Redox reactions
Chapter 6
1
Overview
• Skills learnt in this chapter:
• Calculating oxidation numbers
• Balancing reduction and oxidation half equations
• Balancing redox equations
• Major application:
• Electrochemistry (A2)
2
What are redox reactions? Reducing
agent
It is a reaction that involves electron(s)
oxidation
transfer from one species 0 +2
(atom/ion/molecule) to another Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
+2 0
• In a redox reaction, oxidation and reduction
reduction occur simultaneously. Oxidising
• One element increases in oxidation agent
number
• Another element decreases in oxidation
number
• If none of the oxidation numbers change,
then the reaction is not an oxidation-
reduction reaction
3
Oxidation, reduction, reducing and oxidizing agents
A node Cathode
electrons lost electrons gained
Oxidation Reduction
Oxygen gained by an element Oxygen lost by an element
Oxidation number increases Oxidation number decreases
Reducing agent: Oxidising agent:
• The substance that gets oxidized • The substance that gets reduced
• Decreases the ON of another atom • Increases the ON of another atom
4
Oxidising agents and reducing agents
• An oxidising agent (oxidant) is a substance which brings about
oxidation by removing electrons from another atom or ion.
• A reducing agent (reductant) is a substance which brings about
reduction by donating (giving) electrons to another atom or ion.
In every redox reaction, there must be an oxidising agent and a
reducing agent.
5
Oxidation numbers
if is here cuz it is only a theoretical concept as
oxidation numbers apply to covalent molecules
too, which logically cant have charges
• Oxidation number is the electric charge an atom would have if the
electrons in a bond (bonding electrons) were assigned exclusively to
the more electronegative atom. Example HF F would be 1- charge, H would be +1
• Oxidation numbers can be assigned to atoms in ionic and covalent
compounds.
• Oxidation numbers help us to identify which element is oxidised
and which element is reduced in redox reactions.
6
Oxidation numbers
• The sum of the oxidation numbers in a compound is zero.
• The sum of the oxidation numbers in an ion is equal to the charge
on the ion.
• In either a compound or an ion, the more electronegative element is
given the negative oxidation number.
NO3-
oxidation number of N=x
oxidation number of O=-2
x+3(-2)=-1
x=5
oxidation number of N=+5
7
Oxidation number in ionic compounds
Type Example Oxidation number
Group 1 elements NaH, LiAlH4, NaBH4 +1
Group 2 elements MgO, CaO +2
Oxygen Na2O, MgO, CaO -2
Oxygen in peroxides O22- in H2O2, Na2O2, BaO2 -1 charge=1-
oxidation number=-1
H in metal hydrides NaH, LiAlH4, NaBH4, MgH2 -1
any given ion Fe3+, Cr2+ Same as ionic charge
8
Oxidation number in covalent compounds
Type Example Oxidation number
Atoms in a diatomic molecule Br2, H2, Cl2 0
Non-peroxide oxygen H2O, NO2 -2
Oxygen in F2O F2O +2
Hydrogen HBr +1
Fluorine HF -1
Chlorine NaCl, HCl -1
NaClO
Chlorine in compounds
chlorine oxidation number is:
+1
NaClO3 +5
? (solve)
containing O or F +1
ClF
*Following these rules, all other atoms in a covalent compound must balance out the charge
9
Exercise
Which substance is oxidized and which substance is reduced in each
reaction?
a. 2Mg(s) + O2(g) → 2MgO
Mg is oxidised O2 is reduced
Mg=+2, O=-2
b. 16Fe(s) + 3S8(s) → 8Fe2S3(s)
Fe is oxidised S is reduced
Fe=+3 S=-2
10
Exercise
Which substance is oxidized and which substance is reduced in each
reaction?
