TTooppiicc 99 OOxxiiddaattiioonn aanndd RReedduuccttiioonn AAnnsswweerrss

9.1 Exercises

1. Define oxidation and reduction in terms of

a) electron transfer

Oxidation is the loss of electrons from an atom, ion or molecule. Reduction is the gain of electrons by an atom,

ion or molecule.

b) oxidation number

Oxidation is the increase of the oxidation number of an atom. Reduction is the decrease of the oxidation number

of an atom.

2. Why is oxidation and reduction like lending and borrowing?

Oxidation cannot occur without reduction having also occurred, just as you cannot borrow a book without

someone lending it to you. Both involve a transfer.

3. Assign oxidation numbers of the elements for the following groups of species, and give the rule/s

you have applied in assigning oxidation numbers for each group:

a) 0 0 0 0

O2 H2 Cl2 Na

Rule/s: All of the above species have an oxidation number of 0, as they are all in the elemental form (how they

exist naturally, not combined with any other elements).

b) +1 -1 +2 -2

Na+ Cl

- Ca

2+ S

2 -

Rule/s: The oxidation number is the same as the charge, as these are monatomic ions.

c) +1 -2 -1 +1 -3 +1 -4 +1

H2O HNa NH3 CH4

Rule/s: Three rules are used here. Firstly, hydrogen always has an oxidation of +1 (except in combination with

reactive metals such as Na when it is -1). Secondly, oxygen always has an oxidation state of –2 (except in H2O2

where it is -1). These known values are used first. Finally, as all these molecules have an overall charge of 0,

the sum of the oxidation numbers must also equal zero.

Conventions for showing oxidation numbers The oxidation numbers are assigned for individual atoms, not groups of the same atom within a polyatomic ion. For example the oxidation number of each hydrogen atom in NH3 is +1 and this is what is written above the hydrogen atom. Because there are three hydrogen atoms, +1 x 3 = +3, and this total, added to the oxidation number of nitrogen (–3), gives the overall zero charge for the NH3 molecule.

d) +1 -1

AgCl

Rule/s: This is an ionic compound. Chlorine is in group 7 and thus forms an ion with a charge of -1, which

means its oxidation number is also –1. Silver (Ag) must have a charge and oxidation number of +1 to balance

out the chloride ion.

+1 +1 -2

HClO

1

Rule/s: The common values for hydrogen (+1) and oxygen (-2) are used. The total charge of the molecule is 0,

so the sum of the oxidation numbers of the individual species must equal 0, therefore chlorine has an oxidation

number of +1 in this case.

Oxidation states of halogens Normally halogens have negative oxidation states, but on combination with oxygen or a higher halogen, e.g. fluorine, they have positive oxidation states. This is because halogens in period 3 (or in a higher period) are able to "expand their octet".

+2 -1

MgBr2

Rule/s: This is an ionic compound. Bromine is in Group 7, and forms an ion with charge –1, so its oxidation

number is also –1. Magnesium is in Group 2, and forms an ion with charge +2, so its oxidation number is +2.

Note that the charge and oxidation number balance out to 0.

+2 -1

CuCl2

Rule/s: This is an ionic compound. Chlorine is in Group 7, and forms an ion with charge –1, so its oxidation

number is also –1. Copper must have a charge and oxidation number of +2 to balance out the two chloride ions.

4. Assign oxidation numbers of all elements in the following species:

a) +1 +7 -2

KMnO4

Oxygen has the known oxidation number of –2. Potassium is in Group 1, and thus has an oxidation number of

+1. As the molecule has total charge of 0, the sum of the oxidation numbers must equal 0, so manganese has

an oxidation number of +7.

b) +4 -2

CO2

Oxygen has oxidation number –2, and the molecule has 0 charge. Therefore carbon must have an oxidation

number of +4 making the molecule overall neutral.

c) -2 +1 -2 +1

C2H5OH

The known values for hydrogen and oxygen are used. As the molecule has a charge of 0, the sum of the

oxidation numbers must be 0, and the two carbon atoms must have a combined oxidation number of –4, so

each carbon atom has an oxidation number of –2.

d) +7 -2

ClO4

The molecule has a negative charge, so the sum of the oxidation numbers should be -1. Oxygen has the usual -

2 value and four of them give a combined oxidation number of –8. The overall sum of oxidation numbers must

add to –1, so Cl must be equal to +7.

e) +5 -2

PO43

Oxygen has the usual value of –2 and four of them give a combined oxidation number of –8 so the phosphorus

must have an oxidation number of +5 to give the molecule a net charge of –1.

2

5. Oxidation calculations

a) What is the oxidation number for the elements below in their ionic form?

Na+

+1

F- –1

Mg2+

+2 O2-

–2

Al3+

+3 N3-

–3

b) Calculate the oxidation number of nitrogen in the following:

NH3 -3 N2O5 +5

NH4 -4 N2 0

HNO3 +5 Mg3N2 -3 (Mg is in Group 2, so is +2)

c) Calculate the oxidation number of sulfur in the following:

ZnS -2 H2SO4 +6

SO2 +4 SO3 +6

SO42-

+6 H2S -2

6. State the names of the following compounds using oxidation numbers.

It may be helpful to look in the introduction section for the charges and names of the common groups (NO2-)

and (SO42-

)

a) Fe2O3

Iron(III) oxide

b) FeO

Iron(II) oxide

c) CuCl2

Copper(II) chloride

d) CoCl2

Cobalt(II) chloride

e) Cu(NO2)2

Copper(II) dioxonitrate(III) (the old name, nitrite, is acceptable).

