☰ Contents · Chemistry

Electrolysis

Lessons 20 · 1 lessons · I.R. Asqarov, K. G‘opirov, N.X. To‘xtaboyev. Chemistry Grade 9, revised 4th edition. “O‘zbekiston” NMIU, Tashkent, 2019
20

Electrolysis and its practical importance

Textbook: pp. 85–96
GoalKnow the essence of electrolysis and the rules for melts and solutions; write electrode processes; know the practical uses of electrolysis and calculate with Faraday's law.
New words
electrolysis · elektrolizcathode and anode · katod va anodinert anode · inert anodelectroplating · galvanik qoplash
Explanation

Electrolysis is the redox process that occurs when direct current passes through a solution or melt of an electrolyte: reduction at the cathode (–) and oxidation at the anode (+). In molten NaCl, Na⁺ + e⁻ → Na at the cathode and 2Cl⁻ – 2e⁻ → Cl₂ at the anode, i.e. 2NaCl → 2Na + Cl₂. In solution water also takes part. At the cathode, water rather than the cation is reduced for active metals up to aluminium (2H₂O + 2e⁻ → H₂ + 2OH⁻); metals after hydrogen (Cu, Ag, Au) are deposited; those in the middle (Zn, Fe, Ni, Pb) give both metal and hydrogen. At an inert anode (graphite, Pt) anions of oxygen-free acids (Cl⁻, Br⁻, I⁻, S²⁻) are oxidised, while for anions of oxygen acids (SO₄²⁻, NO₃⁻, PO₄³⁻) water is oxidised and O₂ is released; a soluble anode (Cu, Ni, Zn) dissolves itself. Examples: 2NaCl + 2H₂O → H₂ + Cl₂ + 2NaOH (table-salt solution gives hydrogen, chlorine and alkali); 2CuSO₄ + 2H₂O → 2Cu + O₂ + 2H₂SO₄; in Na₂SO₄ solution water is split: 2H₂O → 2H₂ + O₂. Electrolysis produces Na, Mg, Ca and Al, refines metals and plates objects with nickel, chromium or gold. Faraday's law: m = M·I·t / (n·F), where F = 96500 C/mol and n is the number of electrons. Electrolysis experiments are shown only by the teacher with a low-voltage source; chlorine is a poisonous gas.

Worked examples
CuCl₂ solution (inert electrodes): cathode Cu²⁺ + 2e⁻ → Cu; anode 2Cl⁻ – 2e⁻ → Cl₂. Overall: CuCl₂ → Cu + Cl₂.
Faraday: when 10 A flows for 9650 s, the mass of copper (M = 64, n = 2) is m = 64 · 10 · 9650 : (2 · 96500) = 32 g, because 10 · 9650 = 96500 C, i.e. 1 mol of electrons, which deposits 0.5 mol of Cu.
Class activity

“Electrode diagram”: draw a vessel with two electrodes on the board. For the solutions on the cards (KI, CuSO₄, Na₂SO₄, molten NaCl) teams mark the products at the cathode and anode. The teacher then shows the electrolysis of Na₂SO₄ solution at low voltage with universal indicator (only the teacher operates the equipment).

Practice
1
What is released at the cathode and anode when KI solution is electrolysed with inert electrodes? Write the overall equation.
2
How many moles of H₂ form when 36 g of water is electrolysed (2H₂O → 2H₂ + O₂)? (M(H₂O) = 18)
3
How many grams of copper are deposited at the cathode when 10 A flows through CuSO₄ solution for 9650 s? (m = M·I·t : (n·F), F = 96500, M = 64, n = 2)
4
Why is hydrogen, not sodium, released in the electrolysis of Na₂SO₄ solution?