Redox

Redox

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A redox reaction couples oxidation (loss of electrons, oxidation number rises) with reduction (gain of electrons, oxidation number falls). The species that takes electrons is the oxidizing agent; the species that gives electrons is the reducing agent. Two must-know half-reactions in acidic medium:

• $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
• $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

For any galvanic cell: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ and $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, where n = electrons transferred, F = 96485 C mol⁻¹. High-yield pointers: (1) O is −1 in peroxides (H₂O₂, Na₂O₂) and −½ in superoxides (KO₂), not −2. (2) In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive. (3) A metal higher in the activity series (more negative E°) displaces a metal below it from solution.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Oxidation Number Rules

Assign oxidation states using this priority order: free elements = 0; Group 1 = +1; Group 2 = +2; F = −1; O = −2 (except peroxides −1, superoxides −½, OF₂ +2); H = +1 (except metal hydrides like NaH, CaH₂ where it is −1). In a neutral molecule the sum of oxidation numbers = 0; in a polyatomic ion it equals the ion’s charge. These rules are the foundation for the oxidation number method of balancing redox equations.

Balancing Redox Reactions

For acidic medium, use the ion-electron method: split into oxidation and reduction half-reactions, balance atoms other than O and H, add H₂O for O and H⁺ for H, then balance charge using electrons. Multiply half-reactions so that electrons cancel. For basic medium, add the same number of OH⁻ as H⁺ to both sides after balancing in acid. CUET UG typically gives a half-reaction (often dichromate or permanganate) and asks for the balanced full equation.

Oxidizing and Reducing Agents

Common oxidizing agents (gain electrons, O.N. decreases): KMnO₄, K₂Cr₂O₇, HNO₃, O₂, halogens, H₂O₂. Common reducing agents (lose electrons, O.N. increases): Zn, Fe, Mg, H₂, C, CO, SO₂, H₂S, FeSO₄, Na₂S₂O₃. The agent with the higher reduction potential is the stronger oxidant.

Electrochemical Cell Setup

In a galvanic cell, oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive terminal). A salt bridge (KCl or KNO₃) maintains charge neutrality by allowing ion flow. The standard hydrogen electrode (SHE) with E° = 0.00 V serves as the reference. A more positive E° means a greater tendency to be reduced.

Key Formulas

• $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
• $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ (F = 96485 C mol⁻¹)
• $K = n$-dependent equilibrium constant from E°

Half-reaction	E° (V)	Role
F₂ + 2e⁻ → 2F⁻	+2.87	Strongest oxidant
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51	Strong oxidant
2H⁺ + 2e⁻ → H₂	0.00	SHE reference
Zn²⁺ + 2e⁻ → Zn	−0.76	Strong reductant
Li⁺ + e⁻ → Li	−3.04	Strongest reductant

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Disproportionation and Edge Cases

A disproportionation reaction has the same element simultaneously oxidized and reduced. Examples: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$ (O goes from −1 to −2 and 0) and $2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}$ (Cu⁺ is both oxidised to Cu²⁺ and reduced to Cu⁰). The reverse (comproportionation) is also possible. MnO₄²⁻ in neutral/weakly basic solution disproportionates to MnO₄⁻ and MnO₂.

Linking Redox to Electrolysis

In an electrolytic cell, an external power source drives a non-spontaneous reaction. Unlike a galvanic cell, here the anode is connected to the positive terminal (so it remains the site of oxidation) and the cathode to the negative terminal (site of reduction). Faraday’s first law: $m = \frac{M \cdot I \cdot t}{n \cdot F}$ gives the mass deposited, where M is molar mass, I current, t time, n electrons per ion. This connects redox to quantitative CUET numericals.

Connections to Other Topics

Redox ties into (a) p-Block — interhalogens act as oxidising agents; (b) d-Block — variable oxidation states of Mn, Cr, Fe underpin KMnO₄ and K₂Cr₂O₇ chemistry; (c) Electrochemistry — Nernst equation $E = E^\circ - \frac{0.0591}{n}\log Q$ extends standard potentials to non-standard conditions; (d) Extraction of metals — calcination, roasting, and reduction by C or CO are all redox processes.

Common Mistakes

• Writing oxygen’s oxidation state as −2 inside H₂O₂ and Na₂O₂.
• Forgetting to add H₂O/H⁺ before balancing charge in half-reactions.
• Confusing anode polarity between galvanic and electrolytic cells.
• Treating a metal’s position on the activity series as the sole predictor — it works only for aqueous salt displacement, not for reactions with acids producing H₂.

Worked Example

For the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, given $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34$ V and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V: $E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10$ V. With n = 2, $\Delta G^\circ = -2 \times 96485 \times 1.10 = -2.12 \times 10^5$ J mol⁻¹. The negative ΔG° confirms spontaneity.

Practice Prompts

1. Balance in acidic medium: $\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Fe}^{3+} + \text{Cr}^{3+}$. Identify the oxidant, reductant, and the number of electrons transferred per dichromate.
2. Predict whether $2\text{Fe}^{3+} + \text{Sn}^{2+} \rightarrow 2\text{Fe}^{2+} + \text{Sn}^{4+}$ is spontaneous given $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77$ V and $E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15$ V. Calculate ΔG°.

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