This requires some assumed knowledge that the NO2- ion has a –1 charge. Use a bit of common sense here:

you know that the Cu, a metal, must be positive (i.e. most likely +1 or +2) and that the compound is neutral

overall. The fact that there are two NO2- ions for every one copper should give you a clue that the copper ion

has a +2 charge.

f) Na2SO4

Sodium(I) tetraoxosulfate(VI) (old name: sulfate)

Common oxoanions are given names with the suffixes –ite and –ate. The ion with the larger number of oxygen atoms is given the suffix ‘-ate’ and that with the smaller number of oxygen atoms ‘-ite” Oxoanions of sulfur:

-Sulfates (SO42−

), the salts of sulfuric acid (H2SO4), in which sulfur has an oxidation state of +6

-Sulfites (SO2−

), the salts of sulfurous acid (H2SO3) in which sulfur has an oxidation state of +4

S

O

HO OH

O

SO

OH

OH

Sulfuric acid Sulphurous acid

3

7. Deduce formulas of the inorganic compounds with the following names:

a) sulfuric(VI) acid

H2SO4 (Note that the sulfur has an oxidation number of +6 as shown by the roman numerals and the SO4 group

overall has an oxidation number of –2)

b) sulfurous(IV) acid

(erratum, this should read sulfurous not sulfuric, see previous page for an explanation of the sulfur

oxoanions)

H2SO3 (Note that the sulfur in the SO3 has an oxidation number of +4 as shown by the roman numerals and the

SO3 group overall has an oxidation number of –2)

c) potassium manganate(VII)

KMnO4 (Note that the manganese has an oxidation number of +7)

d) sodium dichromate(VI)

Na2Cr2O7 (Note that chromium is has an oxidation number of +6)

e) copper(I) tetraoxosulfate(VI)/sulphate

Cu2SO4 (Note that the copper has a +1 oxidation state and sulphate group overall has a –2 oxidation state)

f) nitric(V) acid

HNO3 (Note that the nitrogen has an oxidation state of +5)

8. Indicate which of the following reactions are redox reactions. For the redox reactions, identify the

substance oxidised and the substance reduced by the change in oxidation numbers.

Erratum: in part a) Change Cl-(g) to Cl

-(aq) Please note the error in state symbols in the workbook.

a) +1 -1 +1 -1

Ag+(aq) + Cl

-(aq) ���� AgCl(s)

This is not a redox reaction, as none of the elements change oxidation number.

The equation written above merely indicates a precipitation reaction (Ag+ and Cl

- ions associating to form AgCl

and coming out of solution as a solid precipitate. The transformation step of the elements Ag(s) and Cl2(g) to

ions in solution (Ag+ and Cl

-) would have involved redox reactions.

b) +2 +4 -2 +2 -2 +4 -2

CaCO3(aq) ���� CaO(s) + CO2(g)

This is not a redox reaction as none of the elements change oxidation number.

c) -2 +1 -1 -2 +1 - 2 +1 -2 +1 -1

CH3Br(aq)+ OH –(aq)

���� CH3OH(aq) + Br

–(aq)

This is not a redox reaction as none of the elements change oxidation number. This is a substitution reaction.

You will learn about these reactions in topic 10.

d) +5 -2 -1 +1 0 +1 -2

BrO3 – (aq) + 5Br

-(aq) + 6H

+ (aq)

���� 3Br2(l) + 3H2O(l)

This is a redox reaction, as the bromine in BrO3 is reduced (decreases oxidation number), and the bromide ion

is oxidised (increases oxidation number.)

e) 0 +1 -2 +1 -1 0

2F2(g) + 2H2O(l) ���� 4HF(aq) + O2(g)

This is a redox reaction. The fluorine is reduced as its oxidation number decreases from 0 to -1. The oxygen is

oxidised as it increases in oxidation from -2 to 0.

4

f) 0 -1 -1 0

Cl2(g) + 2Br –(aq) ���� 2Cl

– (aq) + Br2(g)

This is a redox reaction. The chlorine is reduced as it decreases in oxidation number from 0 to -1. The bromine

is oxidised as it increases in oxidation number from -1 to 0.

9. For the following reactions, identify the substance oxidised and the substance reduced by the

change in oxidation numbers.

a) reaction that destroys ozone in the stratosphere:

+2 -2 0 +4 -2 0

NO(g) + O3(g) ���� NO2(g) + O2

Nitrogen is oxidised as the oxidation number increases from +2 to +4. One oxygen atom is reduced from ozone

(O3) as the oxidation number decreases from 0 to -2 in the product NO2.

b) formation of rust (simplified equation):

0 0 +3 -2

Fe(s) + O2(g) ���� Fe2O3(s)

The iron is oxidised as the oxidation number increases from 0 to +3. Oxygen is reduced as the oxidation

number decreases from 0 to -2.

c) the reaction of hydrazine with nitrogen tetroxide is utilised in rocket propellant:

-2 +1 +4 -2 0 +1 -2

2N2H4(g) + N2O4(g) ���� 3N2(g) + 4H2O(g)

Nitrogen from N2H4 is oxidised as it increases in oxidation number from -2 to 0. Nitrogen from N2O4 is reduced

as it decreases in oxidation number from +4 to 0.

d) the reaction that occurs when a car battery is discharging:

0 +4 -2 +1 +6 -2 +2 +6 -2 +1 -2

Pb(s) + PbO2(aq) + H2SO4(aq)���� 2PbSO4(aq) + 2H2O(l)

The solid lead is oxidised as the oxidation number increases from 0 to +6. The lead in PbO2 is reduced as the

oxidation number decreases from +4 to +2.

Q 10-11 Theory of Knowledge questions to be discussed with your TOK teacher.

9.2 Exercises

1. What is a redox equation?

A redox equation is a balanced equation describing a redox reaction in which one species is oxidised and one

species is reduced.

2. What is a half equation?

A half equation demonstrates only one half of the overall redox reaction; either the oxidation or the reduction

half. The two half equations (oxidation and reduction) are added together to produce the overall redox reaction.

3. List three different ways to describe a reducing agent.

1. A reducing agent supplies electrons to another species; it reduces the other species.

2. A reducing agent is oxidised during a redox reaction.

3. A reducing agent contains an element whose oxidation number increases during the reaction.

5

4. How can we tell from a chemical equation whether a reaction is a redox equation?

Assign oxidation numbers to all elements. If you see a compound that has an element that has either an

increase or decrease in oxidation number from one side of the equation to the other, then you know it is a redox

reaction.

5. Write half-equations for

a) the reduction of copper(II) ions in aqueous solution to copper metal

Cu2+

(aq) + 2e � Cu(s)

Balance the charge by adding electrons. In this case the copper is reduced as it gains electrons and the

oxidation number decreases from +2 to 0.

b) aluminium metal is oxidised to Al3+

in aqueous solution

Al(s) � Al

3+(aq) + 3e

Note from question 5 a) and b): You can see that when a species is reduced the electrons appear on the reactant side of the half equation and that when a species is oxidised the electrons appear on the product side. This highlights the definition of reduction and oxidation with respect to electrons being either lost or gained by a species.

6. The following species are common oxidising agents. Deduce their reduction half-equations.

a) O2

O2 + 4e � 2O

2- Remember, if a species is an oxidising agent, it must itself be reduced. So look for a decrease

in oxidation number in the element/species. Reduction is gain of electrons. Look at the element which is most

likely to gain electrons in polyatomic species.

b) Cl2

Cl2 + 2e

� 2Cl

c) H2O2/H+

H2O2 + 2H

+ + 2e

� 2H2O

Notice that the oxidation number of oxygen decreases from –1 in the peroxide (H2O2) to –2 in water. Oxygen is reduced, i.e. it gains electrons. If you think logically about this, it is much more likely that the oxygen acts as the oxidant rather than the hydrogen ions acting as the oxidant and being reduced, i.e. hydrogen does not gain electrons easily at all!

d) MnO4-/H

+

MnO4 + 8H

+ + 5e

� Mn

2+ + 4H2O Oxygen has a –2 oxidation state in MnO4

-, ie oxygen cannot be reduced

further, so it must be Manganese, Mn that gains the electrons. Mn has +7 oxidation state in MnO4-, and +2 in

the manganese ions, it has gained electrons and has been reduced.

e) Br2

Br2 + 2e

� 2Br

+ 4H2O

f) ClO-/H

+

ClO + 2H+

+ 2e � 2Cl

+ H2O Chlorine has an oxidation state of +1 in ClO

- and this is reduced to a –1

oxidation state in the chlorine Cl- ions.

6

g) HNO3

HNO3 + H+ + e

� NO2 + H2O Nitrogen has an oxidation state of +5 in HNO3. This is the highest oxidation state

for nitrogen, so it can only be reduced. Nitrogen in HNO3 is reduced to +4 in NO2

7. The following species are common reducing agents. Deduce oxidation half-equations.

a) Na

Na � Na

+ + e

b) Zn

Zn � Zn

2+ + 2e

c) Mg

Mg � Mg

2+ + 2e

d) Fe

Fe � Fe

2+ + 2e

e) Fe2+

Fe2+

� Fe

3+ + e

f) H2O2

H2O2 � O2 + 2H+ + 2e

g) SO2

H2O + SO2 � SO3 + 2H+ + 2e

h) SO32-

H2O + SO3 � SO4 2 + 2H

+ + 2e

8. Complete and balance the following equations, by giving balanced half equations followed by the

overall redox equation if applicable. For the equations that are redox reactions identify the oxidizing

agent. For reactions that are not redox explain why.

a) H2(aq)+ Br2(g) ����

Oxidation: H2 � 2H

+ + 2e

Reduction: Br2 + 2e � 2Br

Overall: Br2 + H2 � 2Br + 2H

+

H2 is the reducing agent and Br2 is the oxidising agent.

It is easy to tell which element will be reduced and which will be oxidised by simply comparing their electronegativities. Br is very electronegative, so likely to gain electrons.

b) Fe2+

(s) + Cl2(g) ����

Oxidation: Fe2+

(s)+ � Fe3+

+ e-

Reduction: Cl2(g) + 2e- � 2Cl

-

Overall: 2Fe2+

(s) + Cl2(g) � 2Fe3+

+ 2Cl- Iron in a +2 oxidation state, Fe(II), is oxidised to Fe(III),

i.e. the +3 oxidation state.

