NEET · Chemistry —

Electrochemistry appeared 2 times across 1 year — 4.4% of Chemistry. This question is from Faraday's Laws of Electrolysis.

Year 2024 Total
Questions 2 2

A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = 63 g mol⁻¹; 1 F = 96487 C mol⁻¹)

Solution & Explanation

Related Formula
w = (M × I × t)/(n × F)
Core Logic

For Cu²⁺ + 2e^- arrow Cu(s), n = 2. I = 1.5 A, t = 10 × 60 = 600 s.

w = (63 × 1.5 × 600)/(2 × 96487) = (56700)/(192974) ≈ 0.2938 g
Step 1: Conclusion

Mass of copper deposited is 0.2938 g.

Pattern Recognition

Apply Faraday's First Law: w = Z I t = (M)/(n F) I t.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Reference Study Guides

More Electrochemistry Previous-Year Questions

Q55 neet_2024_05_may_morning Nernst Equation and Half-Cell Potential
Calculate emf of the half cell given below: Pt(s) H₂(g, 2 atm) HCl (aq, 0.02 M) E°H₂/H^+ = 0 (Given: (2.303 RT)/(F) = 0.059, 2 = 0.3010)
  • A. (1) 0.109 V
  • B. (2) 0.035 V
  • C. (3) -0.035 V
  • D. (4) -0.109 V

Solution

Related Formula
H₂(g) arrow 2H^+(aq) + 2e^- E = E° - (0.059)/(n) [H^+]²PH₂
Core Logic
[H^+] = 0.02 M, PH₂ = 2 atm, n = 2 E = 0 - (0.059)/(2) ((0.02)²)/(2) = -0.0295 (2 × 10⁻⁴) E = -0.0295 ( 2 - 4) = -0.0295 (0.3010 - 4) = -0.0295 (-3.699) = +0.109 V
Step 1: Final Value

Half cell EMF is +0.109 V.

Pattern Recognition

For oxidation half cell H₂ arrow 2H^+ + 2e^-, Nernst equation uses ratio [H^+]² / PH₂.

Chapter Mix

Class 12 Chemistry: Electrochemistry

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)