The reaction of bro3- and br- in acidic solution

Redox reaction is also called Oxidation-Reduction Reaction. Sometimes the equations are easy enough to balance using the most common method we use for balancing the other reactions, but sometimes redox reactions can be complicated and difficult to figure out in 1 step. So we break down the equation and use half reactions to balance the equation.

Here is an example:

Balance the Redox Reaction in acidic solution:      BrO3-(aq)+SN2+(aq)--> Br-(aq)+Sn4+(aq)

First of all, we can split the equation into two half reactions first and get:

BrO3-(aq)--> Br-(aq)

Sn2+(aq)--> Sn4+(aq)

Then we can use water H2O to balance the oxygen atoms O:

BrO3-(aq)--> Br-(aq) + 3H2O(l)

Now we use H+ to balance the above equation, since this is acidic solution (given in question):

BrO3-(aq) + 6H+(aq) --> Br-(aq) + 3H2O(l)

Then we add electrons to the left side in order to balance the charge and get this:

BrO3-(aq) + 6H+(aq) + 6e- --> Br-(aq) + 3H2O(l)   (Half Equation 1)

Now it’s time to fix Sn2+(aq)--> Sn4+(aq) as well, using the same method:

Sn2+(aq)--> Sn4+(aq) + 2e-       

Here in order to cancel the electrons on both hal equations when adding them up, we need to multiply the above equation by 3:

3Sn2+(aq)--> 3Sn4+(aq) + 6e-     (Half Equation 2)  

We are almost done, just add Half Equation 1&2 together:

BrO3-(aq) + 6H+(aq) + 3Sn2+(aq) --> Br-(aq) + 3H2O(l)+ 3Sn4+(aq)    

You see you fast that was? By the way, we use H+ to balance the equation because this reaction takes place in acidic solution. If there’s a question asking to balance a redox reaction in basic solution, then use H+ first, then use OH- to cancel out the H+ later. Some people may ask why use H+ first all the time. But most people’s experience show that using H+ then OH- is faster than just using OH- (it takes more time to think if just use OH- and it will easily mess up the answer.)

rate constant of overall reaction is 4 `sec^(-1)`rate of reaction is independent of the conc. Of acidthe change in pH of the solution will not affect the ratedoubling the conc. Of `H^(+)` ions will increase the reaction rate by 4 times

Solution : Since the rate law contains `[H^(+)]` term. This means that rate changes variation in the value of pH.

• Let us investigate oxidation and reduction reactions that involve the transfer of electrons at a distance.

Laboratory Activity 1A Transfer of Electrons at a Distance

Aim: To investigate oxidation and reduction reactions in terms of PAK 21 Learning Science
electron transfer at a distance. Through Inquiry

Materials: 1.0 mol mpdrmoepl−3adromefd−s)3u,ol0pf.h1suomrdioiculamdcmidh−,y3Hdorf2oSaxOciidd4,ie0f,i.e2NdmapOooHtladssmsoiu−lu3mtoifmoinra.onng(aInI)atseu(lVpIhIa),teK, MFenSOO44 solution
(freshly solution
and 0.5

Apparatus: U-tube, connecting wires with crocodile clips, galvanometer, retort stand, carbon
electrodes, dropper and test tube.
Mind
Procedure: G Challenge Carbon electrode must not
touch the layer of sulphuric
acid, H2SO4. Why?
X Y Carbon electrodes, C

Iron(II) sulphate, Acidified potassium The negative terminal is the electrode
FeSO4 solution manganate(VII), KMnO4 where electrons are released.
Sulphuric acid, H2SO4 solution The positive terminal is the electrode
where electrons are accepted.
U-tube

Figure 1.2

1. CPusoaiurnergf1ua.l0ldymr,opopolpudermr0u−.32notmiflsotuhlledpmhsou−lr3uioctifaoicnriodrn,eH(aIcI2hS)eOssu4tliphnhetoahtetehi,geFhUetS-oOtfu43bsecomulun.ttiiolnhainlftfouallramndXcolafmthpe it vertically.
2. U-tube

3. CaramreYfuollfy,thpeouUr-0tu.1bemuoslindgma−3dorfoapcpiedrifuiendtipl othtaesssoiulumtiomnarnegaacnhaetse(thVeIIh)e, iKgMhtnoOf 34 solution into
cm.

7

Theme 1 Chemical Process

4. Connect the carbon electrodes to the galvanometer using the connecting wires.
5. DcthaierpbcooirnnceueloietfcattshroeshdcoeawrinbnotoinntehFleeigcautcrriodedi1fei.se2d.inptoottahsesiiurmonm(IIa)nsgualnpahtaet(eV, FIIe)S, OK4MsonlOut4isoonluwtihoinlettohceoomthpelrete
6. Observe the direction of the deflection of the galvanometer needle and determine the positive
and negative terminals for each electrode.
7. Leave the apparatus set-up for 30 minutes.
8. Observe the colour change of the iron(II) sulphate, FeSO4 solution and the acidified potassium
9. Rmeacnogradnyaoteu(rVoIbI)s,eKrvMatnioOn4 solution.
in the table below. solution using a dropper and
10. After 30 minutes, draw out a little of the iron(II) sulphate, FeSO4
pour it into a test tube.
11. Add sodium hydroxide, NaOH solution into the test tube to verify the product formed.

Data and observation: Observation Inference
Solution

AIrcoind(iIfIie)dsuplopthaastseiu, FmeSmOa4nganate(VII), KMnO4

Discussion:
1. Based on your observation, write the half equations for electrode X and electrode Y.
2. What are the types of reactions that take place at electrode X and electrode Y?
3. Write the overall ionic equation for the reaction that takes place.
4. Identify the oxidised substance, reduced substance, oxidising agent and reducing agent.
Give reasons for your answers.
5. State the direction of the electron flow in the experiment.
6. State the positive terminal and negative terminal of the experiment.
7. What is the function oUf-stuulbpeh, udrriacwacaidla, bHel2lSeOd 4s?et-up
8. Other than using the of apparatus that can be used for the
experiment to investigate the transfer of electrons at a distance between potassium iodide, KI
solution and chlorine water, Cl2.

Prepare a complete report after carrying out this laboratory activity.

In this laboratory activity, it is found that:

The deflection of the galvanometer needle shows that there is a transfer of electrons
through the connecting wires from the reducing agent (negative terminal) to the
oxidising agent (positive terminal).

The colour change of the solutions also shows that the oxidation and reduction
reactions have taken place.

Iron(II) asurelpsheaptaer,aFteedSOby4 solution and acidified potassium manganate(VII), KMnO4
solution an electrolyte which is sulphuric acid, H2SO4.

