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Chemistry - Redox

Grade 9IGCSE

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

🔑Concepts

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Redox stands for Reduction-Oxidation. These reactions involve the transfer of oxygen, hydrogen, or electrons.

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Oxidation can be defined as: the gain of oxygen (O2O_2), the loss of hydrogen (H2H_2), or the loss of electrons (e−e^-).

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Reduction can be defined as: the loss of oxygen (O2O_2), the gain of hydrogen (H2H_2), or the gain of electrons (e−e^-).

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A useful mnemonic for electron transfer is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

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An Oxidizing Agent is a substance that oxidizes another substance and is itself reduced in the process.

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A Reducing Agent is a substance that reduces another substance and is itself oxidized in the process.

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Oxidation State (Number): A value assigned to an atom in a compound. An increase in oxidation state indicates oxidation, while a decrease indicates reduction.

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Common test for Oxidizing Agents: Aqueous Potassium Iodide (KIKI) turns from colorless to brown (formation of I2I_2).

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Common test for Reducing Agents: Acidified Potassium Manganate(VII) (KMnO4KMnO_4) turns from purple to colorless (reduction of Mn7+Mn^{7+} to Mn2+Mn^{2+}).

📐Formulae

Oxidation: M→Mn++ne−\text{Oxidation: } M \rightarrow M^{n+} + ne^-

Reduction: X+ne−→Xn−\text{Reduction: } X + ne^- \rightarrow X^{n-}

Overall Redox: Zn+Cu2+→Zn2++Cu\text{Overall Redox: } Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu

Oxidation State of neutral element (e.g., Fe,O2,Cl2)=0\text{Oxidation State of neutral element (e.g., } Fe, O_2, Cl_2) = 0

Sum of oxidation states in a neutral compound=0\text{Sum of oxidation states in a neutral compound} = 0

💡Examples

Problem 1:

In the reaction CuO+H2→Cu+H2OCuO + H_2 \rightarrow Cu + H_2O, identify which substance is oxidized and which is reduced based on oxygen transfer.

Solution:

CuOCuO is reduced to CuCu; H2H_2 is oxidized to H2OH_2O.

Explanation:

Copper(II) oxide (CuOCuO) loses oxygen to become Copper (CuCu), which is reduction. Hydrogen (H2H_2) gains oxygen to become water (H2OH_2O), which is oxidation.

Problem 2:

Write the half-equations for the reaction between Magnesium and Chlorine: Mg+Cl2→MgCl2Mg + Cl_2 \rightarrow MgCl_2.

Solution:

Oxidation: Mg→Mg2++2e−\text{Oxidation: } Mg \rightarrow Mg^{2+} + 2e^- Reduction: Cl2+2e−→2Cl−\text{Reduction: } Cl_2 + 2e^- \rightarrow 2Cl^-

Explanation:

Magnesium atoms lose electrons to form Mg2+Mg^{2+} ions (Oxidation). Chlorine molecules gain those electrons to form Cl−Cl^- ions (Reduction).

Problem 3:

Determine the oxidation state of Sulfur (SS) in H2SO4H_2SO_4.

Solution:

2(+1)+x+4(−2)=02+x−8=0x−6=0x=+6\begin{array}{r} 2(+1) + x + 4(-2) = 0 \\ 2 + x - 8 = 0 \\ x - 6 = 0 \\ x = +6 \end{array}

Explanation:

The oxidation state of Hydrogen is +1+1 and Oxygen is −2-2. Since the compound is neutral, the sum must be zero. Solving for xx gives the oxidation state of Sulfur as +6+6.