c. 2Li(s) + O2(g) → Li2O2(s)
lithium is oxidised O2 is reduced
Li=+1 O=-1
d. 2Fe(s) + 3I2(s) → 2FeI3(s)
Fe is oxidised I2 is reduced(ionic compound)
Fe=+3, I=-1
11
Exercise
• Determine the oxidation number of each of the elements in the
following compounds:
1. CaH2 Ca=+2 H=-1 11. N2O N=+1, O=-2
2. CrCl3 Cr=-3 Cl=-1 12. NO N=+2, O=-2
3. Cr2O72- Cr=+6, O=-2 13. NO2- N=+3, O=-2
4. F2O F=-1, O=+2 14. NO3- N=+5, O-2
5. Fe2O3 Fe=+3, O=-2 15. PO42- P=+5, O=-2
6. LiAlH4 Li=+1, Al=+3, H=-1 16. SO42- S=+6, O=-2
7. MnO4- Mn=+7, O=-2 17. SO3 S=6, O=-2
8. NH3 N=-3, H=+1 18. SO32- S=+4, O=-2
9. NH4+ N=-3, H=+1 19. S2O32- S=+2, O=-2
10. NaBH4 Na=+1, B=+3, H=+4 20 S4O62- S=+2.5, O=-2
12
A2-level oxidation numbers
Example: In the compound [ICl2]+[SbCl6]-, the oxidation number of
Nickel, Ni can form complexes with charged (e.g. CO+, chlorine is -1.
OH-, CN-) and neutral (e.g. NH3, H2O) species.
What are the oxidation numbers of I and Sb in the
The oxidation state of nickel is not +2 in compound?
A) [Ni(CO)4] A) I: +1 ; Sb: +5
A
B) [Ni(H2O)4(OH)2] B) I: +1 ; Sb: +7
C
C) [Ni(NH3)6]2+ C) I: +3 ; Sb: +5
D) [Ni(CN)4]2- D) I: +3 ; Sb: +7
13
Balancing half equations
• Half equations are equations which describe either the oxidation or
reduction process in a redox reaction.
• How to balance:
1. Determine change in ON
2. Write out unbalanced equation containing reactants, products and no. of
electrons transferred
3. Balance no. of non-O and non-H atoms on both sides of equation
4. Balance no. of electrons transferred because redox reactions occur in water
5. Balance no. of O atoms by adding H2O to either side of equation
6. Balance no. of H atoms by adding H+ to either side of equation
7. Double check charge on both sides of equation
14
Balancing half equation: example
Br2 → BrO3-
1. Change in ON: Br: 0 to +5 (oxidation)
2. Unbalanced equation with no. of Br2 →2BrO3- + 5e- *2
electrons transferred: so 10 electrons
3. Balance Br on both sides Br2 → 2BrO3- + 5e-
4. Balance no. of electrons transferred Br2 → 2BrO3- + (5x2 = 10)e-
5. Balance O on both sides Br2 + 3H2O → 2BrO3- + 10e-
6. Balance H on both sides Br2 + 3H2O → 2BrO3- + 10e- + 6H+
7. Double check charge on both sides
there's not enough water on the left side
*There are a few mistakes on purpose here. Can you spot it?
15
Exercise: balance the following half-equations
1. Mg → Mg2+ + 2e^-
2. 2Br- → Br2 +2e^-
3. S → S2-
2e^-1 +
16
Exercise: balance the following half equations:
2H2O+ Cl2 →2ClO- + 2e^-1 + 4H^+ 4H^+ 3e^-1 + NO3- → NO + 2H2O
17
Exercise: balance the following half equations:
5H2O+ S2O32- →2SO42- + 10H^+ +4e^- H2O+ AsO33- → AsO43- + 2H^+ + 2e^-1
10H^+ +8e^-1
18
Writing redox equations
Balance & Combine
Half Overall
equations equation
Deduce and Balance
19
Deducing half equations
Write out the reduction and oxidation half equations for the
following reaction:
3CuO + 2NH3 → 3Cu + N2 + 3H2O
Red. half equation:
Ox. half equation:
1. Determine which element is being oxidized and reduced
2. Write out half equations
3. Balance electrons, followed by oxygen and hydrogen (if any)
20
Redox equations
• Redox equation = Reduction half equation + Oxidation half equation
21
How to balance redox equations
• Method 1: stoichiometric coefficient
• Write the unbalanced equation
• Deduce the change in oxidation number – identify oxidizing agent and reducing
agent
• Balance the change of oxidation number on both sides
• Balance number of H and O on both sides