Electrons move from iron(II) ion, Fe2+ to acidified manganate(VII) ion, MnO4− through
the connecting wires.

8

Redox Equilibrium Chapter 1

Iron(II) sulphate, FeSO4 solution G Acidified potassium manganate(VII),
(Reducing agent) e− K(OMxindOis4insoglaugtieonnt)
e−

Iron(II) ion, Fe2+ releases an Ϫ ϩ
Manganate(VII) ion, MnO4− receives electrons
electron to form iron(III) ion, Fe3+ to form manganese(II) ion, Mn2+
Fe2+ → Fe3+ + e− MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
(green (brown (purple (colourless
solution) solution) solution) solution)

Redox: Voltaic cell
on page 27.

Figure 1.3 The set-up of apparatus to investigate the transfer of electrons at a distance

• Do you know how to pwortiatesstihuemhmalaf negqaunaatitoe(nVfIoIr),iKroMn(nIOI)4ssuollpuhtiaotne,aFsesShOo4wsnoliuntFioinguarned1.t3h?e half
equation for acidified

Writing half equation for oxidation:

e green colour of iron(II) sulphate, FeSO4 changes to brown

1 Write the reactant and product Fe2+(aq) → Fe3+(aq)
2 Add the electron to the side
(green) (brown)
that has more positive charges
to balance the charges Fe2+(aq) → Fe3+(aq) + e−
Charge = +2 Charge = +3 Charge = -1
Total charge = +2 Total charge = (+3) + (-1) = +2

Writing half equation for reduction:

e purple colour of acidi ed potassium manganate(VII), KMnO4 solution changes to colourless

→Mn(pOur4p−(lea)q)
1 Write the reactants + H+(aq) Mn2+(aq)
and product
(acidic condition) (colourless)

In acidic condition, the reaction produces water

2 Balance the number of →MnO4−(aq) + 8H+(aq) Mn2+(aq) + 4H2O(l)
atoms for each element
→MChnaOrg4e−(=aq-)1 Mn2+(aq) +
3 Sum up the total charges + 8H+(aq) Charge = +2 C4Hha2Org(el)= 0
of reactants and products Charge = +8
Total charge = (-1) + (+8) Total charge = (+2) + 0
4 Add electrons to the side = +7 = +2
that has more positive →MnO4−(aq)
charges to balance + 8H+(aq) + 5e− Mn2+(aq) + 4H2O(l)
the charges Charge = -5
Total charge = (-1) + (+8) + (-5) Total charge = (+2) + 0
= +2 = +2

9

Theme 1 Chemical Process

• Based on the half Bab 09/1 Video2 B01-10a Bab 09/1 Video3 B01-10b
equation of oxidation Writing Ionic Equationhttp://kubupublication.com.my/Kimia/Tingkatan5/Video2.html Oxidising Agent andhttp://kubupublication.com.my/Kimia/Tingkatan5/Video3.html
and the half equation of https://bit.ly/kpkt5v2 Reducing Agent
reduction, what is the
overall ionic equation https://bit.ly/kpkt5v3
for the redox reaction in
Laboratory Activity 1A?

Half →Fe2+ Fe3+ + e– … ➀ Number of electrons in equations
equations:
→MnO4– + 8H+ + 5e– Mn2+ + 4H2O … ➁ ➀ and ➁ must be the same

→5Fe2+ 5Fe3+ + 5e– … ➀ × 5 Sum up equations
→MnO4– + 8H+ + 5e–
Mn2+ + 4H2O … ➁ ➀ and ➁

→5Fe2+ + MnO4– + 8H+ + 5e– 5Fe3+ + 5e– + Mn2+ + 4H2O

Ionic →5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H2O
equation:

• Table 1.1 shows examples of commonly used oxidising agents, whereas
Table 1.2 shows examples of commonly used reducing agents.

Table 1.1 Examples of oxidising agents

Oxidising agent Half equation
Acidified potassium manganate(VII), MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)
KMnO4
Acidified potassium dichromate(VI), Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l)
K2Cr2O7
Chlorine water, Cl2 Cl2(aq) + 2e− → 2Cl−(aq)
Br2(aq) + 2e− → 2Br−(aq)
Bromine water, Br2 Fe3+(aq) + e− → Fe2+(aq)
H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l)
Iron(III) chloride, FeCl3

Acidified hydrogen peroxide, H2O2

Table 1.2 Examples of reducing agents

Reducing agent Half equation
Potassium iodide, KI 2I−(aq) → I2(aq) + 2e−
2Br−(aq) → Br2(aq) + 2e−
Potassium bromide, KBr Fe2+(aq) → Fe3+(aq) + e−

Iron(II) sulphate, FeSO4 Zn(s) → Zn2+(aq) + 2e−
Reactive metals (e.g. zinc, Zn)

Sulphur dioxide, SO2 SO2(g) + 2H2O(l) → SO42−(aq) + 4H+(aq) + 2e−

10

Redox Equilibrium Chapter 1

Oxidation and Reduction in Terms of Change in Oxidation Number

• Chemists found that the concept of electron transfer in explaining redox reactions is limited
to the reactions associated with the ionic compounds. Hence, the definitions of oxidation and
reduction are extended by using the concept of oxidation numbers in describing redox reactions
to cover all reactions.
Oxidation reaction takes place when the oxidation number of an element increases.
Reduction reaction takes place when the oxidation number of an element decreases.

• An example of the reaction is between carbon, C and oxygen, O2 to produce carbon dioxide, CO2.
Oxidation

(Oxidation number increases)

C(s) + O2(g) → CO2(g)

Oxidation Oxidation Oxidation Oxidation
number = 0 number = 0 number = +4 number = -2 = -2

Reduction
(Oxidation number decreases)

Oxidised substance: Carbon, C Oxidation number of carbon, C increases

Reduced substance: Oxygen, O2 Oxidation number of oxygen, O decreases
Oxidising agent: Oxygen, O2 Oxygen, O2 oxidises carbon, C
Reducing agent: Carbon, C Carbon, C reduces oxygen, O2

• Conclusion:
(a) An oxidising agent is a substance that oxidises other substances and is reduced in
a redox reaction.
(b) A reducing agent is a substance that reduces other substances and is oxidised in
a redox reaction.
A ctivity 1C
Carry out this activity in groups. PAK 21

Table 1.3 shows the chemical equations for two examples of redox reactions.
(a) Discuss why the reactions in Table 1.3 can be
Table 1.3

classified as redox reactions. Chemical equation
(b) For each reaction, identify the following: CuO(s) + H2(g) → Cu(s) + H2O(l)
(i) Oxidised and reduced substances. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2 (g)
(ii) Oxidising and reducing agents.
Present the outcome of your group discussion in your class in a Gallery Walk activity.