The Cl2 reduced, therefore the Cl2 is the oxidising agent.

c) N2(g) + H2(g) ����

Oxiation: 3H2 � 6H+ + 6e

Reduction: N2 + 6e � 2N

3

7

Overall: N2 + 3H2 � 2NH3

N2 is the oxidising agent and H2 is the reducing agent.

d) HCl(aq)+ CaCO3(s) ����

This is a reaction of the form acid + carbonate � salt + CO2 + water

Balanced equation with oxidation states of elements assigned:

+1 -1 +2 +4 -2 +2 -1 +4 -2 +1 -2

2HCl(aq) +CaCO3(s) � CaCl2(s)+ CO2(g) + H2O(l)

This is not a redox reaction as none of the elements change oxidation number.

e) C3H8(g) + O2(g) ����

This is a combustion reaction of the form hydrocarbon + oxygen � water + CO2

Balanced equation:

C3H8 + 5O2 � 3CO2 + 4H2O

For a propane fuel cell under acidic conditions the half equations are as follows:

Oxidation: C3H8 + H2O � 3CO2 + 20H+

+ 20e

Reduction: O2 + 4e + 4H+

+ � 2H2O (x 5 to balance electrons)

Overall: C3H8 + 5O2 � 3CO2 + 4H2O

Carbon in C3H8 is the reducing agent and O2 is the oxidising agent.

You have done well if you have worked this out. In an examination situation you would be asked only to balance

the equation for this combustion equation not provide half equations. You can tell which species is being

oxidised by looking at which has oxygen being added to it, here it is the propane, C3H8 oxygen is being added to

it. Working out the numbers of electrons transferred is a little tricky if you only look how the oxidation number of

carbon (it seems that each carbon in C3H8 has an oxidation number of 8/3 or 2 2/3) so look at the oxygen

instead. The oxidation number of oxygen changes from 0 in O2 on the left of the equation to –2 in CO2 and H2O

on the right. There are 10 oxygen’s on each side, so for 10 oxygen’s to be reduced from a 0 to a –2 oxidation

state, 20 electrons are required.

f) CH3OH(l)+ O2(g) ����

Oxidation: CH3OH + H2O � CO2 + 6H+

+ 6e (x 2)

Reduction: 4e + 4H+

+ O2 � 2H2O (x 3)

Overall: 2CH3OH + 3O2 � 2CO2 + 6H2O

Carbon in CH3OH is the reducing agent and O2 is the oxidising agent.

g)

+1 +6 -2 +1 -1 +1 -2 0

H2SO4(aq) + HI(g) ���� H2S(g) + I2 (g) +

Oxidation: 2HI � I2 + 2H+

+ 2e (x 4)

Reduction: H2SO4 + 8e+ 8H+

� H2S + 4H2O

Overall: H2SO4 + 8HI � H2S + 4I2 + 4H2O

Sulfur in the H2SO4 is the oxidising agent and iodine in HI is the reducing agent.

h) CuO(s) + HCl(aq) ����

This is a reaction of the form acid + base � salt + water

Balanced equation: CuO(s) + 2HCl(aq) � CuCl2 + H2O

This is not a redox reaction as none of the elements change oxidation number.

8

i) Mg(s) + Cu

2+ ����

Oxidation: Mg(s) � Mg2+

+ 2e

Reduction: 2e + Cu2+ � Cu(s)

Overall: Mg (s) + Cu2+ � Mg

2+ + Cu (s)

Mg(s) is the reducing agent and Cu2+

is the oxidising agent.

j) MnO4-(aq) + H

+ (aq) + Fe

2+(aq)����

Oxidation: Fe2+

� Fe3+

+ e (x 5)

Reduction: 8H+

+ 5e+ MnO4 � Mn2+

+ 4H2O

Overall: 8H+

+ 5Fe2+

+ MnO4 � Mn2+

+ 4H2O + 5Fe3+

Fe2+

is the reducing agent and manganese in MnO4− is the oxidising agent.

It may be difficult at first to see what the products are going to be, but remember to look at which elements/species are most likely to be oxidised or reduced as a clue to working it out. Iron, Fe

2+ could either be oxidised to Fe

3+ or reduced to Fe(s). Oxygen

can only either remain in the same oxidation state or be oxidised as it is in its lowest oxidation state of –2 and can now only lose electrons (be oxidised). Mn is in a high oxidation state of +7 in MnO4

- so is most likely reduced. From other examples you

have completed, you may have noticed that MnO4- ions are often reduced to Mn

2+ ions

and water, H2O. To reduce Mn with oxidation state +7 to Mn2+

ions, with oxidation state +2, 5 electrons are required.

9. Why are group 1 and 2 metals such good reducing agents?

Group 1 and 2 metals are highly electropositive and readily lose electrons to become positively charged ions

(cations) to achieve a full outer shell. Oxidation is a loss of electrons (remember OIL RIG), and if the species is

oxidised it must therefore be acting as a reducing agent.

10. Magnesium burns in air, giving a very bright light.

a) write the equation for this reaction.

2Mg(s) + O2(g) � 2MgO(s)

b) write the half reaction for the oxygen.

Reduction: O2 + 4e � 2O

2

c) Describe the transfer of electrons in the reaction and hence identify the oxidising agent and

reducing agent.

The oxygen element gains electrons, so it is reduced and is itself an oxidising agent. The magnesium atom

loses two electrons and is oxidised to Mg2+

. It is the reducing agent as it reduces oxygen atoms to O2-

.

11. Use the change in oxidation numbers to identify which of these reactions are redox reactions. If the

reaction is a redox reaction, name the element reduced, the element oxidized, the reducing agent, and

the oxidizing agent.

a)

0 +1 -2 +1 -2 +1 0

2Na(s) + 2H2O(l) ���� 2NaOH(aq) + H2(g)

This is a redox reaction as the oxidation numbers change. Na is oxidized as the oxidation number increases

from 0 to +1, and thus is the reducing agent. Hydrogen is reduced and is the oxidizing agent.

