Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20199 Apr 2019Evening ShiftChemistryElectrochemistryActual

A solution of N i N O 3 2 is electrolyzed between platinum electrode 0.1 Faraday electricity. How many mole of N i will be deposited at the cathode?

Options

  1. A0.10
  2. B0.15
  3. C0.05
  4. D0.20

Correct answer

C. 0.05

Step-by-step solution

N i + 2 + 2 e - → N i   2   F   charge gives = 1 mole of N i 0.1   F   charge gives = 1 2 × 0.1 = 0.05     m o l e     N i

Practice Electrochemistry on Quantrex Academy →

More from Electrochemistry

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: Salt Concentration (M) Molar conductivity (S cm ^2 mol ⁻ 2026Given at 298 K: E^ _ Fe²⁺/Fe = X Volt; E^ _ Fe³⁺/Fe = Y Volt. The E^ _ Fe³⁺/Fe²⁺ in Volt at 298 K is given by: 2026For a general redox reaction Anode : Red ₁ Ox ₁^ n₁^+ + n₁ e^- Cathode : Ox ₂ + n₂ e^- Red ₂^ n₂^- Which of the following statements is incorrect? 2026Consider the following data. Electrolyte ^ _m (S cm ^2 mol ⁻¹ ) BaCl ₂ x₁ H ₂ SO ₄ x₂ HCl x₃ BaSO ₄ is sparingly soluble in water. If the conductivity of the saturated BaSO ₄ solut 2026One half cell in a voltaic cell is constructed by dipping silver rod in AgNO₃ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO₄ . A v 2026At 298 K , the molar conductivity of x % (w/w) MX solution (aqueous) is 123.5 S cm ^2 mol ⁻¹ . The conductance of same solution is 1.9 10⁻³ S . The value of x is _______ 10⁻² . (Gi 2026An electrochemical cell, consist of the following two redox couples, M^ x+ (aq)/M(s)[E_ red ^ =+0.15 V ] and Fe³⁺(aq)/Fe(s)[E_ red ^ =-0.036 V ] . The cell EMF (E_ cell ) is record 2026Consider the following two half-cell reactions along with the standard reduction potential given: CO₂ + 6H^+ + 6e^- CH₃ OH + H₂ O E°_ red = 0.02 V 1 2 O₂ + 2H^+ + 2e^- H₂ O E°_ red 2026 Full Electrochemistry list All JEE Main PYQs