11

Theme 1 Chemical Process

Determination of Oxidation Number Bab 09/1 Video4 B01-12 Redox in Terms of
http://kubupublication.com.my/Kimia/Tingkatan5/Video4.html
What is oxidation number?
• The oxidation number or oxidation state is the charge of the Oxidation Number
https://bit.ly/kpkt5v4
elements in a compound if the transfer of electrons occurs in
an atom to form chemical bonds with other atoms.
• The oxidation number of an element in an ion or a compound
can be determined based on the following guidelines:

1 The oxidation number for all 2 The oxidation number for a monoatomic
elements is zero. ion is equal to the charge of the ion.

Element Oxidation number Monoatomic ion Oxidation number
Sodium, Na 0 Copper(II) ion, Cu2+ +2
0 +1
Carbon, C 0 Potassium ion, K+ -1
0 Bromide ion, Br− -2
Helium, He 0 +3
Oxide ion, O2−
Oxygen, O2
Chlorine, Cl2 Aluminium ion, Al3+

3 The oxidation number of the element in its compound for Group 1, Group 2 and
Group 13 in the Periodic Table are +1, +2 and +3, respectively.

Compound Oxidation number Compound Oxidation number
Lithium chloride, LiCl +1 Magnesium oxide, MgO +2

Sodium oxide, Na2O +1 Aluminium chloride, AlCl3 +3

4 The sum of oxidation numbers of the element in a compound is zero.

Example: Determine the oxidation number of
The sum of oxidation numbers for sodium chloride, NaCl. sulphur in sulphur trioxide, SO3.
The oxidation number of Na + oxidation number of Cl x + 3(-2) = 0
= (+1) + (-1) = 0 x = +6
The oxidation number of S = +6

5 The sum of oxidation numbers of the element in a polyatomic ion is equal to the
charge of the ion.

Example: Determine the oxidation number of
The sum of oxidation numbers for ammonium ion, NH4+. carbon in carbonate ion, CO32−.
The oxidation number N + Oxidation number 4H x + 3(‒2) = -2
= (‒3) + 4(+1) = +1 x = +4
The oxidation number of C = +4

12

Redox Equilibrium Chapter 1

6 The oxidation number of hydrogen in a compound is normally +1 except in metal
hydrides which is -1.

The oxidation number of hydrogen in hydrogen Determine the oxidation number of hydrogen in
chloride, HCl. sodium hydride, NaH.
x + (‒1) = 0 +1 + x = 0
x = +1 x = -1
The oxidation number of H = +1 The oxidation number of H = -1

7 The oxidation number of oxygen in a compound is normally -2, except in peroxides
which is -1.

The oxidation number of oxygen Determine the oxidation number of oxygen in hydrogen peroxide, H2O2.
in magnesium oxide, MgO. 2(+1) + 2x = 0
(+2) + x = 0 2x = -2
x = -2 x = -1
The oxidation number of O = -2 The oxidation number of O = -1

8 The oxidation number of the elements in Group 17 in a compound is normally -1.
• The oxidation number of fluorine is always -1.
• The oxidation number of chlorine, bromine and iodine is usually -1, except when they are
bound to a more electronegative element, such as oxygen. Thus, their oxidation number
will have a positive value.

The oxidation number of fluorine The oxidation number of chlorine The oxidation number of bromine
in hydrogen fluoride, HF. in potassium chlorate(I), KClO. in bromate(V) ion, BrO3−.
(+1) + x = 0 (+1) + x + (-2) = 0 x + 3 (‒2) = -1
x = -1 x = +1 x = +5
The oxidation number of F = -1 The oxidation number of Cl = +1 The oxidation number of Br = +5

Bab 09/1 Video5 B01-13a Bab 09/1 Nota1 B01-13b
Determining the Oxidation Numberhttp://kubupublication.com.my/Kimia/Tingkatan5/Video5.html Reinforcement Exercises inhttp://kubupublication.com.my/Kimia/Tingkatan5/Nota1.pdf
https://bit.ly/kpkt5v5 Determining the Oxidation Number

https://bit.ly/kpkt5n1

A ctivity 1D 1. MnO4− 3. AlH3 5. Cu2O 7. VO2+
2. Ba(OH)2 4. HOBr 6. C2O42− 8. S8
Determine the oxidation number
for the underlined elements.
Show all workings, if any.

13

Theme 1 Chemical Process

The Oxidation Number and Naming of Compounds According
to the IUPAC Nomenclature

• Transition elements are metals that normally According to the IUPAC nomenclature, metals
exhibit more than one oxidation number in that have only one oxidation number in their
their compounds. compounds do not need Roman numerals in
the naming. For example, Mg(NO3)2 is named
• According to the IUPAC nomenclature, Roman as magnesium nitrate.
numerals are used to indicate the oxidation
number of the metals in their compounds.

• For example, Figure 1.4 shows iron that forms
two types of nitrates.

Iron Fe(NO3)2 The oxidation number of Fe = +2 Roman numeral (II)
Compound Fe(NO3)3 Iron(II) nitrate indicates the oxidation
number of Fe is +2
The oxidation number of Fe = +3
Iron(III) nitrate Roman numeral (III)
indicates the oxidation
number of Fe is +3

Figure 1.4 Naming according to the IUPAC system for iron compound

• Table 1.4 shows the naming of compounds containing metals that have more than one oxidation
number according to the IUPAC nomenclature.

Table 1.4 Compound names according to the IUPAC nomenclature

Compound formula Oxidation number IUPAC name

MnO2 The oxidation number of manganese is +4 Manganese(IV) oxide

KMnO4 The oxidation number of manganese is +7 Potassium manganate(VII)

CrCl3 The oxidation number of chromium is +3 Chromium(III) chloride

K2Cr2O7 The oxidation number of chromium is +6 Potassium dichromate(VI)

A ctivity 1E PAK 21
Carry out this activity in pairs.
1. State the name of the following compounds according to the IUPAC nomenclature.
(a) Cu2O (b) CuO (c) Hg2Cl2 (d) HgCl2 (e) K2CrO4
2. Write the formula for the following compounds.
(a) Vanadium(V) oxide (b) Sodium oxide (c) Lead(II) carbonate

3. State the name of the following two compounds Fe2O3 Al2O3
according to the IUPAC nomenclature and explain
the reason for the difference of those names.

Record the findings in your notebook.

14

Redox Equilibrium Chapter 1

Conversion of Iron(II) Ion, Fe2+ to Iron(III) Ion, Fe3+ and Vice Versa

• Compound of iron(II) ion, Fe2+ and compound of iron(III) ion,
Fe3+ are widely used in our lives. Photograph 1.3 shows iron pills,
which contain iron(II) gluconate, wFeo(mCe6Hn.11O7)2, that are prescribed
to anaemic patients and pregnant
• The presence of iron(II) ion, Fe2+ and iron(III) ion, Fe3+ can be
verified by using sodium hydroxide, NaOH solution, aqueous
asoomrlpumotiotoannsis,ai,upmoNtatHshs3ii,oucmpyaonthaaestxseai,ucKmySanCohNfeexsroarcaluytetai(noIoInIf.)e,rrKa3t[eF(IeI()C, NK)64][Fseo(lCuNtio)6n]
Photograph 1.3
Iron pills iron(II) gluconate,

Fe(C6H11O7)2

Conversion of Iron(II) Ion, Fe2+ to Iron(III) Ion,

Laboratory Activity 1B Fe3+ and Vice Versa

Aim: To study the conversion of iron(II) ion, Fe2+ to iron(III) ion, Fe3+ PAK 21 Learning Science
and vice versa. Through Inquiry

Materials: 0.5 mol mddmmol−−33doomff−iir3rooonnf((sIIoIIId))isucuhmllpohhriyaddteer,,oFFxeeiCSdOel3,4sNosolauOlutiHtoionsn,obl(rufortemioshninl.ye prepared), zinc powder, Zn
0.5 mol water, Br2,
and 2.0

Apparatus: Boiling tube, test tube, boiling tube holder, dropper, spatula,
filter funnel, filter paper, Bunsen burner and test tube rack.

Procedure: Bromine water, Br2

A. Conversion of iron(II) ion, Fe2+ to iron(III) ion, Fe3+ Iron(II) sulphate,
1. (Pforeusrh2lycpmr3epoafr0e.d5)minotlodamb−o3 iolifnigrotnu(bIeI)hseuldlpbhyatae,bFoeilSinOg4 solution FeSO4 solution
tube Figure 1.5
holder as shown in Figure 1.5.
2. bUosiilninggthtuebderoupnptielrn, oadfudrbthroermcionleowuratcehra,nBgre2 while shaking the Zinc powder, Zn
can be observed.
3. Gently heat the mixture in the boiling tube. Iron(III) chloride,
4. Add sodium hydroxide, NaOH solution into the mixture to FeCl3 solution
confirm the presence of iron(III) ion, Fe3+. Figure 1.6
5. Record all observations.

B. Conversion of iron(III) ion, Fe3+ to iron(II) ion, Fe2+
1. Pour 2 cm3 of 0.5 mol dm−3 of iron(III) chloride, FeCl3 solution
into a boiling tube as shown in Figure 1.6.
2. Add half a spatula of zinc powder, Zn into the solution.
3. Gently heat the mixture.
4. Filter the mixture into a test tube.
5. Add sodium hydroxide, NaOH solution into the filtrate
to confirm the presence of iron(II) ion, Fe2+.
6. Record all observations.

Data and observation:
Construct a suitable table to record all observations in section A and section B.

15

Theme 1 Chemical Process

Discussion:
1. For each conversion in sections A and B, identify the substances that are oxidised or reduced.
Give reasons for your answers in terms of:
(a) change in the oxidation number. (b) transfer of electron.
2. What is the role of bromine water, Br2 in section A and zinc powder, Zn in section B?
Give your reasons.
3. Why is freshly prepared iron(II) sulphate, FtheeSOpr4essoelnuctieoonf used for the experiment in section A?
4. Name a reagent that can be used to verify iron(III) ion, Fe3+ in section A and
iron(II) ion, Fe2+ in section B.
5. Suggest another substance that can replace bromine water, Br2 in section A and zinc powder, Zn
in section B.

Prepare a complete report after carrying out this laboratory activity.

• Conversion of iron(II) ion, Oxidation reaction
Fe2+ to iron(III) ion, Fe3+ • e oxidation number of iron increases from +2 to +3
and vice versa in • Iron(II) ion, Fe2+ releases an electron to produce iron(III) ion, Fe3+

Figure 1.7 shows Fe2+ Fe3+
that the change in the Oxidation number = +2 Oxidation number = +3

oxidation number in Reduction reaction
the redox reaction is
associated with the • e oxidation number of iron decreases from +3 to +2
transfer of an electron. • Iron(III) ion, Fe3+ accepts an electron to produce iron(II) ion, Fe2+

Figure 1.7 Conversion of iron (II) ion, Fe2+ to iron(III) ion, Fe3+ and vice versa

Displacement Reaction of Metal from Its Salt Solution

• Displacement of metal is carried out by adding a metal into a salt solution of another metal.
A more electropositive metal is able to displace a less electropositive metal from its salt solution.

• Figure 1.8 shows a piece of zinc plate, Zn is added into a test tube containing copper(II) nitrate,
Cu(NO3)2 solution. The observation and inference for the reaction are as follows:

Observation Inference

The blue colour of • The concentration of copper(II) ions,
csoolpuptieorn(IIb)encoitmraetes,pCaule(rN. O3)2 Cu2+ decreases.

Copper(II) Brown solid is deposited. • Solid copper, Cu is formed.
snoitlruattioe,nCu(NO3)2 The zinc plate, Zn becomes • Copper(II) ion, Cu2+ receives two
Zinc, Zn thinner.
Figure 1.8 electrons and forms copper atom, Cu.
• Copper(II) ion, Cu2+ is reduced.
• Zinc atom, Zn ionises to form

zinc ion, Zn2+.
• Zinc atom, Zn releases two electrons

to form zinc ion, Zn2+.
• Zinc, Zn is oxidised.

16

Redox Equilibrium Chapter 1

• Half equation, ionic equation and chemical equation for the reaction can be constructed by
referring to the observations of and inferences from the reaction that takes place.

→Oxidation half equation: Zn(s) Zn2+(aq) + 2e− (zinc atom, Zn releases electrons)

→Reduction half equation: Cu2+(aq) + 2e− Cu(s) (copper(II) ion, Cu2+ receives electrons)

Ionic equation: →Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Chemical equation:
→Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s)

• Electropositivity is the tendency of atoms to release electrons to form cations.
• Electrochemical series is a series of metal arrangements according to the standard electrode
potential, E0 in the order of most negative to most positive. More electropositive metals
are stronger reducing agents because the E0 value is more negative; so, atoms are easier to
lose electrons. Figure 1.9 shows the Electrochemical Series that is part of the standard
electrode potential.
Redox: Standard electrode potential on page 23.
Redox: Corrosion of metal on page 52.
Thermochemistry: Heat of displacement on page 125.