9

b)

+4 -2 +4 -2

N2O4(g) ���� 2NO2(g)

This is not a redox reaction, as the oxidation numbers do not change.

c)

0 +1 -1 0 +1 -1

Cl2(g) + 2NaBr(aq) ���� Br2(g) + 2NaCl(aq)

This is a redox reaction as the oxidation numbers change. Br is oxidized as the oxidation number increases

from -1 to 0, and thus is the reducing agent. Cl is reduced and is the oxidizing agent.

O2 + 4e � 2O

2

d)

+1 +4 -2 +1 +6 -2 +1 +6 -2 +4 -2 +1 -2

Na2CO3(s) + H2SO4(aq)���� Na2SO4(aq) + CO2(g) + H2O(l)

This is not a redox reaction, as the oxidation numbers do not change.

12. Balance these redox equations:

a) HNO3 +H2S ���� S + NO + H2O

Oxidation: H2S � S + 2H+ + 2e (x 5)

Reduction: HNO3 + 5e +

5H

+ � NO + 3H2O (x 2)

Overall: 2HNO3 + 5H2S � 2 NO + 6H2O + S

b) Cl2(g) + KOH(aq) ���� KClO3(aq) + KCl (aq) + H2O

Oxidation: Cl2 + 6H2O � 2ClO3- + 10e

- + 12H

+

Reduction: Cl2 + 2e � Cl- (x 5 to get lower equation)

Overall: 6Cl2(g) + 12KOH(aq) � 2KClO3(aq) + 10KCl (aq) + 6H2O

Remember, you can use the change in oxidation state of the chlorine to work out how to balance this equation, it is often easier than using the half equations.

13. In question 12, name the compounds that have the atom with the highest oxidation number.

The atoms with the highest oxidation number are chlorine in the compound KClO3 (potassium chlorate) with a

+5 oxidation state and nitrogen in nitric acid, HNO3, with a +5 state. (ClO3- = chlorate ion)

9.3 Exercise

1. What is a reactivity series?

A list of elements in order of their reactivity. Those that are higher on the list are more reactive than the

elements lower. They can also displace elements that are lower on the list from a solution of their ions.

2. Why may hydrogen be included in a metal activity series?

Like metals, it can form positive ions, e.g. H+. Comparison to hydrogen allows us to define an element as

electropositive or electronegative. Elements higher in the series than H2 are electropositive and elements lower

than H2 are electronegative.

10

3. Use the reactivity series below to complete the following equations. For any reactions that will not

occur write “no reaction”.

K > Na > Al > Fe > H2 > Cu and F > Cl > I

A displacement reaction can only occur if the metal of the elemental species is higher on the reactivity series

than the metal/hydrogen ion of the ionic species.

a) 2Al + 3H2SO4 � Al2(SO4)3 + 3H2

b) Cl2 + 2KI � 2KCl + I2

c) Cu + FeSO4 �

No reaction, as copper is the elemental species and it is lower on the reactivity series than iron.

d) 2Na + 2H2O � 2NaOH + H2

4. Answer the following questions and for each “yes” response, write a balanced reaction equation:

a) Can H2 reduce Cu2+

ions in an aqueous solution to copper metal?

Yes. H2(g) + Cu2+

(aq) � Cu(s) + 2H+

(aq)

b) calcium metal reduce Pb2+

ions in an aqueous solution to lead metal?

Yes. Ca(s) + Pb2+

(aq) � Pb(s) + Ca2+

(aq)

c) Can lead metal reduce Fe3+

ions in an aqueous solution to Fe2+

?

Yes. Pb(s) + 2Fe3+

(aq) � Pb2+

(aq) + 2Fe2+

(aq)

See the IB Chemistry Data Booklet for a standard electrode potential table. Pb is above Fe

3+ for the reaction Fe

3+ � Fe

2+ + e so this reaction occurs.

Would Pb reduce Fe2+

to Fe(s)?

5. Group l metals react violently with water.

a) Write an equation for the reaction of sodium with water.

2Na(s) + H2O(l) � 2NaOH(aq) + H2(g) A vigorous reaction!

b) Why does reactivity with water increase down group 1?

All group 1 metals react violently with water. The reactivity of the metals increase down the group as the size

and number of shells of electrons of the metal increases. The more shells an atom has, the less tightly held the

outer electrons are and the more reactive they become.

c) Why are group 1 metals such good reducing agents?

Group 1 metals only have one outer shell electron which they lose readily in order to obtain a full outer shell of

electrons. As they easy lose electrons (are oxidised) this makes them excellent reducing agents.

6. Zinc reacts with acids to produce a salt and hydrogen.

a) Write an equation for this reaction

Zn(s) + HCl(aq) � ZnCl2(s) + H2(g)

7. Balance equations for reactions

a) Write balanced chemical equations, with state symbols, for these reactions. If a reaction does

not occur write “no reaction”

2Al(s) + 6HCl(aq) � 2AlCl3(aq) + 3H2(g)

Sn + NaNO3 � No reaction

11

Cu + ZnSO4 � No reaction

2K(s) + 2H2O(l) � 2KOH(aq) + H2(g)

b) Now explain why you have written no reaction where you have.

Sn is the elemental species and cannot displace Na from the NaNO3 salt as Sn is lower on the reactivity series

than Na.

Cu is the elemental species and cannot displace Zn from the ZnSO4 salt as Cu is lower on the reactivity series

than Zn.

8. A student studying the reactivity of four metals in displacement reactions, listed the results as these

ionic equations:

Fe(s) + Cu2+

(aq) ���� Fe2+

(aq) + Cu(s)

Cu(s) + 2Ag+(aq) ���� Cu

2+(aq) + 2Ag(s)

Mg(s) + Fe2+

(aq) ���� Mg2+

(aq) + Fe(s)

a) Deduce a reactivity series based on the behaviour of the metals in the above equations

Iron can displace (or reduce) copper, so it is higher on the reactivity list. Copper can reduce silver. Magnesium

can reduce iron. Therefore in order of most reactive we have the series:

Mg > Fe > Cu > Ag

b) Define oxidation in terms of electron transfer, using one of the equations above as an example

Oxidation is defined as the loss of electrons. In the first equation Fe(s) is being oxidised to Fe2+

(aq) as it is

losing electrons.

c) Define reduction in terms of electron transfer, using one of the equations above as an example

Reduction is defined as the gain of electrons. In the first equation Cu2+

is being reduced to Cu(s) as it is gaining

electrons.

d) Which is the strongest reducing agent and why?

Magnesium is the strongest reducing agent as it is highest on the reactivity series. It is most easily oxidised

metal in the this table, i.e. magnesium most easily loses electrons to form Mg2+

(aq).

e) Which is the strongest oxidising agent and why?

Silver is the strongest oxidising agent as it is lowest on the reactivity series. It is most easily reduced ion, i.e.

silver most easily gains electrons to form Ag(s).

f) Deduce whether a silver coin will react with aqueous magnesium chloride.

No, as the silver is the solid elemental species and is lower on the reactivity series than magnesium, and thus

could not displace it.

9.4 Exercises

1. Define the following terms:

a) electrode

A conductor that collects electrons as they are produced by oxidation reactions in a half cell and then supplies

them to the other half cell where they are used for a reduction reaction.

b) anode

The electrode at which oxidation occurs. In the voltaic cell it is the negatively charged electrode.

c) cathode

The electrode at which reduction occurs. In the voltaic cell it is the positively charged electrode.

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d) electrolyte

The liquid that conducts electricity as a result of the presence of positive or negative ions. Electrolytes are

molten ionic compounds or solutions containing ions, i.e. solutions of ionic salts or compounds that ionise in

solution.

e) half cell

Half of a voltaic cell. It is comprised of an electrode in a solution of ions.

f) salt bridge

An electrical connection between two half cells that completes the electrical circuit and keeps the cells

electrically neutral.

g) voltaic cell

Comprised of two half cells, a voltaic cell produces electrical energy from spontaneous redox reactions.

h) fuel cell

An electrochemical energy conversion device, for example the hydrogen cell. Fuel cells differ from batteries.

They produce electricity from external sources of fuel, which needs to be continually replenished. The hydrogen

cell uses hydrogen as a fuel and oxygen as the oxidant.

2. Primary and secondary cells are both types of voltaic cells. What is the difference between a

primary cell and a secondary cell?

A primary cell is a voltaic cell in which the chemical reaction producing the electricity is not reversible and the

cell cannot be recharged by application of a charging current. For example, non-rechargeable batteries are

primary fuel cells. A secondary cell is a voltaic cell in which the chemical reaction producing the electricity is

reversible and the cell can be recharged. Rechargeable car batteries are made from secondary cells.

3. How does the salt bridge maintain charge balance in the solutions of the half cells?

The salt in the salt bridge contains positive and negative ions in solution. The negative ions move to the anode

to balance the charge from the positive ions produced by oxidation. Similarly the positive ions from the salt

move to the cathode to balance the charge due to the consumption of positive ions which are reduced to the

solid metal.

4. Draw a diagram for a voltaic cell with zinc and copper as the two electrodes and dilute sulfuric acid

as the electrolyte. This is known as the Daniell Cell.

Magnesium is more reactive (is higher in the reactivity series) than iron. This means that magnesium is a

stronger reducing agent, and thus is oxidised to produce Mg2+

cations and electrons. As electrons are produced

by the oxidation of magnesium, the magnesium half-cell forms the anode. Therefore the iron forms on the

cathode, and is reduced from the aqueous cations into the solid metal.

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a) Write balanced half-equations for the reactions occurring at each electrode.

Anode: Zn(s) � Zn2+

(aq) + 2e

Cathode: Cu2+

(aq) + 2e � Cu(s)

b) On your diagram:

i) label each electrode as either the anode or the cathode

ii) show the movement of the ions in the salt bridge

iii) assign a charge to each electrode

c) State which of the two metals is more reactive, give a reason for your choice.

Zinc is higher on the reactivity series than copper and therefore more reactive. It undergoes oxidation from Zn(s)

to Zn2+

.

d) What would you expect to happen if

i) A piece of zinc metal was placed in an aqueous solution of Cu2+

ions

Since zinc is more reactive zinc would undergo oxidation: Zn(s) � Zn2+

(aq) + 2e and the copper ions would be

reduced to copper metal: Cu2+

(aq) + 2e � Cu(s). Overall: Zn(s) + Cu2+

(aq) � Zn2+

(aq) + Cu(s)

ii) A piece of copper metal was placed in an aqueous solution of Zn2+

ions

The copper is not reactive enough to reduce the Zn2+

ions from the solution, so there would be no reaction.

e) What is the main energy conversion in the Daniell cell?

The spontaneous reactions between the electrodes and electrolytes produces a potential difference between

the two electrodes and this causes the flow of electrons between them. This is electricity! The energy

conversion is chemical energy to electrical energy.

