Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 1
Higher Chemistry: Unit 2 – Chemical Changes and Structure
Revision of Redox Equations and Titrations
The lesson will help you revise the following topics
1. Oxidising and Reducing Agents
2. Complex Redox Equations
3. Redox Titrations
You will have been successful in this lesson if you:
1. Read the summary given (do not copy these notes – you already have them)
2. Watch the Recorded Lesson – Link provided
3. Complete questions provided
4. EXTENSION: There is a further reading section to help you gain more depth of
understanding for this section. There are also suggested questions for you to try from the
blue book of revision questions.
If you have any questions about the content of this lesson, you should ask your class teacher
either through your class MS team or via email. The teams will be monitored through the
week and someone will get back to you as soon as they can.
You may wish to revise the following to help you understand this lesson:
Higher chemistry – Unit 3 – Redox Reactions
Learning Outcomes
Success Criteria
Links to Prior Knowledge
You may wish to have a copy of the data booklet handy for this lesson. Download or
print a copy of the Higher Chemistry Data Booklet from MS Teams or the SQA website -
https://www.sqa.org.uk/sqa/files_ccc/ChemistryDataBooklet_NewH_AH-Sep2016.pdf
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 2
Redox Equations and Titrations
1. Oxidising and Reducing Agents
Redox Revision from Nat 5
OILRIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
Oxidation and Reduction reactions always occur together.
These types of reactions are called redox reactions.
Ion-electron equations for the oxidation and reduction in a typical redox reaction are (from
data booklet page 12):
Cu → Cu2+ + 2e- Oxidation
Ag+ + e- → Ag Reduction
Cu + 2Ag+ → 2Ag + Cu2+ Balanced Redox Equation
Points to Note:
- Electrons and spectator ions are not included in balanced redox equations.
- For the reaction above, the silver equation had to be multiplied by 2 to balance the
electrons in the redox equation.
Do not copy the notes below – these are a summary of the class notes you
already have for this unit.
WATCH: Click on the link for a recorded lesson for this topic:
RECORDED LESSON from Ms Hastie
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 3
Oxidising Agents
An oxidising agent is a substance which accepts electrons.
In the above reaction, the silver ion, Ag+ is the oxidising agent: Ag+ + e- → Ag
The oxidising agent causes the oxidation of another reactant and in the process is reduced.
Elements with a high electronegativity tend to form ions by gaining electrons (reduction)
and so often act as oxidising agents. Strong oxidising agents are found at the BOTTOM LEFT
in the Electrochemical Series (ECS), page 12 of your Data Booklet.
Group 7 elements, especially fluorine and chlorine are strong oxidising agents. Common
oxidising agents also include:
Acidified dichromate ions (Cr2O72-)
Acidified permanganate ions (MnO4-)
Hydrogen peroxide (H2O2)
Reducing Agents
A reducing agent is a substance which donates electrons.
In the above reaction, copper is the reducing agent: Cu → Cu2+ + 2e-
In donating electrons, a reducing agent causes the reduction of another reactant, and in the
process, is oxidised.
Elements with low electronegativities tend to form ions by losing electrons (oxidation) and
so often act as reducing agents. Strong reducing agents are found at the TOP RIGHT in the
Electrochemical Series (ECS), page 12 of your Data Booklet.
Group 1 metals are strong reducing agents. Another example of a reducing agent is carbon
monoxide (CO).
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 4
Identifying Oxidising and Reducing Agents
In the reaction: Cu + 2AgNO3 → 2Ag + Cu (NO3 )2
The oxidising agent is the one being reduced (or the one that is causing the oxidation). This
is the silver ion, Ag+
The reducing agent is the one being oxidized. This is the copper atom, Cu
Nitrate, NO3- , is the spectator ion
Everyday Examples of Oxidising Agents
Oxidising agents are widely used in cleaning products because of the
effectiveness with which they can kill fungi and bacteria, and can
inactivate viruses.
The oxidation process is also an effective means of breaking down
coloured compounds making oxidising agents ideal for use as
‘bleach’ for clothes and hair.