Metal ion Metal
(Oxidising agent) (Reducing agent)
⇌K+ + e−
⇌Ca2+ + 2e− K E0 = -2.92 V More electropositive metals are located at the
higher position in the electrochemical series.
Ca E0 = -2.87 V
⇌Na+ + e−e strength as oxidisng agent increases e reaction that takes place involves the transfer
⇌Mg2+ + 2e− e strength as reducing agent increasesNa E0 = -2.71 Vof electrons from metal atom to metal ion.

⇌Al3+ + 3e− Mg E0 = -2.38 V e more electropositive the metal, the easier
it is for the metal atom to release electrons.
⇌Zn2+ + 2e− Al E0 = -1.66 V Metals at the top of the electrochemical series
are the stronger reducing agents.
⇌Fe2+ + 2e− Zn E0 = -0.76 V In contrast, the lower the position of the metal
ions in the electrochemical series, the easier it
⇌Sn2+ + 2e− Fe E0 = -0.44 V is for the metal ions to receive electrons.
Metal ions that are located at the bottom of
⇌Pb2+ + 2e− Sn E0 = -0.14 V the electrochemical series are the stronger
oxidising agents.
⇌2H+ + 2e− Pb E0 = -0.13 V
⇌Cu2+ + 2e−
⇌Ag+ + e− H2 E0 = 0.00 V
Cu E0 = +0.34 V

Ag E0 = +0.80 V

Figure 1.9 Electrochemical Series

• By using a suitable metal, can you explain the redox reaction that takes place in the displacement
of silver, Ag from its salt solution?

Mind What can you observe when a piece of copper, Cu is added into a test tube
Challenge containing magnesium sulphate, MgSO4 solution? Does displacement reaction
take place? Explain your answer.

17

Theme 1 Chemical Process

Laboratory Activity 1C Displacement of Metal from Its Salt Solution

Aim: To investigate a redox reaction in the displacement of metal from its PAK 21 Learning Science
salt solution. Through Inquiry

Materials: Magnesium ribbon, Mg, lead plate, Pb, copper plate, Cu, 0.5 mol dm−3 of lead(II)
nitrate, Pmbo(lNdOm3)−23 solution, 0.5 mol dm−3 of magnesium nitrate, Mg(NO3)2 solution
and 0.5 of copper(II) nitrate, Cu(NO3)2 solution.

Apparatus: Test tube, test tube rack and sandpaper.

Procedure:
1. Using the metal plates and salt solutions provided, plan an experiment to investigate
a redox reaction in the displacement of metal from its salt solution.
2. Write clear steps of the investigation.
3. Record all observations.

Results:
Construct a suitable table to record all observations and inferences.

Discussion:
1. For each set of experiment:
(a) Write the oxidation half equation, reduction half equation and overall ionic equation.
(b) Identify:
(i) Oxidised substance. (iii) Oxidising agent.
(ii) Reduced substance. (iv) Reducing agent.

(c) Give reasons for your answers in (b)(i), (ii), (iii) and (iv).

Prepare a complete report after carrying out this laboratory activity.

Displacement Reaction of Halogen from Its Halide Solution

• The atom of halogen elements has the tendency to receive electrons to form halide ions. Therefore,
halogens undergo reduction reactions and act as oxidising agents.

The reactivity of halogen • Halogens become more difficult to receive electrons.
decreases down Group 17 Therefore, the strength of these halogens as oxidising
agents decreases.

• Conversely, it is easier for the halide ions to release their
electrons, and the strength of halide ions as reducing
agents increases.

• The displacement of a halogen is carried out by adding a halogen into a solution of another
halide. The halogen at the top of Group 17 (more reactive) can displace the halogen at the bottom
(less reactive) from its halide solution.

• Figure 1.10 shows ochf alonrdinienwfearteenrc, eCsl2friosmadtdheedrienatcotiaotnesatrteuabse containing potassium bromide, KBr.
The observations follows.

18

Redox Equilibrium Chapter 1

Chlorine Observation Inference
water, Cl2 The colour of potassium bromide, • BBrroommiidnee,iBonr2, is formed.
KBr changes from colourless to • Br− releases electrons
Potassium brown. to form bromine mis oolxeicduilsee,dB.r2.
bromide, KBr • Bromide ion, Br−
solution
Figure 1.10 The colour of chlorine wyealtleorw, Ctol2 • Chlorine mtoofloercmulec,hClol2riadceceiopnts, Cl−.
changes from greenish electrons
colourless. • Chlorine, Cl2 is reduced.

• The half equation, ionic equation and chemical equation for the reaction can be constructed by
referring to the observations of and inferences from the reaction that takes place.

→Oxidation half equation: 2Br−(aq) Br2(aq) + 2e− (bromide ion, Br− releases electrons)

→Reduction half equation: Cl2(aq) + 2e− 2Cl−(aq) (chlorine molecule, Cl2 accepts electrons)

Ionic equation: 2Br−(aq) + OCxli2d(iasiqn)g → Br2(aq) + 2Cl−(aq)

Reducing agent
agent

Electron transfer

Chemical equation: →2KBr(aq) + Cl2(aq) Br2(aq) + 2KCl(aq)
Colourless Greenish
solution yellow Brown Colourless
solution

Halogens Halide ions e more reactive the halogen, the easier it
(Oxidising agent) (Reducing agent) is for the halogen to accept an electron.

e strength as oxidisingF2 + 2e− 2F− Halogens that are located at the top of
agent increasesCl2 + 2e− 2Cl− Group 17 are stronger oxidising agents.
Br2 + 2e− 2Br− In contrast, the lower the position of the
e strength as reducingI2 + 2e−2I− halide ions in Group 17, the easier it is for
agent increasesAt2 + 2e−2At− the halide ions to release an electron.
Halide ions that are located at the
bottom of Group 17 are stronger
reducing agents.

Figure 1.11 Group 17 elements

• Using a suitable halogen, can you Mind What can be observed when iodine, I2
solution is added into a test tube
explain the redox reaction that takes Challenge containing potassium bromide, KBr?
place in the displacement of iodine, I2
from its halide solution? Does a displacement reaction take
place? Explain your answer.

19

Theme 1 Chemical Process

Laboratory Activity 1D Displacement of a Halogen from Its Halide Solution

Aim: To investigate the redox reaction in the displacement PAK 21 Learning Science
of a halogen from its halide solution. Through Inquiry

Materials: Chlorine wsoaltuetri,oCn,l20, .b5rmomolindemw−a3 toefr,pBorta2,ssium Precautionary steps
icohdloinried, eI,2 KCl solution, 0.5 mol dm−3 of potassium
Chlorine, bromine and iodine
bromide, KBr solution, 0.5 mol dm−3 of potassium are poisonous. Use chlorine
iodide, KI solution and organic solvent water, bromine water and iodine
solution with extreme care.

