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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 · 2018 · Paper I

Q3(c)(ii). Transport number of Ag⁺ by the Hittorf method

  • 10 marks
  • Answered
Question

A silver nitrate solution containing 0.00739 gm of AgNO₃ per gm of water is electrolysed between silver electrodes. During the experiment 0.078 gm of Ag was deposited on the cathode. At the end of the experiment anode solution contained 23.14 gm of H₂O and 0.236 gm of AgNO₃. What is the transport number of Ag⁺? After electrolysis, the anode solution has (a) 23.14 gm water and (b) 0.236 gm AgNO₃.

Answer in brief

t(Ag⁺)=0.471 and t(NO₃⁻)=0.529.

Key points

  • Before electrolysis the same 23.14 g of water carried 23.14×0.00739=0.1710 g AgNO₃ =(0.1710)/(169.87)=1.0067×10⁻³ mol
  • (0.236)/(169.87)=1.3893×10⁻³ mol Gain =0.3826×10⁻³ mol
  • Faradays passed = Ag deposited =(0.078)/(107.87)=0.7231×10⁻³ mol. The same amount of Ag dissolved at the anode.
  • Without migration the anode solution would have gained 0.7231×10⁻³ mol. Ag⁺ that migrated out: 0.7231×10⁻³-0.3826×10⁻³=0.3405×10⁻³ mol

The full model answer

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

Principle: the silver anode dissolves (Ag → Ag⁺ + e⁻), adding one mole of Ag⁺ per faraday, while t_+ mole of Ag⁺ migrate out towards the cathode. Net gain of AgNO₃ in the anode compartment =(1-t_+) per faraday =t₋. The amount of Ag on the cathode measures the faradays passed (it acts as a coulometer).

  1. Before electrolysis the same 23.14 g of water carried 23.14×0.00739=0.1710 g AgNO₃ =(0.1710)/(169.87)=1.0067×10⁻³ mol
  2. After electrolysis: (0.236)/(169.87)=1.3893×10⁻³ mol Gain =0.3826×10⁻³ mol
  3. Faradays passed = Ag deposited =(0.078)/(107.87)=0.7231×10⁻³ mol. The same amount of Ag dissolved at the anode.
  4. Without migration the anode solution would have gained 0.7231×10⁻³ mol. Ag⁺ that migrated out: 0.7231×10⁻³-0.3826×10⁻³=0.3405×10⁻³ mol
  5. t(Ag⁺)=(Ag⁺ migrated)/(faradays passed)=(0.3405)/(0.7231) t(Ag⁺)=0.471
  6. t(NO₃⁻)=1-0.471= 0.529. Check: the gain equals t₋× faradays =0.529×0.7231×10⁻³=0.3826×10⁻³ mol, as found.
  7. The nitrate ion carries slightly more of the current than Ag⁺: it moves a little faster. The repeated data in the last sentence of the question add nothing new.

Application. transport numbers are needed to correct the e.m.f. of concentration cells with transference (E=2t_∓RT/Fln(a₂)/(a₁)) and to split molar conductivities into ionic conductivities; the moving-boundary method checks the Hittorf values.

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Model answer to Q3(c)(ii), 2018 Paper I: solution page 20
Solution page 20 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 3