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# Chapter 01: Mole Concept, Stoichiometry & Redox

A-Level H2 Chemistry: Chapter 1
Mole Concept, Stoichiometry and Redox Worked Example 1:

A mixture of 35.0g of hydrogen and 27.0 g of oxygen is allowed to react and produce water. Which is the limiting reagent?

Solution:

1. Writing out the balanced chemical reaction
2H₂O + O₂ → H₂O

2. Calculate the amount of H₂ and O₂
n(H₂) = 35.0/2.0 = 17.5 mol
n(O₂) = 27.0/32.0 = 0.8437 mol

3. Compare which produces a smaller amount of H₂O
- If H₂ is the limiting reagent, n(H₂O) produced is 17.5mol
- If O₂ is the limiting reagent, n(H₂O) produced is 0.8437 x 2 = 1.687mol

Answer: Since O₂ produces the smallest amount of H₂O, it is the limiting reagent.

## Balancing Redox Equations

Most students tend to face difficulty when balancing redox equations. By following the tips and tricks below, you will be able to solve such questions in a breeze.

1. Always ensure that the equation satisfies 2 criterias:
Mass Balance -
Same number of reactants of each kind must be reflected as reactants and products
Charge Balance - Sums of actual charges on both sides of the equation must be equal

2. In most cases, we are required to balance redox equations in acidic medium. We assume the medium to be acidic if the question does not specify.

We will revise how to tackle such questions with worked example 2. 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.

Step-By-Step Walkthrough

Step #1: Construct both oxidation and reduction half-equations by applying the ‘KOHe⁻’ method.

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

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

Step #2: Balance the electron transfer by multiplying the half-equations using integers so that the number of electrons in both half-equations are equal. When both half-equations 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 half-equations 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 4-6), which we will go through in worked example 3 to show you how to balance half-equations 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 half-equations by applying the ‘KOHe⁻’ method.

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

K O Identify and balance the key element that is  oxidised/reduced in the half equation CrO₄²⁻→ CrO₂⁻ S²⁻→S Balance the number of O atoms by adding H₂O CrO₄²⁻ → CrO₂⁻ + 2H₂O S²⁻→ S 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 Balance the overall charge on each side with electrons  (Tip: Check that the electrons added to the 2 half-equations 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 half-equations are equal. When both half-equations 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 half-equations together and eliminate any common terms on both sides to obtain the final balanced equation.

6e+ 8H⁺ + 2CrO₄²⁻ + 3S²⁻→ 2CrO₂⁻ + 4HO + 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 8HO + 2CrO₄²⁻+ 3S²→ 2CrO₂⁻+ 4HO + 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 iodine-starch complex)

1. Why add the starch indicator?

Ans:
The starch indicator is added as colour change from pale yellow to colourless is not distinctive hence starch is added to make colour change easier to recognise, preventing inaccuracy

2. Why add the starch indicator when the solution is pale yellow and not at the start of the experiment?

Ans:
The starch indicator is only added when solution is pale yellow so as to prevent the formation of iodine-starch complex which allows for lesser amounts of free I₂ for titration, causing volume of titrant to be lower than expected.

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 SO₃²⁻ions reduces 4 moles of Cl₂ molecules. What is the sulfur-containing product of this reaction?

A S
B SO₂
C SO₄²⁻
D SO₆²⁻

Step#1

Represent the unknown oxidation number with a symbol, let’s say x. Write out the known half-equation that is either given in the question, or can be derived on your own.

Let x be the oxidation number of sulfur in the sulfur-containing product. The known half-equation would be:
Cl₂ + 2e⁻ → 2Cl⁻

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 half-equation 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 (SO₃²⁻), 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 sulfur-containing product is therefore:

x = +2 +4
x = +6

Final Answer: C, since o.s. of S in SO₄²⁻ is +6.

Continue reading on Chapter 2 - Atomic Structure here