1,1,1-trichloroethane, CH3CCl3.
Apparatus: Test tubes, measuring cylinder and test tube rack.

Procedure:

A. The colour of a halogen in the aqueous solution and organic

solvent 1,1,1-trichloroethane, CH3CCl3
1. APodudr22ccmm33oofftchheloorrignaenwicastoelrv, eCnlt2 into a test tube.
2. SChHa3kCeCthl3einmtioxttuhreetevsigt oturobuesalsy 1,1,1-trichloroethane,
shown in Figure 1.12.
3. and leave the test tube for Halogen in
30 seconds at the test tube rack. aqueous solution
4. Observe and record the colour of the aqueous layer and Halogen in
the layer of 1,1,1-trichloroethane, CwHat3eCr,CBl3r.2 1,1,1-trichloroethane,
5. Repeat steps 1 to 4 using bromine and Figure 1.12 CH3CCl3
iodine, I2 solution.

B. The displacement reaction of a halogen from its halide solution

1. Pour 2 cm3 of potassium chloride, KCl solution into a test tube.
2. Add 2 cm3 of bromine wobatseerr,vBe ra2niyntcootlohuertecshtatnugbee..
3. Shake the mixture and
4. Add 2 tchme3moifx1t,u1r,e1-vtirgicohroloursolyetahnadnele, aCvHe 3tCheCtl3e,sitnttuobtehfeotre3st0tsuebceo.nds.
5. Shake
6. Observe and record the colour of the aqueous layer and the layer of 1,1,1-trichloroethane,
7. RCeHp3eCaCt sl3t.eps 1 to 6 by using halide solutions and halogens as shown in the following table:

Halide solution Halogen
Potassium chloride, KCl
Bromine water, Br2
Potassium bromide, KBr Iodine, I2 solution
Chlorine water, Cl2
Potassium iodide, KI Iodine, I2 solution
Chlorine water, Cl2
Bromine water, Br2

Result:
Construct a suitable table to record all observations in section A and section B.

20

Redox Equilibrium Chapter 1

Discussion: CH3CCl3 in this experiment?
1. What is the function of 1,1,1-trichloroethane,
2. State the halogen that can:
(a) dddeaiiissscppphlllaaahcccaeeelocbioghrdeolonimnrdieinni,seIep,2,lCfBarclro2e2mfmfrrooepmnmottaprpseoosatitacuatssmissoiiunuiommtdhicabdhtreloo,omcKrciIdiudsereo,r,elKuKdCt:Biolrsnso.olulutitoionn. .
(b)
(c)
3. For
(a) write the half-equations for the oxidation reaction and reduction reaction.
(b) write the overall ionic equation for the redox reaction.
(c) identify the oxidised substances, reduced substances, oxidising agents and reducing agents.
Give reasons for your answers.
4. Based on the observation from this laboratory activity:
(a) arrange chlorine, Cl2, bromine, Br2 and iodine, I2 in an ascending order of strength as
oxidising agents.
(b) deduce the relationship between the strength of a halogen as an oxidising agent and its
position in Group 17.
(c) arrange chloride ion, Cl−, bromide ion, Br− and iodide ion, I− in an ascending order of
strength as reducing agents.

Prepare a complete report after carrying out this laboratory activity.

Bab 09/1 Video6 B01-21a Bab 09/1 Video7 B01-21b
Displacement ofhttp://kubupublication.com.my/Kimia/Tingkatan5/Video6.html Displacement ofhttp://kubupublication.com.my/Kimia/Tingkatan5/Video7.html
Metal Reaction Halogen Reaction
https://bit.ly/kpkt5v6 https://bit.ly/kpkt5v7

1.1

1. What is the meaning of a redox reaction?
2. Table 1.5 shows the equations for several redox reactions.

Reaction Table 1.5
I
Chemical equation
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

II 2Pb(s) + O2(g) → 2PbO(s)
III 2Al(s) + 3Cl2(g) → 2AlCl3(s)
IV Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)

For each of the above redox reaction:
(a) write the half equations for oxidation and reduction.
(b) identify the oxidised substance, reduced substance, oxidising agent and reducing agent.
Explain your answer in terms of the transfer of electrons.

21

Theme 1 Chemical Process

1.2 STANDARD ELECTRODE POTENTIAL

Mei, why do you Oh, this is the apparatus Standards
dip the zinc plate set-up for a half-cell. The zinc
into the aqueous plate is known as electrode. Pupils are able to:
solution containing 1.2.1 describe the standard
Zn2+ ions?
electrode potential.
1.2.2 determine oxidising agent

and reducing agent based
on their value of standard
electrode potentials.

Yes… Nabihah. The potential
difference produced in the half-cell is
known as electrode potential of zinc.

Figure 1.13 Understanding electrode potential

• Based on the conversation among the pupils in Figure 1.13, what do you understand about
electrode potential? How can we determine the standard electrode potential?
Electrode potential is the potential difference produced when an equilibrium is established
between metal M and the aqueous solution containing metal Mn+ ions in a half-cell.

• The electrode potential of a cell cannot be measured directly. Therefore, the electrode potential
value of an electrode system is determined based on the difference of electrode potential between
two half-cells.

• The electrode potential of an electrode system can be measured by pairing up the electrode to the
standard reference electrode system. The international consensus is that the standard hydrogen
electrode is selected as reference electrode to measure the value of standard electrode potential.
• The standard electrode potential, E0 of the cell is measured at standard conditions, namely:
(i) concentration of ions in aqueous solutions is 1.0 mol dm−3.
(ii) gas pressure of 1 atm or 101 kPa.
(iii) temperature at 25 °C or 298K.
(iv) platinum is used as an inert electrode when a half-cell is not a metal electrode.

Standard Hydrogen Electrode

• The diagram of standard hydrogen electrode half-cell is shown in Figure 1.14.

Hydrogen gas, H2 at 1 atm Platinum wire, Pt The function of platinum(IV)

Acid containing souxridfaec,ePatOre2aisotfopilnactrineuamse, the
1.0 mol dm−3 of Pt to
hydrogen ion, H+ athdastohrbydhryodgreongemnogleacs,uHle2s; so
Platinum electrode, Pt are in
coated with platinum closer contact with hydrogen
oxide, PtO2 ions, H+ in the solution.

