UPSC Civil Services (Mains), Chemistry optional · 2025 · Paper I
Q1(i). Time to plate a gold layer
Question
The surface area of an object to be gold plated is 49.8 cm², and the density of gold is 19.3 g/cm³. A current of 3.25 A is applied to a solution that contains gold in the +3 oxidation state. Calculate the time required to deposit an even layer of gold, 1 × 10⁻³ cm thick, on the object. (Given: Molecular mass of gold = 196.97 g/mol)
Answer in brief
t = 435 s (about 7.2 minutes).
Key points
- At the cathode Au³⁺ + 3e⁻ → Au, so each mole of gold needs three moles of electrons (z = 3); Faraday’s law gives Q = nzF.
- Volume of layer = 49.8 × (1 × 10⁻³) = 0.0498 cm³; mass = 0.0498 × 19.3 = 0.961 g; n = 4.880 × 10⁻³ mol.
- Charge Q = (4.880 × 10⁻³)(3)(96485) = 1412 C.
- Time t = Q/I = 1412 C / 3.25 A = 435 s.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversFaraday's laws of electrolysiselectroplatinggold platingcharge and currentelectrodeposition
All parts of Question 1
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- 1(b) Born–Haber cycle for NaCl at 298 K5 marks
- 1(c) Raising the conductivity of Ge and Si by trace elements5 marks
- 1(d) Heat conduction between two copper sheets through air5 marks
- 1(e) Superheating of liquids and the Kelvin equation5 marks
- 1(f) Ascending order of dipole moment: NH₃, NF₃, H₂O5 marks
- 1(g) Heat of combustion of naphthalene from bomb calorimetry5 marks
- 1(h) Sealed sample: ethane, neon or a mixture5 marks
- 1(i) Time to plate a gold layer5 marks
- 1(j) Maximum work of a steam turbine5 marks