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5. Consider the following diagram of a voltaic cell:

anode cathode

electron flow

a) On the diagram, label the anode and cathode, indicate the direction of electron flow through the

wire and the movement of ions in the salt bridge

b) Write half equations for the anode and cathode and an overall cell reaction equation.

Anode: Fe (s) � Fe2+

(aq) + 2e-

Cathode: Ag2+

(aq) + 2e- � Ag(s)

Overall: Fe(s) + Ag2+

(aq) � Fe2+

(aq) + Ag(s)

6. Draw a diagram to show an electrochemical cell using a strip of magnesium in a solution of

magnesium nitrate, and an iron rod in a solution of iron(ll) nitrate.

anode cathode

electron flow

iron electrodemagnesium electrode

magnesiumnitrate solution

iron(ll) nitrate solution

a) Write half equations for the anode and cathode and an overall cell reaction equation.

Anode: Mg(s) � Mg2+

(aq) + 2e-

Cathode: Fe2+

(aq) + 2e- � Fe(s)

Overall: Mg(s) + Fe2+

(aq) � Mg2+

(aq) + Fe(s)

b) Label the anode, the cathode, the positive electrode, the negative electrode and the salt bridge.

c) Give the direction of flow for:

i) electrons in external connecting wire.

in a voltaic cell electrons always flow from anode to cathode.

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ii) nitrate ions in the magnesium nitrate solution.

as more Mg2+

(aq) form, the NO3- will remain in the LHS half cell.

iii) positive ions in the salt bridge if this contains KCl(aq)

move to the cathode the replenish the loss of Fe2+

ions as they are reduced to Fe(s).

iv) iron(ll) ions in the iron(ll) nitrate solution.

Fe2+

are converted to Fe(s) by reduction. The Fe(s) is deposited on the cathode.

d) Where do the electrons flowing in the external wire come from? Where are they going?

The electrons come from the Mg(s) anode as Mg(s) is oxidised to Mg2+

. The electrons move to the cathode

where they are used in the reduction of Fe2+

to Fe(s).

7. The following cell was constructed using two metals X and Y and a solution of their salts, X2+

and

Y2+

. The electron flow when the cell was operating is shown on the diagram.

anode (-)cathode (+)

Write a half equation for the reaction occurring at each electrode.

Anode: Y(s) � Y2+

(aq) + 2e-

Cathode: X2+

(aq) + 2e- � X(s)

a) On the diagram:

i) Show the movement of ions in the salt bridge

ii) Label each electrode as the anode or the cathode and assign a charge

b) Which of the two metals is more reactive?

Y is the more reactive metal as it can reduce the other metal, X2+

to X(s)

c) What would you expect to observe if

i) A piece of metal X was placed in an aqueous solution of Y2+

ions No reaction

ii) A piece of metal Y was placed in an aqueous solution of X2+

ions

The Y(s) metal will reduce X2+

ions to X(s) and would be reduced in size as X(s) deposited.

8. Write the appropriate half-equations for the anode and cathode reactions for the following cells

given the overall reaction:

a) nickel-cadmium cell: Ni2O3(s) + Cd(s) + 3H2O ���� Cd(OH)2(s) + 2Ni(OH)2(s)

Anode: Cd(s) � Cd2+

+ 2e-

Cathode: Ni3+

1e- � Ni

2+

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b) car battery (during discharge): PbO2(s) + Pb(s) + 4H

+(aq) + 2SO

2-4 (aq) ���� 2PbSO4(s) + 2H2O(l)

Anode: Pb(s) � Pb2+

+ 2e-

Cathode: Pb4+

+ 2e- � Pb

2+

9. Which of the following statements is correct?

A: In a voltaic cell the anode is the positive electrode.

B: Increasing the size of the electrode increases the cell voltage.

C: In a voltaic cell the electrons travel in the external wire to the cathode from the anode

D: Within a voltaic cell the anions migrate towards the cathode

Answer: D The positive ions are attracted to the electrons on the cathode.

A: the anode is negative, the electrons are moving out. B: size only increases the cell’s operating time, not its

voltage. C: the electrons travel from anode to cathode.

9.5 Exercises

1. Voltaic and electrolytic cells are both types of electrochemical cells. What is the difference between

a voltaic cell and an electrolytic cell?

In a voltaic cell spontaneous redox reaction generate electricity between two half-cells that contain ions which

are allowed to flow between the two solutions through the salt bridge. Two different metals must be used and

they must be separated from each other. It is the different reaction potentials of the two metals which generates

the electricity.

An electrolytic cell uses electricity from an outside source to generate a non-spontaneous redox reaction.

Another difference is that in the one cell the anode and the cathode are both suspended in the same solution

and they may be made from the same metal.

The suffix –lysis comes from the Greek word meaning to loosen or split up.

2. In a voltaic cell, the anode is negatively charged, it is the negative electrode. In an electrolytic cell,

the anode is positively charged, it is the positive electrode. Explain. (Hint, what types of ions always

flow to the anode/cathode in both a voltaic and electrolytic cells and what is their origin)

Oxidation occurs at the anode and reduction at the cathode in both cases. In the voltaic cell the oxidation

process leaves an excess of electrons at the anode, making it negatively charged. In an electrolytic cell the

anode is positive and the cathode negative, due to the external power supply attached. The cations are

attracted to the cathode (negative electrode) and the power supply provides enough potential for these ions to

be reduced. The anions are attracted to the anode (positively charged) and these are oxidised (gain electrons)

3. Define the terms:

a) Electrolysis

The process by which a current from an external source generates a non-spontaneous redox reaction.

b) Electroplating

The coating of a metal onto the surface of the cathode as the metal ions are reduced at the cathode into the

solid metal.

