.• Redox chemistry is involved in;

– Bone healing, batteries, metal extraction, keeping swimming pools clean

What is oxidation – reduction?• Early chemists described the combining with oxygen using the term

oxidation– Eg combustion of propane (write the equation)– Eg burning of Iron (III) in air (write the equation)

Reactions involving the decomposition of a compound, with the loss of oxygen is called reduction (reduced to something simpler)

– Eg Copper (II) oxide may be reduced to copper by hydrogen (write the equation)– Eg Iron (II) oxide is reduced to iron by carbon monoxide in a blast furnace (write

the equation)

As one reactant is reduced, the other reactant is oxidised– Eg Magnesium burns in steam to form Magnesium oxide and hydrogen (write the

equation)

What is oxidation – reduction?• In any oxidation – reduction

reaction– The oxidant is the species which

causes oxidation and is itself reduced

– The reductant is the species which causes reduction and is itself oxidised

Write the equation for the oxidation of propane and label the processes, the oxidant and the reductant.

Oxidation / Reduction basics

OIL - RIG

• Oxidation• Is• Loss of Electrons• Reduction• Is• Gain of electrons

AN ELECTRON TRANSFER VIEW Redox reactions are electron transfer

reactions A substance that is oxidised is one that loses

electrons A substance that is reduced is one that gains

electrons The reductant loses electrons (is oxidised) The oxidant gains electrons (is reduced)

AN ELECTRON TRANSFER VIEW The oxidation of Mg in O2 to form MgO involves the

oxidation of Mg (write the equation and label the oxidant and reductant)

Now add electron transfer to your labels MgCl is formed by the combustion of Mg in Cl – no O2

involved (write the equation and label the oxidant and reductant)

Now add electron transfer to your labels Both above products are ionic substances and so contain

Mg2+. In the process of oxidation, each atom of Mg has lost two electrons

In general, a substance that is oxidised is one that loses electrons and is therefore an electron donor

In general, a substance that is reduced is one that gains electrons and is therefore an electron acceptor.

AN ELECTRON TRANSFER VIEW Write the equation for steel wool burning in

chlorine gas (Iron III)

Fe Cl2

Donates electrons Accepts electrons

Is oxidised Is reduced

Is the reductant Is the oxidant

REVIEW Complete question 1 page 369

Oxidation numbers• Oxidation is an increase in the oxidation number of an

atom• Reduction is a decrease in the oxidation number of an

atom• Electrochemical reactions involve the transfer of electrons.

Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction.

• Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules

• Oxidation numbers video

Oxidation Numbers

• Oxidation number is defined as the imaginary charge an atom would have if it is existed as an ion in a compound.

– Eg. Na +1Cl -1 Cu+2SO4-2

Rules for determining oxidation numbersFor each element or molecule that is involved in a reaction we need to follow these simple rules.

• Rule 1. All pure substances have an oxidation number of zero. This applies to any pure substance whether it is a diatomic gas like O2 or a piece of pure metal like Iron (Fe). Examples of

• Rule 2. In compounds, elements that usually have an ionic charge imparted by their position in a particular group have that same oxidation number. An example is Cl which is usually in the form Cl- in compounds; this will have an oxidation number of -1 in compounds.

• Rule 3. When two or more usually negatively charged ions are involved in a compound, the one with the highest electronegativity value is given its ionic charge as the oxidation number; the others are worked out normally. An example is OF2. F is more electronegative, and so it is assigned the value of -1.

• Rule 4. Oxygen in a compound always has an oxidation number of -2• Rule 5. Hydrogen in compounds always has an oxidation number of +1 except in the rare case

of Metal Hydrides where it has a value of -1. • Rule 6. The oxidation numbers in the compound or molecule must total to the overall charge

of that compound or molecule. For example CO2 has no overall charge and so the oxidation numbers must tally to zero. The sulfate ion, SO42-, has an overall charge of -2, so the oxidation numbers must tally to -2

Rules for determining oxidation numbers1. The oxidation number of an element is zero.– Eg. N2 oxidation number (O.N.) of N = 0

Fe O.N. of Fe = 0

2. The oxidation number of the simple ion is the charge on the ion.– Eg. Al2O3 O.N. of Al3+ = +3 and O.N. of O2- = -2

Cu2+ O.N. of Cu2+ = +2

3. The oxidation number of hydrogen is +1 in its compounds with non-metals. In metal hydrides, the oxidation number of hydrogen is -1.

– Eg. NH3 O.N.of H = +1 CaH2 O.N.of H = -1 HCl O.N. of H = +1

4. The oxidation number of oxygen is usually -2, except in peroxides. In peroxides, it is -1.– Eg. NO2 O.N.of O = -2 BaO2 O.N.of O = -1

H2O O.N. of O = -2 H2O2 O.N.of O = -1

5. In a neutral compound, sum of all oxidation numbers is zero.– Eg. MgCl2 Sum of O.N. = +2+(-1)x2 = 2-2 = 0

6. In a polyatomic ion, sum of all oxidation numbers must be equal to the charge on the ion.– Eg. CO3

2- Sum of O.N. = +4+(-2)x3 = -2

Redox Reactions: In terms of oxidation numbers

• Oxidation numbers are used to track the electron transfer in redox reactions.

• In a redox reaction,1. If oxidation number is increased, oxidation has occurred.2. If oxidation number is decreased, reduction has occurred.

– Eg. 2Mg(s) + O2 (g) 2MgO(s)

Oxidation number 0 0 +2 and -2

Oxidation number of Mg has increased from 0 +2 Mg is oxidised.

Oxidation number of O2 has decreased from 0 -2 O2 is reduced.

Using Oxidation Numbers• Assign oxidation numbers to

Mg(OH)2

• Note that the sum of the oxidation numbers must equal zero• Mg + (2 x O) + (2 x H) = +2 + 2x(-2) + 2x(+1)

= 2 – 4 + 2= 0

• The oxidation number of oxygen and hydrogen must be multiplied by two because each formula unit contains two of each of these atoms.

+2- 2 for each O atom

+1 for each H atom

Review

• Work through the sample problems pages 370, 371

• Complete the revision questions 2, 3,4, 5 pages 371, 372

Identifying Redox Reactions

You can use oxidation numbers to determine whether or not a reaction is a redox reaction.

If the oxidation number of an element in a reacting species changes, then that element has undergone either oxidation or reduction

If the oxidation number increases it is _____ If the oxidation number decreases it is _____

Review Work through the sample problems

pages 372, 373 Complete the revision questions 6 – 8

page 373

Half - equationsRedox reactions can be written as two half-equationsHalf-equations are a useful way of understanding the processes involved in a redox reaction.When an iron nail is placed in a blue copper sulfate solution, the nail becomes coated with a metallic copper and the blue colour of the solution fades.A redox reaction has taken place as electrons have been transferred from the iron nail to the copper ions in the solution, allowing solid copper to form.The full equation for the reaction is as follows:Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

View the Copper Sulfate video

Half - equationsWrite the Ionic equation from the full equation

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)Fe(s) + Cu2+(aq) + SO4

2-(aq) Fe2+ (aq) +SO42-

(aq) + Cu(s)

Write the Net Ionic Equation (no spectator ions)

Fe(s) + Cu2+(aq) Fe2+ (aq) + Cu(s)

Half - equationsWrite the ionic and net ionic equations for Iron (II) with Copper SulfateConjugate redox pairs are made up of two species that differ by a certain number of electrons. Each has its own half-equationOxidation conjugate pair

Fe(s) Fe2+(aq)

Reduction conjugate pairCu2+(aq) Cu(s)

Half - equationsIn order to balance the conjugate pair to produce proper half-equations, electrons need to be shown.

• Complete the half-equation for oxidation– Fe(s) Fe2+(aq) + ________

• Complete the half-equation for reduction– Cu2+(aq) + __________ Cu(s)

• These half-equations are balanced with respect to both atoms and charge.

• Combining these two half-equations will give you the ionic equation for the reaction as a whole.

ReviewComplete questions 9 and 10 pages 374, 375

Writing balanced half-equations for ions in aqueous solution

1. Write the half-equations for oxidation and reduction showing conjugate pairs

2. Balance all elements except hydrogen and oxygen3. Balance oxygen atoms, where needed, by adding water4. Balance hydrogen atoms, where needed, by adding H+5. Balance the charge by adding electrons6. Multiply each half-equation by factors that will lead to

the same number of electrons in each half-equation7. Add the half-equations and omit the electrons

Writing balanced half-equations for ions in aqueous solution

• The statement MnO- is reduced to Mn2+ seems incorrect because according to the charges on the ions, there appears to be a loss, rather than a gain of electrons• MnO4

-(aq) Mn2+(aq)• It is only when the entire half-equation for the

change is written that its true nature as reduction becomes obvious, with five electrons being accepted by each permangate ion:

• MnO4-(aq) + 8H+(aq) + 5e- Mn2+ (aq) + 4H2O(l)

Writing balanced half-equations for ions in aqueous solution

• Balance and complete the half-equation (balance atoms and electrons) for;a. Br2(l) Br-(aq)b. Fe3+ (aq) Fe2+ (aq)

Are the above reactions oxidation or reduction?

