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 |
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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 half-equations 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 |
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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 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 |
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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 half-equations 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 |
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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 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⁻ + 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 |
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- 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) 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 sulfur-containing 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 half-equation that is either given in the question, or can be derived on your own. |
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Let x be the oxidation number of sulfur in the sulfur-containing product. The known half-equation 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 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 (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 sulfur-containing 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