Figure 1.14 Standard hydrogen electrode

22

Redox Equilibrium Chapter 1

• The standard hydrogen electrode consists of a platinum electrode, Pt dipped into an acid
containing 1.0 mol dm−3 of hydrogen ions, H+ and hydrogen gas, H2 at a pressure of 1 atm
channelled into the acid.
• Half equation of the hydrogen half-cell:

⇌ 2H+(aq) + 2e− H2(g) Bab 09/1 Video9 B01-23a Standard
http://kubupublication.com.my/Kimia/Tingkatan5/Video9.html
• The standard hydrogen electrode potential, E0 is given
the value of 0.00 V: Hydrogen
Electrode
⇌ H+(aq) + e− 12 H2(g) E0 = 0.00 V https://bit.ly/kpkt5v9

Standard Electrode Potential Value, E0

How is the standard electrode potential value, E0 of zinc determined?

• Figure 1.15 shows the standard electrode potential, E0 zinc is obtained when a half-cell consisting
of a zinc electrode, Zn is dipped into a solution containing 1.0 mol dm−3 of zinc ions, Zn2+ while
connected to a standard hydrogen electrode and a salt bridge.

Zinc 0.76 V Mind The salt bridge contains
electrode, Zn Challenge a saturated solution of
Salt bridge Hydrogen gas, H2 1 atm potassium nitrate. What is
the function of a salt bridge?

Solution Platinum electrode, Pt Bab 09/1 Video8 B01-23b
containing Solution containing 1.0 mol dm−3 http://kubupublication.com.my/Kimia/Tingkatan5/Video8.html
1.0 mol dm−3 of hydrogen ion, H+ Standard
of zinc ion, Zn2+ Electrode
Potential
Figure 1.15 Apparatus set-up to determine standard electrode https://bit.ly/
potential, E0 of zinc kpkt5v8

• Since the standard hydrogen potential is 0.00 V, the voltmeter reading of 0.76 V shows the electrode
potential of zinc. Zinc, Zn has a greater tendency to release electrons compared to hydrogen. Hence,
zinc, Zn becomes the negative terminal. Electrons move from zinc electrode, Zn (negative terminal)
to platinum electrode, Pt (positive terminal) through the connecting wires.

• Half equation at the negative terminal • Half equation at the positive terminal
(oxidation reaction) (reduction reaction)
Zn(s) → Zn2+(aq) + 2e− 2H+(aq) + 2e− → H2(g)

• A standard electrode potential for zinc half-cell is written as reduction.

⇌Zn2+(aq) + 2e− Zn(s) E0 = -0.76 V

• The negative symbol shows that zinc electrode is the negative terminal when connected to
the standard hydrogen electrode.

23

Theme 1 Chemical Process

Oxidising Agents and Reducing Agents Based on the Value of Standard
Electrode Potential

• Table 1.6 shows a part of the standard electrode potential series of half-cells that are arranged in an
ascending order of the standard electrode potential value from the most negative to the most positive.

• The standard electrode potential, E0 is also known as the standard reduction potential. All half-cell
equations are written as reduction.

• E0 value is a measure of the tendency of a substance to accept or donate electrons.

⇌ Oxidising agent + electron Reducing agent

Table 1.6 Standard electrode potential series

Strength as oxidising agents increase Half-cell equations E0 / V (298 K)
Strength as reducing agents increase -3.04
Li+(aq) + e− ⇌ Li(s) -2.92
K+(aq) + e− ⇌ K(s) -2.87
Ca2+(aq) + 2e− ⇌ Ca(s) -2.71
Na+(aq) + e− ⇌ Na(s) -2.38
Mg2+(aq) + 2e− ⇌ Mg(s) -1.66
Al3+(aq) + 3e− ⇌ Al(s) -0.76
Zn2+(aq) + 2e− ⇌ Zn(s) -0.44
Fe2+(aq) + 2e− ⇌ Fe(s) -0.25
Ni2+(aq) + 2e− ⇌ Ni(s) -0.14
Sn2+(aq) + 2e− ⇌ Sn(s) -0.13
Pb2+(aq) + 2e− ⇌ Pb(s) 0.00
C2Hu2++((aaqq))++22ee−− ⇌⇌ H C2(ug()s) +0.34
IFO2e(23s(+)g()a+q+2)e2+−H e⇌2−O ⇌ ( l2) I + −F(ae4q2e+)−( a ⇌q) 4OH−(aq) +0.40
Ag+(aq) + e− ⇌ Ag(s) +0.54
HMCBCFS22rrlO2(n222Og((OO8gl)22))−7(4+2(+a−+−a(q(2aq2a)2eqe)qe−+)− )+− + 2+ ⇌⇌2⇌H8e1 H −+4 (222 Ha+⇌CBF(q+a−rl )(−( q−a2(a(+)aqaSq q+q)O )2))+e542−e−6 −( e a⇌ −⇌q ) ⇌ 2M H2nC2O2r+(3(+al()qa)q+) +4H7H2O2O(l)(l) +0.77
+0.80
+1.07
+1.33
+1.36
+1.52
+1.77
+2.01
+2.87

• Based on Table 1.6, a comparison of standard electrode potential, E0 value is used to determine
whether silver, Ag or magnesium, Mg is an oxidising agent or reducing agent.

E0 value of Ag is more positive, E0 value of Mg is more negative,
• Silver ion, Ag+ on the left side is a stronger • Magnesium atom, Mg on the right side is a

oxidising agent. stronger reducing agent.
• It is easier for Ag+ ion to receive electrons and • It is easier for magnesium atom, Mg to release

undergo reduction. electrons and undergo oxidation.
• Conversely, silver atom, Ag on the right side is • Conversely, magnesium ion, Mg2+ on the left

difficult to release electrons. side is difficult to accept electrons.

24

Redox Equilibrium Chapter 1

• To summarise, the value of E0 can be used to determine the substance that will undergo oxidation
or reduction, along with the strength of the oxidising agents or reducing agents as shown in
Figure 1.16.

Has a greater tendency Has a greater tendency
to accept electrons. to release electrons.

Easier to undergo Molecules or ions Atoms or ions with a Easier to undergo
reduction reaction. with a more positive more negative or less oxidation reaction.
or less negative positive standard
standard electrode electrode potential
potential value E0. value, E0.

A stronger A stronger
oxidising agent. reducing agent.

Figure 1.16 The relationship between E0 value to the
strength of a substance as an oxidizing agent and a reducing agent

• By referring to the standard electrode potential value in Table 1.6, how do you identify the
oxidising agent and reducing agent in Photographs 1.4 (a) and 1.4 (b)?

Copper wire, Cu in (i) The standard electrode potential value, E0 is arranged from the
silver nitrate, AgNO3
solution ⇌⇌most negative to the most positive.

