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Redox Reactions - Redox Reactions in Terms of Electron Transfer Reactions

Grade 11CBSEChemistry

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

🔑Concepts

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Redox reactions are defined as reactions involving the transfer of electrons from one chemical species to another.

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Oxidation is the process involving the loss of electrons by an atom, molecule, or ion. For example: Na→Na++e−Na \rightarrow Na^+ + e^-.

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Reduction is the process involving the gain of electrons by an atom, molecule, or ion. For example: Cl+e−→Cl−Cl + e^- \rightarrow Cl^-.

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An Oxidizing Agent (Oxidant) is a species that accepts electrons and undergoes reduction itself.

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A Reducing Agent (Reductant) is a species that dones/loses electrons and undergoes oxidation itself.

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Redox reactions can be split into two half-reactions: the oxidation half-reaction and the reduction half-reaction. The sum of these half-reactions gives the overall net ionic equation.

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In a balanced redox reaction, the total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.

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Competitive electron transfer reactions demonstrate the relative tendency of metals to lose electrons. For example, ZnZn has a greater tendency to lose electrons than CuCu, as seen in the reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s).

📐Formulae

M→Mn++ne− (Oxidation half-reaction)M \rightarrow M^{n+} + ne^- \text{ (Oxidation half-reaction)}

X+ne−→Xn− (Reduction half-reaction)X + ne^- \rightarrow X^{n-} \text{ (Reduction half-reaction)}

Reducing Agent→Oxidized Product+ne−\text{Reducing Agent} \rightarrow \text{Oxidized Product} + ne^-

Oxidizing Agent+ne−→Reduced Product\text{Oxidizing Agent} + ne^- \rightarrow \text{Reduced Product}

💡Examples

Problem 1:

Identify the species undergoing oxidation and reduction in the following reaction: 2Na(s)+Cl2(g)→2NaCl(s)2Na(s) + Cl_2(g) \rightarrow 2NaCl(s).

Solution:

NaNa is oxidized to Na+Na^+ and Cl2Cl_2 is reduced to Cl−Cl^-.

Explanation:

In this reaction, each sodium atom loses one electron: Na→Na++e−Na \rightarrow Na^+ + e^- (Oxidation). Each chlorine atom in the Cl2Cl_2 molecule gains one electron: Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^- (Reduction). Therefore, NaNa acts as the reducing agent and Cl2Cl_2 acts as the oxidizing agent.

Problem 2:

Explain the electron transfer in the reaction between Zinc and Copper(II) ions: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s).

Solution:

Oxidation: Zn(s)→Zn2+(aq)+2e−Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-; Reduction: Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s).

Explanation:

Zinc loses two electrons to form Zn2+Zn^{2+} ions, which is an oxidation process. Copper(II) ions gain those two electrons to form solid copper, which is a reduction process. ZnZn is the reductant and Cu2+Cu^{2+} is the oxidant.