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Electrochemistry - Electrolytic Cells and Electrolysis

Grade 12CBSEChemistry

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 electricity through its aqueous solution or fused state.

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In an Electrolytic Cell, electrical energy is converted into chemical energy. The anode is the positive electrode (oxidation occurs) and the cathode is the negative electrode (reduction occurs).

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

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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 substances produced at the electrodes are directly proportional to their chemical equivalent weights: \frac{w_1}{w_2} = rac{E_1}{E_2}.

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Electrochemical Equivalent (ZZ): It is the mass of the substance deposited by 11 Coulomb of charge. Z=E96500Z = \frac{E}{96500}, where EE is the equivalent weight.

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The products of electrolysis depend on the nature of the material being electrolyzed and the type of electrodes used (inert electrodes like PtPt or AuAu vs. reactive electrodes).

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In the electrolysis of aqueous NaClNaCl, H2H_2 gas is liberated at the cathode instead of NaNa because the reduction potential of H2OH_2O (−0.83 V-0.83\ V) is higher than that of Na+Na^+ (−2.71 V-2.71\ V).

📐Formulae

Q=I×tQ = I \times t

w=Z×I×tw = Z \times I \times t

Z=Molar Massn×FZ = \frac{\text{Molar Mass}}{n \times F}

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

1 F=96487 C mol−1≈96500 C mol−11\ F = 96487\ C\ mol^{-1} \approx 96500\ C\ mol^{-1}

Equivalent Weight (E)=Molar MassValency factor (n)\text{Equivalent Weight } (E) = \frac{\text{Molar Mass}}{\text{Valency factor } (n)}

💡Examples

Problem 1:

A solution of CuSO4CuSO_4 is electrolyzed for 1010 minutes with a current of 1.51.5 amperes. What is the mass of copper deposited at the cathode? (Atomic mass of Cu=63.5 uCu = 63.5\ u)

Solution:

Given: I=1.5 AI = 1.5\ A, t=10×60=600 st = 10 \times 60 = 600\ s. Charge Q=I×t=1.5×600=900 CQ = I \times t = 1.5 \times 600 = 900\ C. The cathode reaction is Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s). To deposit 1 mol1\ mol of CuCu, 2 F2\ F (2×96500 C2 \times 96500\ C) is required. Mass w=M×Qn×F=63.5×9002×96500=0.296 gw = \frac{M \times Q}{n \times F} = \frac{63.5 \times 900}{2 \times 96500} = 0.296\ g.

Explanation:

We first calculate the total charge QQ in Coulombs. Using the stoichiometry of the reduction reaction (n=2n=2), we apply Faraday's Law to find the mass deposited.

Problem 2:

During the electrolysis of molten Al2O3Al_2O_3, how many Coulombs are required to produce 40.5 g40.5\ g of AlAl? (Atomic mass of Al=27 uAl = 27\ u)

Solution:

The reaction at the cathode is Al3++3e−→Al(s)Al^{3+} + 3e^- \rightarrow Al(s). Moles of Al=40.527=1.5 molAl = \frac{40.5}{27} = 1.5\ mol. Since 1 mol1\ mol of AlAl requires 3 F3\ F of charge, 1.5 mol1.5\ mol requires 1.5×3 F=4.5 F1.5 \times 3\ F = 4.5\ F. Total charge Q=4.5×96500=434250 CQ = 4.5 \times 96500 = 434250\ C.

Explanation:

The number of moles of electrons required is calculated based on the valency of Aluminum (n=3n=3). Multiplying the moles of substance by nn and Faraday's constant gives the total charge.

Electrolytic Cells and Electrolysis Revision - Class 12 Chemistry CBSE