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Redox Processes - Oxidation and reduction

Grade 12IBChemistry

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

🔑Concepts

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Oxidation is defined as the loss of electrons, the loss of hydrogen, or an increase in oxidation state. Reduction is the gain of electrons, the gain of hydrogen, or a decrease in oxidation state (OIL RIGOIL\ RIG).

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Oxidation State Rules: Atoms in elements are 00; monoatomic ions equal their charge; Oxygen is usually −2-2 (except in peroxides like H2O2H_2O_2 where it is −1-1); Hydrogen is +1+1 (except in metal hydrides where it is −1-1).

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Oxidizing Agents are species that undergo reduction and gain electrons, causing another species to be oxidized. Reducing Agents are species that undergo oxidation and lose electrons.

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A disproportionation reaction occurs when the same element is simultaneously oxidized and reduced, such as 2H2O2(aq)→2H2O(l)+O2(g)2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g).

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The Activity Series ranks metals by their ability to lose electrons. A metal higher in the series will displace the ions of a metal lower in the series from a solution (e.g., Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)).

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Voltaic (Galvanic) Cells convert chemical energy from spontaneous redox reactions into electrical energy. Oxidation occurs at the Anode (negative) and reduction at the Cathode (positive).

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Electrolytic Cells use electrical energy to drive non-spontaneous redox reactions. Oxidation occurs at the Anode (positive) and reduction at the Cathode (negative).

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The Winkler Method is used to measure Biochemical Oxygen Demand (BODBOD). It follows a ratio where 1 mol O2≡2 mol I2≡4 mol S2O32−1 \text{ mol } O_2 \equiv 2 \text{ mol } I_2 \equiv 4 \text{ mol } S_2O_3^{2-}.

📐Formulae

Ecellθ=Ecathodeθ−EanodeθE^{\theta}_{cell} = E^{\theta}_{cathode} - E^{\theta}_{anode}

ΔGθ=−nFEcellθ\Delta G^{\theta} = -nFE^{\theta}_{cell}

Q=I×tQ = I \times t

n=QFn = \frac{Q}{F}

Oxidation State Sum (Neutral)=0\text{Oxidation State Sum (Neutral)} = 0

Oxidation State Sum (Ion)=Charge of Ion\text{Oxidation State Sum (Ion)} = \text{Charge of Ion}

💡Examples

Problem 1:

Determine the oxidation state of Chromium in the dichromate ion, Cr2O72−Cr_2O_7^{2-}.

Solution:

+6+6

Explanation:

Let the oxidation state of CrCr be xx. Oxygen is typically −2-2. The sum of oxidation states must equal the charge of the ion: 2(x)+7(−2)=−22(x) + 7(-2) = -2. Solving for xx: 2x−14=−2⇒2x=12⇒x=+62x - 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6.

Problem 2:

Calculate the standard cell potential (EcellθE^{\theta}_{cell}) for a voltaic cell consisting of a Zn/Zn2+Zn/Zn^{2+} half-cell (Eθ=−0.76 VE^{\theta} = -0.76\ V) and a Cu/Cu2+Cu/Cu^{2+} half-cell (Eθ=+0.34 VE^{\theta} = +0.34\ V).

Solution:

+1.10 V+1.10\ V

Explanation:

The half-cell with the more positive reduction potential (CuCu) acts as the cathode. Ecellθ=Ecathodeθ−Eanodeθ=0.34 V−(−0.76 V)=1.10 VE^{\theta}_{cell} = E^{\theta}_{cathode} - E^{\theta}_{anode} = 0.34\ V - (-0.76\ V) = 1.10\ V.

Problem 3:

Balance the following redox equation in acidic solution: MnO4−(aq)+Fe2+(aq)→Mn2+(aq)+Fe3+(aq)MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq).

Solution:

MnO4−(aq)+5Fe2+(aq)+8H+(aq)→Mn2+(aq)+5Fe3+(aq)+4H2O(l)MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)

Explanation:

The reduction half-reaction is MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O. The oxidation half-reaction is Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-. To balance electrons, multiply the oxidation half-reaction by 55. Combining them gives the balanced equation.

Oxidation and reduction - Revision Notes & Key Formulas | IB Grade 12 Chemistry