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Electrochemistry - Electrolysis

Grade 12IGCSEChemistry

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

🔑Concepts

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Electrolysis is the process of decomposition of an electrolyte (an ionic compound in molten or aqueous state) by the passage of an electric current.

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The Anode is the positive electrode where oxidation occurs (loss of electrons). Anions move toward the anode. General half-equation: Xn−→X+ne−X^n- \rightarrow X + ne^-

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The Cathode is the negative electrode where reduction occurs (gain of electrons). Cations move toward the cathode. General half-equation: Mn++ne−→MM^{n+} + ne^- \rightarrow M

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Selective Discharge in Aqueous Solutions: At the cathode, the ion of the less reactive element is discharged (e.g., Cu2+Cu^{2+} is discharged before H+H^+). At the anode, halide ions (Cl−Cl^-, Br−Br^-, I−I^-) are discharged preferentially over OH−OH^-. If no halides are present, OH−OH^- is discharged to form O2O_2 and H2OH_2O.

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Faraday's First Law states that the mass (mm) of a substance produced at an electrode is directly proportional to the quantity of electricity (QQ) passed through the electrolyte.

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The Faraday constant (F≈96500 C mol−1F \approx 96500\text{ C mol}^{-1}) represents the charge of one mole of electrons: 1 mole e−=96500 C1\text{ mole } e^- = 96500\text{ C}.

📐Formulae

Q=I×tQ = I \times t

n(e−)=QFn(e^-) = \frac{Q}{F}

m=Q⋅Mrz⋅Fm = \frac{Q \cdot M_r}{z \cdot F}

n=I⋅tz⋅Fn = \frac{I \cdot t}{z \cdot F}

💡Examples

Problem 1:

During the electrolysis of aqueous copper(II) sulfate (CuSO4CuSO_4) using inert electrodes, a current of 2.0 A2.0\text{ A} is passed for 3030 minutes. Calculate the mass of copper deposited at the cathode. (Given: ArA_r of Cu=63.5Cu = 63.5, F=96500 C mol−1F = 96500\text{ C mol}^{-1})

Solution:

  1. Convert time to seconds: t=30×60=1800 st = 30 \times 60 = 1800\text{ s}.
  2. Calculate total charge: Q=I×t=2.0×1800=3600 CQ = I \times t = 2.0 \times 1800 = 3600\text{ C}.
  3. Determine moles of electrons: n(e−)=360096500≈0.0373 moln(e^-) = \frac{3600}{96500} \approx 0.0373\text{ mol}.
  4. Use the half-equation Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu to find moles of CuCu: Since z=2z = 2, n(Cu)=0.03732≈0.01865 moln(Cu) = \frac{0.0373}{2} \approx 0.01865\text{ mol}.
  5. Calculate mass: m=n×Ar=0.01865×63.5≈1.18 gm = n \times A_r = 0.01865 \times 63.5 \approx 1.18\text{ g}.

Explanation:

The mass of copper is determined by relating the total charge passed (QQ) to the stoichiometry of the reduction half-reaction at the cathode, where 2 moles of electrons are required to deposit 1 mole of copper metal.

Problem 2:

Predict the products at the anode and cathode for the electrolysis of concentrated aqueous sodium chloride (NaClNaCl).

Solution:

Cathode: H2H_2 gas is produced (2H++2e−→H22H^+ + 2e^- \rightarrow H_2). Anode: Cl2Cl_2 gas is produced (2Cl−→Cl2+2e−2Cl^- \rightarrow Cl_2 + 2e^-).

Explanation:

In aqueous NaClNaCl, both Na+Na^+ and H+H^+ ions migrate to the cathode. Since H+H^+ is lower in the reactivity series, it is preferentially reduced. At the anode, both Cl−Cl^- and OH−OH^- migrate; because the solution is concentrated, the halide ion (Cl−Cl^-) is discharged preferentially over OH−OH^-.

Electrolysis - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry