A voltmeter containing silver trioxonitrate(V) solution was connected in series to another voltmeter containing copper(II)tetraoxosulphate(VI)…

JAMB Chemistry 2024 95 views

A voltmeter containing silver trioxonitrate(V) solution was connected in series to another voltmeter containing copper(II)tetraoxosulphate(VI) solution. when a current of 0.200 ampere was passed through the solution, 0.780g of silver was deposited. Calculate the mass of copper that would be deposited in the copper that would be deposited in the copper voltmeter[Cu = 63.5; Ag = 108; 1F = 96500C]

  1. 0.23g ✓
  2. 0.33g
  3. 0.43g
  4. 0.46g

Explanation

This question uses Faraday’s law of electrolysis. When two voltameters (electrolytic cells) are connected in series, the same quantity of electric charge passes through both solutions. This means we can use the amount of silver deposited to find how much copper is deposited.

Step 1: Find the moles of silver deposited.
Moles of Ag = Mass ÷ Molar mass
Moles of Ag = 0.780 ÷ 108 = 0.007222 mol

Step 2: Write the half-equations for each metal.
Silver: Ag⁺ + e⁻ → Ag (1 electron needed per atom)
Copper: Cu²⁺ + 2e⁻ → Cu (2 electrons needed per atom)

Step 3: Find the moles of electrons transferred.
Since each silver atom needs 1 electron, moles of electrons = moles of Ag = 0.007222 mol

Step 4: Find the moles of copper deposited.
Each copper atom needs 2 electrons. The same electrons flow through both cells.
Moles of Cu = 0.007222 ÷ 2 = 0.003611 mol

Step 5: Find the mass of copper deposited.
Mass = Moles × Molar mass
Mass = 0.003611 × 63.5 = 0.2293g ≈ 0.23g

Therefore, 0.23g of copper would be deposited.

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