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Redox Reactions

Master electron transfer, oxidation states, and redox properties through interactive visualizations and practice.

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Topic Overview

Redox reactions are processes of electron transfer in chemical reactions, involving core concepts such as oxidizing and reducing agents, oxidation and reduction, and changes in valency. This knowledge often appears in the form of substance transformation judgment, oxidation product/reducer identification, and elemental property analysis, which requires students to be proficient in the calculation of valency and electron transfer analysis.

Key Points

The basis for determining oxidizing and reducing agents
Relationship between changes in valency and electron transfer
Identification of oxidation and reduction products
Comparison of the strength of redox properties of common substances
Balancing redox reactions using half-reaction method
Understanding standard reduction potentials

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Quick Tip

Remember: LEO says GER! Loss of Electrons is Oxidation, Gain of Electrons is Reduction.

1 Basic Redox Identification

Easy

In the reaction: 2Na + Cl₂ → 2NaCl, which element is oxidized?

Electron Transfer Visualization

(0)

Cl₂

(0)

→

Na⁺

(Oxidation)

→

-1

Cl⁻

(Reduction)

LEO says GER

🦁 LEO

Loss of Electrons = Oxidation

🦍 GER

Gain of Electrons = Reduction

Na loses 1 electron → Oxidation

Cl gains 1 electron → Reduction

Sodium (Na)

Chlorine (Cl)

Both are oxidized

Neither is oxidized

Hint: Oxidation means loss of electrons. Which element loses electrons in this reaction?

Step-by-Step Solution

1. Identify oxidation states before and after reaction:

2. Na: From 0 (elemental) to +1 in NaCl → OXIDATION (loss of electron)

3. Cl: From 0 (Cl₂) to -1 in NaCl → REDUCTION (gain of electron)

4. Sodium (Na) is oxidized because it loses electrons

5. Chlorine (Cl) is reduced because it gains electrons

6. Therefore, the oxidized element is sodium

Answer: A

2 Identifying Oxidizing Agent

Medium

In the reaction: Zn + CuSO₄ → ZnSO₄ + Cu, what is the oxidizing agent?

Single Displacement Reaction

(0)

Cu²⁺

(+2)

↓

Zn²⁺

(+2)

(0)

Electron flow:

Zn → Cu²⁺

Activity Series Analysis

Li (Strongest reducer)

Zn ←

H₂

Cu ←

Au (Strongest oxidizer)

Zn is above Cu in activity series → Zn reduces Cu²⁺

CuSO₄ (Cu²⁺ ions)

ZnSO₄

Hint: Oxidizing agent causes oxidation of another substance and gets reduced itself.

Step-by-Step Solution

1. Analyze oxidation state changes:

2. Zn: 0 → +2 (oxidation, loses 2 electrons)

3. Cu: +2 → 0 (reduction, gains 2 electrons)

4. Oxidizing agent is the substance that gets reduced

5. Cu²⁺ gains electrons → it is reduced

6. Therefore, CuSO₄ (source of Cu²⁺) is the oxidizing agent

7. Zn is the reducing agent (gets oxidized)

Answer: A

3 Complex Redox Identification

Hard

In the reaction: 2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O, which element undergoes reduction?

Oxidation State Analysis

Element

Oxidation State

Mn in KMnO₄

Mn in MnSO₄

Fe in FeSO₄

Fe in Fe₂(SO₄)₃

→

Reduced

→

Oxidized

Electron Transfer Balance

Reduction half-reaction:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Mn: +7 → +2 (gains 5 electrons)

Oxidation half-reaction:

Fe²⁺ → Fe³⁺ + e⁻

Fe: +2 → +3 (loses 1 electron)

Total electrons: 5Fe²⁺ → 5Fe³⁺ + 5e⁻

Mn undergoes reduction (+7 → +2)

Fe undergoes oxidation (+2 → +3)

Manganese (Mn)

Iron (Fe)

Sulfur (S)

Oxygen (O)

Hint: Calculate oxidation states of Mn and Fe in reactants and products. Reduction means decrease in oxidation state.

Step-by-Step Solution

1. Calculate oxidation states:

2. In KMnO₄: K = +1, O = -2 (each), so Mn = +7

3. In MnSO₄: Mn = +2 (SO₄²⁻ has S = +6, O = -2)

4. Mn changes from +7 to +2 → REDUCTION (gain of 5 electrons)

5. In FeSO₄: Fe = +2 (SO₄²⁻)

6. In Fe₂(SO₄)₃: Fe = +3 (each SO₄²⁻)

7. Fe changes from +2 to +3 → OXIDATION (loss of 1 electron)

8. Sulfur remains +6, oxygen remains -2 in all compounds

9. Therefore, manganese (Mn) undergoes reduction

Answer: A

Redox Complete!

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