Chapter 01: Mole Concept, Stoichiometry & Redox
ALevel H2 Chemistry: Chapter 1
Mole Concept, Stoichiometry and Redox
Worked Example 2:
When copper is added to concentrated nitric(V) acid, a pale blue solution of Cu²⁺ is formed. Brown fumes of nitrogen dioxide, NO₂ are produced. Write the redox equation for this reaction.
StepByStep Walkthrough
Step #1: Construct both oxidation and reduction halfequations by applying the ‘KOHe⁻’ method.
K  Identify and balance the key element that is oxidised/reduced in the half equation 

Cu → Cu²⁺ NO₃⁻ → NO₂ 

O  Balance the number of O atoms by adding H₂O 
Cu → Cu²⁺ NO₃⁻ → NO₂ + H₂O 

H  Balance the number of H atoms by adding H⁺ (Note: Always balance number of O atoms before balancing H atoms) 
Cu → Cu²⁺ 2H⁺ + NO₃⁻ → NO₂ + H₂O 

e⁻  Balance the overall charge on each side with electrons (Tip: Check that the electrons added to the 2 halfequations is on different sides) 
Cu → Cu²⁺+ 2e⁻ e⁻ + 2H⁺ + NO₃⁻ → NO₂ + H₂O 
K  Identify and balance the key element that is oxidised/reduced in the half equation  Cu → Cu²⁺ NO₃⁻→ NO₂ 

O  Balance the number of O atoms by adding H₂O  Cu → Cu²⁺ NO₃⁻ → NO₂ + H₂O 
H  Balance the number of H atoms by adding H⁺ (Note: Always balance number of O atoms before balancing H atoms) 
Cu → Cu²⁺ 2H⁺ + NO₃⁻ → NO₂ + H₂O 
e⁻  Balance the overall charge on each side with electrons (Tip: Check that the electrons added to the 2 halfequations is on different sides) 
Cu → Cu²⁺+ 2e⁻ e⁻ + 2H⁺ + NO₃⁻ → NO₂ + H₂O 
Step #2: Balance the electron transfer by multiplying the halfequations using integers so that the number of electrons in both halfequations are equal. When both halfequations are added together, the electrons are fully cancelled.
Cu → Cu²⁺ + 2e⁻
e⁻ + 2H⁺ + NO₃⁻ → NO₂ + H₂O (x2)
⇒ 2e⁻ + 4H⁺ + 2NO₃⁻ → 2NO₂ + 2H₂O
Step #3: Add the halfequations together and eliminate any common terms on both sides to obtain the final balanced equation.
Cu + 2e⁻ + 4H⁺ + 2NO₃⁻ → Cu²⁺+ 2e⁻ + 2NO₂ + 2H₂O
Final Answer: Cu + 4H⁺ + 2NO₃⁻ → Cu²⁺ + 2NO₂ + 2H₂O
However, in certain cases, the question will indicate that the reaction took place in the basic medium. There are 3 additional steps (steps 46), which we will go through in worked example 3 to show you how to balance halfequations in basic medium.
Worked Example 3:
By writing half equations, give the full balanced equation of the following reaction described.
CrO₄²⁻+ S²⁻ → CrO₂⁻ + S (in basic medium)
Step #1: Construct both oxidation and reduction halfequations by applying the ‘KOHe⁻’ method.
K  Identify and balance the key element that is oxidised/reduced in the half equation 

CrO₄²⁻→ CrO₂⁻ S²⁻→S 

O  Balance the number of O atoms by adding H₂O 
CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻→ S 

H  Balance the number of H atoms by adding H⁺ (Note: Always balance number of O atoms before balancing H atoms) 
4H⁺ + CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻ → S 

e⁻  Balance the overall charge on each side with electrons (Tip: Check that the electrons added to the 2 halfequations is on different sides) 
3e⁻ + 4H⁺ + CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻ → S + 2e⁻ 
K  Identify and balance the key element that is oxidised/reduced in the half equation  CrO₄²⁻→ CrO₂⁻ S²⁻→S 

