krit.club logo

Chemical Reactions - Redox reactions

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Oxidation is defined as the loss of electrons, the gain of oxygen, or the loss of hydrogen. It leads to an increase in the oxidation state of an element (e.g., MgMg2++2eMg \rightarrow Mg^{2+} + 2e^-).

Reduction is defined as the gain of electrons, the loss of oxygen, or the gain of hydrogen. It leads to a decrease in the oxidation state of an element (e.g., Cl2+2e2ClCl_2 + 2e^- \rightarrow 2Cl^-).

A Redox reaction is a chemical process in which both oxidation and reduction occur simultaneously. The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) helps remember electron transfer.

Oxidizing Agents (oxidants) are substances that accept electrons from other species. In the process, the oxidizing agent is itself reduced.

Reducing Agents (reductants) are substances that donate electrons to other species. In the process, the reducing agent is itself oxidized.

Oxidation Number (or State) is a value assigned to an atom to represent the number of electrons lost or gained. Rules include: elements in their native state are 00, oxygen is usually 2-2, and hydrogen is usually +1+1.

The sum of oxidation numbers in a neutral compound is 00. In a polyatomic ion, the sum equals the charge of the ion (e.g., in SO42SO_4^{2-}, the sum is 2-2).

Disproportionation is a specific type of redox reaction where the same element is simultaneously oxidized and reduced from a single oxidation state to two different states.

📐Formulae

Oxidation:XXn++neOxidation: X \rightarrow X^{n+} + ne^-

Reduction:Yn++neYReduction: Y^{n+} + ne^- \rightarrow Y

Net Redox Equation: Red1+Ox2Ox1+Red2\text{Net Redox Equation: } Red_1 + Ox_2 \rightarrow Ox_1 + Red_2

Oxidation Numbers (Compound)=0\sum \text{Oxidation Numbers (Compound)} = 0

Oxidation Numbers (Ion)=Charge\sum \text{Oxidation Numbers (Ion)} = \text{Charge}

💡Examples

Problem 1:

Identify which species is oxidized and which is reduced in the reaction: Mg(s)+2HCl(aq)MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g).

Solution:

MgMg is oxidized and H+H^+ is reduced.

Explanation:

The oxidation state of MgMg increases from 00 in Mg(s)Mg(s) to +2+2 in MgCl2MgCl_2 (loss of electrons). The oxidation state of HH decreases from +1+1 in HClHCl to 00 in H2(g)H_2(g) (gain of electrons).

Problem 2:

Determine the oxidation state of Phosphorus (PP) in the phosphate ion PO43PO_4^{3-}.

Solution:

Oxidation state of P=+5P = +5.

Explanation:

Let the oxidation state of PP be xx. Oxygen is 2-2. The sum of oxidation states must equal the charge of the ion: x+4(2)=3x + 4(-2) = -3. Solving for xx: x8=3x=+5x - 8 = -3 \Rightarrow x = +5.

Problem 3:

In the reaction Zn(s)+CuSO4(aq)ZnSO4(aq)+Cu(s)Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s), identify the oxidizing agent.

Solution:

Cu2+Cu^{2+} (or CuSO4CuSO_4) is the oxidizing agent.

Explanation:

Cu2+Cu^{2+} gains two electrons to become Cu(s)Cu(s), meaning it is reduced. Since it causes the oxidation of ZnZn, it is the oxidizing agent.

Redox reactions - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry