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Electrochemistry - Electrolysis and laws of electrolysis

Grade 12ICSEChemistry

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 by the passage of electric current through its aqueous solution or molten state. It occurs in an electrolytic cell where electrical energy is converted into chemical energy.

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At the Cathode (Negative Electrode): Reduction takes place. Cations move toward the cathode and gain electrons, e.g., Na++e−→NaNa^+ + e^- \rightarrow Na.

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At the Anode (Positive Electrode): Oxidation takes place. Anions move toward the anode and lose electrons, e.g., 2Cl−→Cl2+2e−2Cl^- \rightarrow Cl_2 + 2e^-.

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Faraday's First Law of Electrolysis: The mass of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte (w∝Qw \propto Q).

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Faraday's Second Law of Electrolysis: When the same quantity of electricity is passed through different electrolytes connected in series, the masses of the substances produced are proportional to their chemical equivalent weights (w∝Ew \propto E).

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Electrochemical Equivalent (ZZ): It is the mass of the substance deposited by passing 11 Coulomb of charge or 11 Ampere of current for 11 second.

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Preferential Discharge Theory: If more than one type of ion is attracted to an electrode, the ion with the lower discharge potential (or higher reduction potential for cations) is discharged first. For example, in aqueous NaClNaCl, H+H^+ is discharged at the cathode instead of Na+Na^+ because EH+/H2∘>ENa+/Na∘E^\circ_{H^+/H_2} > E^\circ_{Na^+/Na}.

📐Formulae

Q=I×tQ = I \times t

w=Z×I×tw = Z \times I \times t

Z=Equivalent Weight96500=Molar Massn×96500Z = \frac{\text{Equivalent Weight}}{96500} = \frac{\text{Molar Mass}}{n \times 96500}

w=M⋅I⋅tn⋅Fw = \frac{M \cdot I \cdot t}{n \cdot F}

w1w2=E1E2\frac{w_1}{w_2} = \frac{E_1}{E_2}

F=NA×e≈96500 C mol−1F = N_A \times e \approx 96500 \text{ C mol}^{-1}

💡Examples

Problem 1:

Calculate the mass of magnesium deposited when a current of 0.50.5 A is passed through molten MgCl2MgCl_2 for 22 hours. (Atomic mass of Mg=24Mg = 24 g/mol)

Solution:

Given: I=0.5I = 0.5 A, t=2×60×60=7200t = 2 \times 60 \times 60 = 7200 s. The reaction at cathode is Mg2++2e−→MgMg^{2+} + 2e^- \rightarrow Mg. Here, n=2n = 2. Molar mass M=24M = 24. Equivalent weight E=Mn=242=12E = \frac{M}{n} = \frac{24}{2} = 12. Using formula w=E⋅I⋅tFw = \frac{E \cdot I \cdot t}{F}: w=12×0.5×720096500≈0.447w = \frac{12 \times 0.5 \times 7200}{96500} \approx 0.447 g.

Explanation:

We first convert time to seconds and identify the valency (n=2n=2) for magnesium ions to find the equivalent weight. Applying Faraday's first law gives the total mass.

Problem 2:

A constant current of 1.51.5 A was passed through an electrolytic cell containing AgNO3AgNO_3 solution until 1.451.45 g of silver was deposited. How long did the current flow? (Atomic mass of Ag=108Ag = 108 g/mol)

Solution:

Reaction: Ag++e−→AgAg^+ + e^- \rightarrow Ag, so n=1n = 1. w=1.45w = 1.45 g, I=1.5I = 1.5 A, M=108M = 108. From w=M⋅I⋅tn⋅Fw = \frac{M \cdot I \cdot t}{n \cdot F}, we get t=w⋅n⋅FM⋅I=1.45×1×96500108×1.5≈863.7t = \frac{w \cdot n \cdot F}{M \cdot I} = \frac{1.45 \times 1 \times 96500}{108 \times 1.5} \approx 863.7 s.

Explanation:

The time tt is calculated by rearranging Faraday's first law formula, using the electrochemical equivalent of silver.

Electrolysis and laws of electrolysis Revision - Class 12 Chemistry ICSE