O  Balance the number of O atoms by adding H₂O  CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻→ S 
H  Balance the number of H atoms by adding H⁺ (Note: Always balance number of O atoms before balancing H atoms) 
4H⁺ + CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻ → S 
e⁻  Balance the overall charge on each side with electrons (Tip: Check that the electrons added to the 2 halfequations is on different sides) 
3e⁻ + 4H⁺ + CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻ → S + 2e⁻ 
Step #2: Balance the electron transfer by multiplying the half equations using appropriate integers so that the number of electrons in both halfequations are equal. When both halfequations are added together, the electrons are fully cancelled.
3e⁻ + 4H⁺ + CrO₄²⁻ → CrO₂⁻ + 2H₂O (x2)
S²⁻ → S + 2e⁻ (x3)
⇒ 6e⁻ + 8H⁺ + 2CrO₄²⁻ → 2CrO₂⁻ + 4H₂O
⇒ 3S²⁻ → 3S + 6e⁻
Step #3: Add the halfequations together and eliminate any common terms on both sides to obtain the final balanced equation.
6e⁻ + 8H⁺ + 2CrO₄²⁻ + 3S²⁻→ 2CrO₂⁻ + 4H₂O + 3S + 6e⁻
Step#4: Since this reaction is in basic medium, add the required number of OH⁻ needed to “neutralise” the H⁺ ions to both sides of the equation.
8OH⁻ +8H⁺ + 2CrO₄²⁻ + 3S²⁻→ 2CrO₂⁻ + 4H₂O + 3S + 8OH⁻
Step #5: Convert the H⁺ and OH⁻ on the same side of the equation to H₂O.
8H₂O + 2CrO₄²⁻ + 3S²⁻→ 2CrO₂⁻ + 4H₂O + 3S + 8OH⁻
Step #6: Cancel out the H₂O on both sides of overall equation
4H₂O 8H₂O + 2CrO₄²⁻+ 3S²⁻→ 2CrO₂⁻+ 4H₂O + 3S + 8OH⁻
Final Answer: 4H₂O + 2CrO₄²⁻ +3S²⁻→ 2CrO₂⁻ + 8OH⁻+ 3S
Important Summarised Redox Titrations to note:
Manganate (VII) titrations 

 Acidified KMnO₄⁻ is a strong oxidising agent used to analyse reducing agents such as Fe²⁺, I⁻ & H₂O₂  Such titrations should be carried out in an acidic medium (use sulfuric acid)  Equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O EXPERIMENT TIPS:  Sufficient acid must be added + constant swirling to prevent incomplete reduction to brown MnO₂  End point reached when purple KMnO₄ reduced to pale pink Mn²⁺ 
Dichromate (VI) titrations 
 Dichromate is a less powerful oxidising agent than potassium manganate(VII) AND DOES NOT oxidise Cl⁻ hence it can be used to analyse reducing agents in presence of chloride ions  Equation: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O EXPERIMENTS TIPS:  Orange Cr₂O₇²⁻ is reduced to green Cr³⁺ ions  Indicator is required as it is practically impossible to judge endpoint from colour change as green predominates long before reduction is complete 
Iodine  thiosulfate titrations 
 Na₂S₂O₃ (reducing agent) reduces iodine to iodide ions and itself is being oxidised in the process to form Na₂S₄O₆, sodium tetrathionate  Equation: 2S₂O₃²⁻ (aq) + I₂ (aq) → S₄O₆²⁻ (aq) + 2I⁻ (aq) EXPERIMENT TIPS:  When brown colour of iodine fades to pale yellow as endpoint approaches, a little starch solution is added which will give an intense blue colour (representing the iodinestarch complex) COMMONLY ASKED QUESTIONS: 1. Why add the starch indicator? 2. Why add the starch indicator when the solution is pale yellow and not at the start of the experiment? 
Determination of Oxidation Number:
This is also another part of this chapter where students commonly find themselves unable to deal with questions involving deduction of oxidation number of an element in a compound. In this section, we will be showing you a simple and easy method.
To find Oxidation State (o.s.) in either reactant or product OR find change in oxidation state (o.s.)
Let’s apply this concept in a sample question below:
Worked Example 4:
1 mole of S₂O₃²⁻ions reduces 4 moles of Cl₂ molecules. What is the sulfurcontaining product of this reaction?
A S
B SO₂
C SO₄²⁻
D S₄O₆²⁻
Step#1 Represent the unknown oxidation number with a symbol, let’s say x. Write out the known halfequation that is either given in the question, or can be derived on your own. 

Let x be the oxidation number of sulfur in the sulfurcontaining product. The known halfequation would be: 
Step#2: Find the mole ratio. Work out no. of electrons gained/lost by the known reactant. 
S₂O₃² [R.A.] : Cl₂ [O.A] : e⁻ 1 : 4 1 : 2 1 : 4 : 8 Firstly, the question mentioned the ratio between the reducing agent (S₂O₃²⁻) and oxidising agent (Cl₂) to be 1:4. In the known halfequation written above, the ratio between the known reactant (Cl₂) and the electrons gained is 1:2. 
Step#3: Find the number of moles of electrons gained/lost by 1 mole of the unknown reactant and solve for the value of unknown x. 
Therefore, using the above mole ratio table, we can deduce the number of electrons lost by 1 mole of the unknown reactant (S₂O₃²⁻), which is 8 moles of e⁻.
1 mole of S₂O₃²⁻ has 2 moles of S atoms which means that 2 moles of S atoms lose 8 moles of e⁻. Thus, 1 mole of S atoms lose 4 moles of e⁻. Oxidation number of S in S₂O₃²⁻ is +2. Since 1 mole of S atoms lose 4 moles of e⁻, oxidation number of S in the sulfurcontaining product is therefore: x = +2 +4 Final Answer: C, since o.s. of S in SO₄²⁻ is +6. 
Continue reading on Chapter 2  Atomic Structure here