A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
The mass of Ni deposited at the cathode is 1.825 g.
Step 1: Calculate the charge passed Current (I) = 5 amperes Time (t) = 20 minutes = = 1200 seconds Charge (Q) = Current × Time = 5 A × 1200 s = 6000 C
Step 2: Write the electrode reaction At the cathode: Ni²⁺(aq) + 2e⁻ → Ni(s) This shows that 2 moles of electrons (2F) are required to deposit 1 mole of Ni.
Step 3: Calculate mass of Ni deposited Using the formula from the context: Mass of Ni = (Molar mass × Charge) / (n × F) Mass of Ni = (58.7 g mol⁻¹ × 6000 C) / ( C mol⁻¹) Mass of Ni = 352200 / 192974 = 1.825 g
Explanation
This question applies Faraday's laws of electrolysis. The context provides Example 2.10 which demonstrates the exact calculation method for CuSO₄ electrolysis. The same approach is applied here for Ni(NO₃)₂. The key steps are: (1) calculate total charge using Q = I × t, (2) identify the stoichiometry of the cathode reaction (Ni²⁺ requires 2 electrons), and (3) apply the mass formula using Faraday's constant (96487 C/mol). The atomic mass of Ni (58.7 g/mol) is used for the calculation.
Solution Steps
Step 1: Convert time to seconds and calculate charge: t = = 1200 s; Q = = 6000 C
Step 2: Identify electrode reaction stoichiometry: Ni²⁺ + 2e⁻ → Ni (requires 2F per mole of Ni)
Step 3: Apply mass formula: Mass = ()/() = 1.825 g