Photograph 1.4(a) Cu2+ + 2e− Cu E0 = + 0.34 V
(Source: Quora.com, 2018) Ag+ + e− Ag E0 = + 0.80 V
(ii) E0 value of Cu is more negative or less positive than E0 of Ag:
Does a reaction occur • Copper, Cu is a stronger reducing agent compared to
between copper, Cu and
silver nitrate, AgNO3? silver, Ag.
• Therefore, copper atom, Cu has a greater tendency to

release electrons to form copper(II) ions, Cu2+.
• Copper, Cu undergoes oxidation reaction.
(iii) The E0 value of Ag is more positive than E0 of Cu:
• The silver ion, Ag+ is a stronger oxidising agent than

copper(II) ions, Cu2+.
• Therefore, silver ion, Ag+ has a greater tendency to receive

electrons to form silver atom, Ag.
• Silver ion, Ag+ undergoes reduction.

The displacement reaction occurs between the strong reducing
agent that is copper, Cu and the strong oxidising agent that is silver
ion, Ag+.
Displacement reaction equation:

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

25

Theme 1 Chemical Process

Copper wire, Cu in (i) The standard electrode potential value, E0 is arranged from the
magnesium nitrate,
Mg(NO3)2 solution ⇌⇌most negative to the most positive.

Photograph 1.4(b) Mg2+ + 2e− Mg E0 = - 2.38 V
(Source: Quora.com, 2018) Cu2+ + 2e− Cu E0 = + 0.34 V
(ii) The E0 value of Cu is more positive than E0 of Mg:
Does a reaction • Copper, Cu is a weaker reducing agent than magnesium, Mg.
between copper, Cu • Therefore, copper atom, Cu has a lower tendency to release
and magnesium nitrate,
Mg(NO3)2 take place? electrons to form copper(II) ion, Cu2+.
• Copper, Cu does not undergo oxidation.
(iii) The E0 value of Mg is more negative than E0 of Cu:
• Magnesium ion, Mg2+ is a weaker oxidising agent than

copper(II) ions, Cu2+.
• Therefore, magnesium ions, Mg2+ are difficult to receive

electrons to form magnesium atom, Mg.
• Magnesium ion, Mg2+ does not undergo reduction.

The displacement reaction does not occur between a weak
reducing agent that is copper, Cu and a weak oxidising agent that is
magnesium ion, Mg2+.

A ctivity 1F PAK 21

Carry out a Think-Pair-Share activity.
1. Refer to page 24 for the standard electrode potential cell value, E0.
2. Discuss with your friends whether the following reactions occur:

(a) Cr2O72−(aq) + 14H+(aq) + 6Cl−(aq) → 2Cr3+(aq) + 3Cl2(g) + 7H2O(l)
(b) H2O2(aq) + 2Br−(aq) + 2H+(aq) → Br2(aq) + 2H2O(l)
3. Present the results of your discussion in class.

1.2

1. Table 1.7 shows the standard electrode potential values of half-cells for some metals.
(a) Arrange the atoms or ions in Table 1.7 in an
⇌Table 1.7 E0 = +0.34 V ascending order of the strength of oxidising
E0 = -2.38 V agents and reducing agents.
Cu2+(aq) + 2e− Cu(s) E0 = +0.80 V
E0 = -0.76 V (b) Based on your answer in (a), explain if the
⇌Mg2+(aq) + 2e− Mg(s) reaction will occur for the following reactants:
⇌Ag+(aq) + e− Ag(s)
⇌Zn2+(aq) + 2e− Zn(s) (i) Mg(s) + Cu2+(aq).
(ii) Mg(s) + Zn2+(aq).
(iii) Cu(s) + Zn2+(aq).

26

Redox Equilibrium Chapter 1

1.3 VOLTAIC CELL

• Do you know that when Standards
two different pieces of
metals are inserted into Pupils are able to:
potatoes, an electrical 1.3.1 explain redox reaction
energy is generated?
in voltaic cell through
experiment.

Redox: Electrolyte and
non-electrolyte on page 31.

Photograph 1.5 Voltaic cell from potatoes
as a source of electricity

What is a simple Two di erent metal plates are dipped into an electrolyte
chemical cell? and connected with connecting wires.
Also known as voltaic cell or galvanic cell.

Converts chemical energy to electrical energy.

A redox reaction occurs in the cell that causes the ow of electrons. e potential
di erence detected by the voltmeter indicates the presence of electrical current.

Redox Reaction in a Voltaic Cell

• Figure 1.17 shows an example of a simple chemical cell for a pair of Mg/Cu metals.
• The E0 value of magnesium is more negative, therefore, magnesium becomes the negative terminal.
• The reaction that occurs at the negative terminal is oxidation.
• The E0 value of copper, Cu is more positive; therefore, e‒ V e‒

copper, Cu becomes the positive terminal. Magnesium Copper plate, Cu
• The reaction at the positive terminal ribbon, Mg
is reduction reaction.
Copper(II)
Negative terminal: Mg(s) → Mg2+(aq) + 2e− chloride, CuCl2
solution
Positive terminal: Cu2+(aq) + 2e− → Cu(s)
Figure 1.17 Example of a simple chemical cell

• Electrons flow from the negative terminal to the Flow of electron
positive terminal; while the current flows from the
positive terminal to the negative terminal.
• Anode are electrodes where oxidation
occurs. In a voltaic cell, the negative Negative terminal Positive terminal
terminal is also known as anode. (Anode) (cathode)
• Cathodes are electrodes where reduction
Oxidation process Reduction process

occurs. In a voltaic cell, the positive
terminal is also known as cathode.
Flow of current

27

Theme 1 Chemical Process

• A chemical cell can also be constructed by combining two half-cells with different E0 values.
• A Daniell cell is an example of a voltaic cell where zinc, Zn metal and copper, Cu metal are used

as electrodes, which are then dipped into their respective ionic salt solutions.
• Both salt solutions are connected by a salt bridge or separated by a porous pot. What is the

function of the salt bridge or the porous pot?
• Figure 1.18 (a) shows a Daniell cell that uses a salt bridge and Figure 1.18 (b) shows a Daniell cell

that uses a porous pot.

e‒ V e‒

e‒ V e‒ Zinc Copper
Salt bridge electrode, Zn electrode, Cu
Zinc Copper
electrode, electrode, Cu
Zn
Zinc Copper(II) Porous pot Copper(II)
sulphate, sulphate,
ZsonluSOtio4 n CsouluStOio4n Zinc sulphate, sulphate, CuSO4
sZonluSOtio4 n solution

Figure 1.18(a) A Daniell cell that uses a salt bridge Figure 1.18(b) A Daniell cell that uses a porous pot

When BrO3 − ion reacts with Br ion in acid medium Br2 is liberated the equivalent mass of Br2 in the reaction is?

What is the properly balanced half reaction for the reduction of a bromine molecule to bromide ions?

What is the equivalent weight of Br2?