2. Complex Redox Equations
Example 1. Write an ion–electron equation for the conversion of bromine to bromate, BrO3¯
Step 1. Balance the main element
(2 x BrO3¯ because Br2 on left hand side, LHS ) Br2 → 2BrO3¯
Step 2. Balance oxygen by adding water molecules
(6 x H2O because 2BrO3¯ = 6 oxygens ) Br2 + 6H2O → 2BrO3¯
Step 3. Balance hydrogen by adding hydrogen ions
(12 x H+ because 6 x H2O) Br2 + 6H2O → 2BrO3¯ + 12H+
Step 4. Balance the charge by adding electrons
Total charge LHS = Br2 + 6H2O = 0+0 = 0
Total charge RHS = 2BrO3¯+ 12H+ =(-2)+(+12)= +10
10e¯ added to RHS to reduce charge to 0
Br2 + 6H2O → 2BrO3¯ + 12H+ + 10e¯
Ion-electron equation: Br2 + 6H2O → 2BrO3¯ + 12H+ + 10e¯
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 5
Example 2. Acidified potassium permanganate (KMnO4) oxidises Fe2+ ions to Fe3+ ions. During
this reaction the MnO4– ions are reduced to Mn2+ ions:
(a) Write an ion-electron equation for the reduction reaction occurring:
Step 1. Balance the main element
(Mn in this case, already balanced) MnO4¯ → Mn2+
Step 2. Balance oxygen by adding water molecules
(4 x H2O because MnO4 = 4 oxygens) MnO4¯ → Mn2+ + 4H2O
Step 3. Balance hydrogen by adding hydrogen ions
(8 x H+ because 4 x H2O) MnO4¯ + 8H+ → Mn2+ + 4H2O
Step 4. Balance the charge by adding electrons
Total charge on LHS = MnO4¯+ 8H+ = (-1)+(+8)= +7
Total charge on RHS = Mn2+ + 4H2O = (+2)+0 = +2
5e¯ added to LHS to reduce charge to 2+
MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O
Ion-electron equation: MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O
Note: The equation above shows that, permanganate, MnO4¯ needs hydrogen ions, H+
in order for this reaction to proceed. The H+ ions are provided by added an acid to
the solution. The term “acidified” is often used when describing oxidising agents eg
acidified dichromate.
(b) Write a balanced redox equation for the overall reaction:
REDUCTION: MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O
OXIDATION: Fe2+ → Fe3+ + e¯
To make a balanced redox equation with no electrons left over, you must multiply to make
the electrons the same on both sides: 5Fe2+ → 5Fe3+ + 5e¯
BALANCED REDOX EQUATION: MnO4¯ + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 6
3. Redox Titrations
Example 1. A standard solution of acidified potassium permanganate (oxidising agent) was used to find the concentration of a solution of iron (II) ions (reducing agent). BALANCED REDOX EQUATION: 5Fe2+ + MnO4
- + 8H+ → Mn2+ + 4H2O + 5Fe3+ The end point for this set up is when the solution in the conical flask turns from colourless to purple showing the MnO4
- is in excess. Therefore no indicator is needed for this titration. Titration results:
Titre Volume MnO4
- / cm3
Start volume End volume Titre volume
1 0.0 15.0 15.0
2 15.0 27.1 12.1
3 27.1 39.0 11.9
TRY THIS QUESTION – WORKED SOLUTION IS ON THE NEXT PAGE Question:
Using the information above find the concentration of iron(II) ions in the solution.
0.1 mol l-1
acidified
permanganate
solution (purple)
25 cm3 Fe2+
solution
(colourless)
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 7
WORKED SOLUTION: Question: Using the information above find the concentration of iron(II) ions in the solution. (a) Find the average titre value
Average titre value = 12.1+11.9
2 = 12.0 cm3
Rough titre of 15.0 cm3 is ignored because it is not concordant, ie not within 0.2 cm3
(b) Find the unknown concentration
Step 1 – find the number of moles of the 0.1 moll-1 permanganate solution: n = c x v
= 0.1 x 0.012
= 0.0012 moles
Step 2 – find the number of moles of the iron(II) solution Mole ratio from balanced redox equation MnO4
- : 5Fe2+ 0.0012 moles MnO4
- : 0.0012 x 5 Fe2+ = 0.006 moles Fe2+ ions Step 3 – find the concentration of the 25 cm3 iron(II) solution c = n/v
= 0.006 / 0.025
= 0.24 moll-1
EXTENSION QUESTION – TRY THIS QUESTION – ANSWER AT THE END OF THE LESSON: The iron(II) solution was made by dissolving five iron tablets in water. The solutions and rinsings were added to a 250 cm3 standard flask and made up to the graduation mark with distilled water. Find the mass of iron(II) in one iron tablet. WATCH: Ms Adams Chemistry – Redox (18 mins) https://youtu.be/vxEavA8nPNY
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 8
You should have now revised:
1. Oxidising and Reducing Agents
2. Complex Redox Equations
3. Redox Titrations
To learn more about proteins, try the following online resources:
BBC Bitesize: https://www.bbc.co.uk/bitesize/guides/z2r44wx/revision/1
Read revision pages (optional) and TRY THE END TOPIC TEST
https://www.bbc.co.uk/bitesize/guides/ztkdd2p/test
TRY THE END TOPIC TEST (also includes chromatography)
Scholar: Log in through GLOW
Higher Chemistry Chemical Changes and Structure Topic 6: Oxidising and
Reduction Agents
Read through the exercises and TRY THE END TOPIC TEST
ALSO Higher Chemistry Chemistry in Society Topic 8: Volumetric Analysis
Read through the exercises and TRY THE END TOPIC TEST
Evans2 chem web: https://www.evans2chemweb.co.uk/login/index.php#
Username: snhs password: giffnock
Select any teacher revision material CfE Higher Unit 1: Chemical
Changes and Structure Oxidising and Reducing Agents
ALSO
Unit 3: Chemistry in Society Chemical Analysis
Learning Outcomes
Further Reading
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 9
ANSWER TO EXTENSION QUESTION: The iron(II) solution was made by dissolving 5 iron tablets in water. The solutions and rinsings were added to a 250 cm3 standard flask and made up to the graduation mark with distilled water. Find the mass of iron(II) in 1 iron tablet. Step 1 – scale up the number of moles in 25 cm3 to total number of moles in 250 cm3 iron(II) solution number of moles of iron(II) in 25 cm3 = 0.006 (from question above)
number of moles of iron(II) in 250 cm3 = 0.006 x 10 = 0.06 moles Fe2+ ions
Step 2 – find the mass of iron in 5 tablets mass = n x gfm =
= 0.06 x 55.8 = 3.348 g in 5 tablets
Step 3 – find the mass of iron in 1 tablet 1.348 g in 5 tablets, therefore 1 tablet = 3.348 / 5 = 0.6696g or 670mg
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 10
1.
2.
3.
4.
Check your understanding – Answers the questions below in you class jotter
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 11
5.
6.
7.
8.
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 12
9.
10.
11.
12.
(a)
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 13
(b)
(c)
13.
14.
(a)
(b)
Higher Chemistry St. Ninian’s High School
CIS: Redox Revision of Redox Reactions Page 14
Use the online learning link above if you would like to extend your knowledge on this topic.
For more practise questions, use your Revision Questions for Higher Chemistry “Blue book”:
Redox reactions page 105 Q1-5
Writing ion-electron equations page 107 Q1-2
Oxidising and reducing agents page 108 Q1-3
Redox titrations page 111 Q1-9
Answer to Redox Revision Questions
1. A
2. D
3. A
4. C
5. D
6. D
7. C
8. B
9.
10.
11.
EXTENSION WORK
ANSWERS CAN BE FOUND ON THE NEXT PAGES FOR YOU TO CHECK YOUR
WORK AS YOU GO ALONG.