VKC ScienceChemistry for written exams Telegram

UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper I

Q3(d). Electrolysis of aqueous CuSO₄ between Pt electrodes

  • 20 marks
  • Answered
Question

800 milliampere of current is passed through an aqueous solution of CuSO₄ using two Pt electrodes for a period of 20 minutes. Predict the products at anode and cathode. Estimate the quantity of products. (At. weight of Cu is 63.6)

Answer in brief

Cathode: copper, 0.3164 g. Anode: oxygen, 0.0796 g (55.7 mL at STP). Sulfuric acid, 4.97 mmol, remains in solution.

Key points

  • Cu²+ (E°=+0.34 V) is reduced far more easily than water, so copper is deposited: Cu²+ + 2e⁻ → Cu
  • Pt is inert and SO₄²- is very hard to oxidise, so water is oxidised and oxygen is evolved: 2H₂O → O₂ + 4H⁺ + 4e⁻
  • 2CuSO₄ + 2H₂O → 2Cu + O₂ + 2H₂SO₄. The blue colour fades and the solution becomes acidic as Cu²+ is replaced by H⁺.
  • Charge passed: Q=It=0.800×1200=960 C

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Given: I=800 mA =0.800 A; t=20 min =1200 s; F=96 500 C mol⁻¹; Cu 63.6 g mol⁻¹.

Products

  1. Cathode (reduction): Cu²+ (E°=+0.34 V) is reduced far more easily than water, so copper is deposited: Cu²+ + 2e⁻ → Cu
  2. Anode (oxidation): Pt is inert and SO₄²- is very hard to oxidise, so water is oxidised and oxygen is evolved: 2H₂O → O₂ + 4H⁺ + 4e⁻
  3. Overall: 2CuSO₄ + 2H₂O → 2Cu + O₂ + 2H₂SO₄. The blue colour fades and the solution becomes acidic as Cu²+ is replaced by H⁺.

Quantities

  1. Charge passed: Q=It=0.800×1200=960 C
  2. Electrons: n(e⁻)=Q/F=(960)/(96 500)=9.948×10⁻³ mol
  3. Copper (2 electrons per atom): n(Cu)=4.974×10⁻³ mol; m=4.974×10⁻³×63.6 m(Cu)=0.3164 g
  4. Oxygen (4 electrons per molecule): n(O₂)=9.948×10⁻³4=2.487×10⁻³ mol; m=2.487×10⁻³×32.0 m(O₂)=0.0796 g (55.7 mL at STP; 60.8 mL at 25°C, 1 atm)
  5. Acid formed: n(H₂SO₄)=n(Cu)=4.974×10⁻³ mol (0.488 g).
  6. Check by Faraday's second law: masses deposited by the same charge are in the ratio of equivalent masses: 0.3164/0.0796=3.97=31.8/8.0.

The figures assume 100 % current efficiency and enough Cu²+ for 20 minutes; once Cu²+ is used up, hydrogen appears at the cathode.

Application. the same cathode reaction, with a copper anode that dissolves instead of giving oxygen, is the electrolytic refining of copper; with Pt electrodes the deposition is used for electrogravimetric estimation of copper.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q3(d), 2019 Paper I: solution page 15
Solution page 15 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 3