• Method 2: balance half equations
• Identify oxidising and reducing agent
• Write the unbalanced half equations
• Balance the half equations
• Multiply and combine half equations such that no. of electrons cancel out on both
sides
22
Examples of redox reactions
• Combustion of metal
• Metal + water
• Metal + acid
• Displacement
• Disproportionation
23
Examples of redox reactions
• Combustion of metal Na (s) + O2 (g) → Na2O (s)
• Metal + water ON of Na: 0 → +1
• Metal + acid
ON of O: 0 → –2
• Displacement +1
• Disproportionation
Method 1: 2Na (s) + O2 (g) → Na2O (s)
-2
Method 2:
Ox. half-eq.: 2 Na → 2 Na+ + 2 e–
Red. half-eq.: O2 + 4 e– → 2 O2–
Overall eq.: 4 Na (s) + O2 (g) → 2 Na2O (s)
24
Examples of redox reactions
• Combustion of metal Na (s) + H2O (l) → NaOH (aq) + H2 (g)
• Metal + water ON of Na: 0 → +1
• Metal + acid
ON of H: +1 → 0 +1
• Displacement
• Disproportionation
Method 1: Na (s) + H2O (l) → NaOH (aq) + H2 (g)
-1
25
Examples of redox reactions
• Combustion of metal Na (s) + H2O (l) → NaOH (aq) + H2 (g)
• Metal + water ON of Na: 0 → +1 ON of H: +1 → 0
• Metal + acid
• Displacement
Method 2:
• Disproportionation
Ox. half-eq.: Na → Na+ + e–
Red. half-eq.: H2O + e– → OH– + H2
H2O + 2e- → OH- + H2
Alkaline conditions: H2O + H+ + (OH-) + 2e- → OH- + H2 + (OH-)
2 H2O + 2 e– → 2 OH– + H2
Overall eq.: 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
26
Examples of redox reactions
• Combustion of metal Na (s) + HCl (aq) → NaCl (aq) + H2 (g)
• Metal + water ON of Na: 0 → +1 ON of H: +1 → 0
• Metal + acid
Ox. half-eq.: Na → Na+ + e–
• Displacement
Red. half-eq.: 2 H+ + 2 e– → H2
• Disproportionation
Overall eq.: 2 Na(s) + 2 H+(l) → 2 Na+(aq) + H2(g)
OR
2Na (s) + 2HCl (aq) → 2NaCl (aq) + H2 (g)
*Cl- acts as a spectator ion
27
Examples of redox reactions
• Combustion of metal CuSO4 (aq) + Fe (s) → FeSO4 (aq) + Cu (s)
• Metal + water
• Metal + acid
Ox. half-eq.: Fe (s) → Fe2+ (aq) + 2 e–
• Displacement
• Disproportionation
Red. half-eq.: Cu2+ (aq) + 2 e– → Cu (s)
*SO42- acts as a spectator ion
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Examples of redox reactions
• Combustion of metal 2H2O2 → 2H2O + O2
• Metal + water
• Metal + acid
Ox. half-eq.: H2O2 → O2 + 2H+ + 2e-
• Displacement
• Disproportionation
• A reaction which Red. half-eq.: H2O2 + 2H+ + 2e- → 2H2O
involves the
simultaneous oxidation
and reduction of the • H2O2 act both as oxidizing
same element in a
compound agent and reducing agent
29
Exercise
When 0.635 g of copper (relative atomic mass, Ar = 63.5) is added to an excess of
silver nitrate solution, 2.158 g of silver (Ar = 107.9) form. The ionic equation for the
reaction is
A) Cu(s) + Ag2+(aq) → Cu2+(aq) + Ag(s)
B) Cu(s) + Ag+(aq) → Cu+(aq) + Ag(s)
C) 2Cu(s) + Ag2+(aq) → 2Cu2+(aq) + Ag(s)
D) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
30
Exercise
n (Tl+NO3-) =
n (NH4VO3) =
ON of V in VO3- =
31
Exercise
n (SnCl2) =
Reduction half eqn:
Oxidation half eqn:
Overall eqn:
32
Exercise
Decide whether the reactions below are redox reactions. Identify the
oxidizing and reducing agents and write down the change in ON
where applicable.
1) 2NaOH + Cl2 → NaCl + NaClO + H2O
2) 14H+ + Cr2O72– + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
3) 2MnO4– + 4H2O + 6I– → 2MnO2 + 8OH– + 3I2
4) KOH + HCl → KCl + H2O
33
Summary
• Concepts:
• Oxidation number
• Reduction
• Oxidation
• Oxidising agent
• Reducing agent
• Disproportionation
34
Example (ans: B)
35