Writing balanced half-equations for ions in aqueous solution

• To balance half-equations; • balance the elements, • add water molecules, • balance by adding hydrogen ions, and• balance the difference in charge with

electrons

Writing balanced half-equations for ions in aqueous solution

• Try; • O2 (g) H2O2 (l)• Balance by adding H+

• O2 (g) + 2H+ H2O2 (l)• Balance the difference in charge• O2 (g) + 2H+ + 2e- H2O2 (l)

• Is this oxidation or reduction?

Writing balanced half-equations for ions in aqueous solution

Step One• Write the half-equations for oxidation and

reduction showing conjugate pairs. Use oxidation numbers to identify which substance has been oxidised and which has been reduced.• Reduction MnO4

-(aq) Mn2+(aq)• Oxidation Cl-(aq) Cl2(aq)

• Calculate the oxidation number for Mn in MnO4-

• Calculate the oxidation number for Cl-

Writing balanced half-equations for ions in aqueous solution

Step Two

• Balance all elements except hydrogen and oxygen, which will be balanced later• MnO4

-(aq) Mn2+(aq)• 2Cl-(aq) Cl2(aq)

Writing balanced half-equations for ions in aqueous solution

Step Three• Balance oxygen atoms, where needed, by

adding water• MnO4

-(aq) Mn2+(aq) + 4H2O• 2Cl-(aq) Cl2(aq)

Writing balanced half-equations for ions in aqueous solution

Step Four• Balance hydrogen atoms, where needed, by

adding H+

• MnO4-(aq) +8H+ (aq) Mn2+(aq) + 4H2O(l)

• 2Cl-(aq) Cl2(aq)

Writing balanced half-equations for ions in aqueous solution

Step Five• Balance the charge of each half-equation by

adding electrons• MnO4

-(aq) + 8H+ (aq) Mn2+(aq) + 4H2O(l)

Charges -1 +8 +2 0

Net Charge +7 +2

Writing balanced half-equations for ions in aqueous solution

Step Five cont.• Since the net charge on the left-hand side is greater

than that of the right-hand side, it is necessary to add five electrons to the left-hand side to make the net charge on both sides equal at +2.

• The reduction half-equation becomes:• MnO4

-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O(l)

Net Charge +7 + 5e- = +2 +2

Writing balanced half-equations for ions in aqueous solution

Step Five cont.• 2Cl-(aq) Cl2(aq)

• Since the net charge of the left-hand side is less than that of the right-hand side, it is necessary to add two electrons to the right-hand side to make the net charge on both sides equal at -2.

• The oxidation half-equation becomes:• 2Cl-(aq) Cl2(aq) + 2e-

Charges 2 x -1 0

Net Charge -2 0

Writing balanced half-equations for ions in aqueous solution

Step Five cont.• Each half-equation is now balanced. • The net charge on each side of the reduction

equation is +2 and the net charge on each side of the oxidation equation is -2

• MnO4-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O

2Cl-(aq) Cl2(aq) + 2e-

Writing balanced half-equations for ions in aqueous solution

Step Six• MnO4

-(aq) + 8H+ (aq) + 5e- Mn2+(aq) + 4H2O(l)

2Cl-(aq) Cl2(aq) + 2e-

• The previous step has resulted in an uneven number of electrons, which will not cancel if the half-equations are added.

• Overcome this by multiplying by each half-equation that will lead to the same number of electrons in each half-equation

• In the reduction half-equation by 2• In the oxidation half equation by 5

Writing balanced half-equations for ions in aqueous solution

Step Six cont.• The reduction half-equation becomes:

• 2MnO4-(aq) + 16H+ (aq) + 10e- 2Mn2+(aq) + 8H2O(l)

• The oxidation half-equation becomes:• 10Cl-(aq) 5Cl2(aq) + 10e-

Writing balanced half-equations for ions in aqueous solution

Step Seven• Add the half-equations:• 2MnO4

-(aq) + 16H+ (aq) + 10e- 2Mn2+(aq) + 8H2O(l)

• 10Cl-(aq) 5Cl2(aq) + 10e-

• 2MnO4-(aq) + 10Cl-(aq) + 16H+ (aq) + 10e- 2Mn2+(aq) 5Cl2(aq) + 8H2O(l) + 10e-

• Then cancel the electrons• 2MnO4

-(aq) + 10Cl-(aq) + 16H+ (aq) 2Mn2+(aq) 5Cl2(aq) + 8H2O(l)

Balancing Half-Equations - summary

1. Identify conjugate redox pairs.2. Write the half-equations for oxidation and reduction

showing conjugate pairs.3. Balance all elements except hydrogen and oxygen.4. Balance oxygen atoms, where needed, by adding water.5. Balance hydrogen atoms, where needed, by adding H+.6. Balance the charge by adding electrons.7. Multiply each half-equation by factors that will lead to

the same number of electrons in each half-equation.8. Add the half-equations and omit the electrons.

Balancing Half-Equations• Try;

• IO3- (aq) I- (aq) – conjugate pair

• What is the oxidation number of I in IO3-?

• Balance all elements (except hydrogen and oxygen)• IO3

- (aq) I- (aq)• Add water to balance oxygen atoms• IO3

- (aq) I- (aq) + 3H2O• Add H+ to balance hydrogen atoms• IO3

- (aq) + 6H+ (aq) I- (aq) + 3H2O• Balance the charge by adding electrons• IO3

- (aq) + 6H+ + 6e-(aq) I- (aq) + 3H2O• Oxidation or Reduction?

Balancing Half-Equations• Try;

• MnO4- (aq) Mn2+ (aq) – conjugate pair

• What is the oxidation number of Mn in MnO4- ?

• Balance all elements (except hydrogen and oxygen)• MnO4

- (aq) Mn2+ (aq)• Add water to balance oxygen atoms• MnO4

- (aq) Mn2+ (aq) + 4H2O (l)• Add H+ to balance hydrogen atoms• MnO4

- (aq) + 8H+ Mn2+ (aq) + 4H2O (l)• Balance the charge by adding electrons• MnO4

- (aq) + 8H+ + 5e- Mn2+ (aq) + 4H2O (l)• Oxidation or Reduction?

Balancing Half-Equations• Try;

• Cr3+ (aq) Cr2O72- (aq) – conjugate pair

• What is the oxidation number of Cr in Cr2O72- ?

• Balance all elements (except hydrogen and oxygen)• 2Cr3+ (aq) Cr2O7

2- (aq)• Add water to balance oxygen atoms• 2Cr3+ (aq) + 7H2O (l) Cr2O7

2- (aq)• Add H+ to balance hydrogen atoms• 2Cr3+ (aq) + 7H2O (l) Cr2O7

2- (aq) + 14H+ (aq)• Balance the charge by adding electrons• 2Cr3+ (aq) + 7H2O (l) Cr2O7

2- (aq) + 14H+ (aq) + 6e-

• Oxidation or Reduction?

Review• Work through the sample problem page 377

• Complete the revision questions 11 and 12 page 378

Reactivity Series of Metals

Reactivity Series of Metals

Displacement Reaction• When zinc is placed in copper

sulphate solution a reaction occurs, causing copper metal to form on the zinc.

• Zinc removes copper from the solution and, as a result, the deep blue colour of the solution pales.

• If you place a copper strip in a solution of zinc sulphate, no reaction occurs.

• Zinc is more reactive than copper.

Review

• Work through the sample problem page 380

• Complete the revision question 13, 14 page 380

Electrochemical Cells• Zinc is a better reductant than copper and is therefore

oxidised (increases oxidation number) more readily than copper.

• As zinc is oxidised, electrons flow from the zinc metal to the copper ions.

• Energy is released in displacement reactions• Chemical energy can be converted into electrical energy in

an electrochemical cell.• The salt bridge provides ions to balance ions consumed or

produced in each cell• Electrochemical cell

Electrochemical Cells

With the addition of a wire and a salt bridge, a simple electrochemical cell, which converts chemical energy into electrical energy is constructed.