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4. How is the current conducted during electrolysis?

The current is produced by an external power supply and the electron flow is from anode to cathode.

5. For the electrolysis of molten sodium chloride, draw a fully labelled diagram to show how the ions

move when the circuit is complete, and give the electrode reactions.

The previous page shows a diagram for a general purpose electrolytic cell. For electrolysis of molten NaCl the cell must be heated, the electrodes inert (usually graphite) and the products Cl(g) and liquid Na(l) protected from air, water and themselves. In industry the molten sodium floats up from the cathode and the chlorine gas is kept separate by the shield as shown in the diagram below.

cathode: Na

+(l) + 1e

- � Na(l)

anode: 2Cl-(l) � Cl2(g) + 2e

-

b) Write half equations to show the reactions at each electrode when fused magnesium chloride is

electrolysed.

Anode: 2Cl-(l) � Cl2(g) + 2e

-

Cathode: Mg2+

(l) + 2e- � Mg(l)

Fused in this context means heated until molten. No electrolysis will occur until the salt is molten, i.e. melted until liquid.

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6. A metallic object, to be plated with copper, is one electrode in an electrolytic cell: a bar of pure

copper is the other electrode.

a) State whether the object to be plated is the anode or cathode during the plating process, and

give your reason.

The object to be plated must be the cathode, as reduction always occurs at the cathode so the copper ions will

be reduced at the cathode to solid copper.

b) Name a solution that could be used as the electrolyte.

Copper sulfate, CuSO4(aq).

CuSO4(aq) provides mobile copper ions for transport of current and also for plating. The SO4

2-(aq) are stable and the pure copper anode will react by forming Cu

2+ ions

because it is above water, the other species present that could react, in an activity series.

c) Give half equations for the electrode reactions.

In this case the chlorine anion will be oxidised to chlorine gas at the anode. Reduction of the copper ions into

solid copper will occur at the cathode.

anode: Cu(s) � Cu2+

(aq) + 2e (the electrode itself reacts)

cathode: Cu2+

(aq) + 2e � Cu(s)

7. The Edison cell is a storage battery that may receive hard treatment and yet gives good service for

years. It may be left uncharged indefinitely and still be recharged. It provides 1.3 V. Its electrolyte is

21% potassium hydroxide, KOH(aq), to which a small amount of lithium hydroxide, LiOH, is added.

The reaction on discharge is:

Fe(s) + 2Ni(OH)3 ���� Fe(OH)2 + Ni(OH)2

a) Give the electrode reactions during discharge and charging.

discharge

cathode: Ni3+

(aq) + e- � Ni

2+(aq)

anode: Fe(s) � Fe2+

(aq) + 2e-

Fe is higher on the reactivity series than Ni3+

(aq) so is able to reduce Ni3+

ions to Ni2+

charging

cathode: Fe2+

(aq) + 2e- � Fe(s)

anode: Ni2+

(aq) � Ni3+

+ e-

The external power supply forces the iron ions to be reduced back to Fe(s) and the Ni2+

oxidised to Ni3+

.

The Edison cell The Edison cell is a rechargeable battery. When it is discharging (i.e. being used) it acts as a voltaic cell. The spontaneous redox reactions that occur as a result of the different reactivities of the metals and their ions generate electricity. When the battery is being charged however, it acts as an electrolytic cell. The external power supply forces the redox reactions to happen in the opposite direction so that it is ready for use again.

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b) For discharge, which electrode is the anode? Is this the same as for the charging? Explain.

The iron electrode is the anode. It is being oxidised. No. During charging the nickel ions, Ni2+

(aq), are being

oxidised so the electrode in their solution is the anode, not the Fe.

Oxidation occurs at the anode and reduction at the cathode, for both discharge AND for charging (i.e. when acting as voltaic or electrolytic cells). The difference is which metal or ion is being reduced. When discharging, the most reactive metal is oxidised at the anode in a spontaneous redox reaction. When charging the most reactive metal ion is reduced at the cathode by the external power supply.

8. Electrolysis occurs during the recharging of a car battery.

During discharge the overall reaction in a car battery is:

PbO2(s) + Pb(s) + 4H+(aq) + 2SO4

2- (aq) ���� 2PbSO4(s) + 2H2O(l)

a) Rewrite this equation to show the overall reaction during charging.

2PbSO4(s) + 2H2O(l) � PbO2(s) + Pb(s) + 4H+(aq) + 2SO4

2- (aq)

b) Write half-equations for the anode and cathode reactions occurring during charging.

anode: Pb(s) � Pb2+

+ 2e-

cathode: PbO2 + 4H+(aq) + 2e

- � Pb

2+(aq)

+ 2H2O

9. Why do solid salts not conduct electricity? What are two methods by which salt can be used to

conduct electricity?

Solid salts do not conduct electricity, as the ions are not able to physically move when they are held in the solid

state in the crystal lattice, and there can be no movement of electrons. Aqueous salts and molten salts are both

able to conduct electricity because the ions and electrons are able to move freely.

10. What products are discharged at the anode and cathode during the electrolysis of the following

molten salts:

a) KBr

Cathode: K(l)

Anode: Br2(g)

b) NaCl

Cathode: Na(l)

Anode: Cl2(g)

c) MgCl2

Cathode: Mg(l)

Anode: Cl2(g)

The metals are molten because of the high temperatures required to keep the salts in the liquid state. These active metals must not be allowed to come into contact with air or moisture or with the elemental halogens, so they are stored under paraffin oil.

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