Electrochemical cell animation

Metals in aqueous solutions simulation

Redox reactions at the electrodes of Zn/Cu cell

1.First, oxidation of zinc at the anode

Zn (s) Zn2+ (aq) + 2e

2.These electrons flow through zinc strip and connecting wire towards the copper strip, producing an electric current.

3.Reduction of copper at the cathodeCu2+ (aq) + 2e Cu (s)

The overall reaction is _____________________________________

Reductant

Oxidant

Electrochemical cell animation

Redox reactions at the electrodes of Zn/Cu cell • Anode - Zinc strip;

• Negative charge as electrons are produced; OxidAtion occurs. • Cathode - Copper strip;

• Positive charge; ReduCtion occurs.

• Salt bridge contains an electrolyte (KNO3 or KCl). • Its role is :

• To complete the circuit by providing continuous path for movement of ions.

• To maintain a balance of charges • NO3

- ions towards anode (to balance build up of cations)• anions towards the anode• and K+ towards cathode (to balance build up of anions) • cations towards the cathode

Redox reactions at the electrodes of Zn/Cu cell

• Electrons carry the current in the wire• The flow of ions completes the circuit• It is important that the ions in the salt

bridge do not react with chemicals in the beakers

Electrochemical Cells

• A simple electrochemical cell consists of :• Two half-cells, containing two electrodes

(anode and cathode) and two electrolytes• A conducting wire• A salt bridge, containing another electrolyte

• The more reactive metal gives electrons to the ions of the less reactive metal

Redox reactions at the electrodes of Zn/Cu cell

• In these half-cells, where is oxidation occuring? • Oxidation is occuring at the anode. • The most reactive metal is the zinc (giving electrons)• Where is reduction occuring?• Reduction is occuring at the cathode• The least reactive metal is the copper (receiving

electrons)• Electrons move from the more reactive to the less

reactive

Redox reactions at the electrodes of Zn/Cu cell

• Write the reaction taking place in the zinc half-cell

• Zn (s) Zn2+ (aq) + 2e-

• Write the reaction taking place in the copper half-cell

• Cu2+ (aq) + 2e- Cu (s)• As zinc is oxidised, electrons flow from the

zinc to the copper ions

• Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)2e-

Review

• Work through the sample problem page 382

• Complete the revision questions 15 – 17 page 383

Predicting Redox Reactions• If two metals (potential reductants) and their

corresponding cations (potential oxidants) are put together, only one reaction will occur.

• The reaction is always between a reductant and an oxidant, and takes place between the stronger oxidant and the stronger reductant.

• The electrochemical series is useful when trying to predict a reaction in a galvanic cell.

Predicting Redox Reactions– Eg. What reaction will take place when

chlorine gas is bubbled into KI solution?

– Step 1: Write down all species present in solution.

K+, I-, Cl2, H2OCl2

KI

Increasing oxidising strength

Predicting Redox Reactions– Step 2: Locate the species from ECS and write down all possible half

reactions as in the order of the electrochemical series.Cl2 (g) + 2e- 2 Cl- (aq) Eo = + 1.36VO2 (g) + 4H+ (aq) + 4e- 2 H2O (l) Eo = + 1.23VI2 (g) + 2e- 2 I- (aq) Eo = + 0.54V2 H2O (l) + 2e- H2 (g) + 2OH- (aq) Eo = - 0.83VK+ (aq) + e- K (s) Eo = -2.93V

Increasing reducing strength

Predicting Redox Reactions– Step 3: Identify the strongest oxidant and strongest reductant.

Possible oxidants are Cl2, H2O and K+

Strongest oxidant is Cl2 (most positive Eo)

Possible reductants are H2O and I-

Strongest reductant is I- (most negative Eo)

Cl2 (g) + 2e- 2 Cl- (aq) Eo = + 1.36V

I2 (g) + 2e- 2 I- (aq) Eo = + 0.54V

The strongest oxidant will react with the strongest reductant.

Predicting Redox Reactions

– Step 4: Write half-equations and redox equation for the predicted reaction.

Oxidation at anode 2 I- (aq) I2 (g) + 2e-

Reduction at cathode Cl2 (g) + 2e- 2 Cl- (aq)

Cl2 (g) + 2I- (aq) 2 Cl- (aq) + I2 (g)

Review• Read the chapter summary

• Complete the multiple choice questions 1 – 28 pages 394 – 396

• Work through a sample of review questions on each section pages 396 – 399

• Review questions on Corrosion and Corrosion protection – 35, 36, 39, 44