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Electricity will pass through a solution of copper(ii) sulphate (concentration 1.0 mol/dm³) using copper electrodes. Calculate the mass of copper deposited when 0.5 a of electricity is passed through the solution for 4 hours. (relative Atomic Mass (ram) of copper is 64) (1 Faraday of electricity is 96 500 coulombs this is the charge carried by 1 mole of electrons)
Accepted Answer
Here's how to calculate the mass of copper deposited:
1. Calculate the total charge passed:
- Current (I) = 0.5 A
- Time (t) = 4 hours = 4
3600 seconds = 14400 seconds
- Charge (Q) = I
t = 0.5 A
14400 s = 7200 C
2. Determine the moles of electrons transferred:
- 1 Faraday (96500 C) carries 1 mole of electrons
- Moles of electrons = Charge / Faraday = 7200 C / 96500 C/mol = 0.0746 mol
3. Write the half-reaction for copper deposition:
- Cu²⁺(aq) + 2e⁻ → Cu(s)
- This shows that 2 moles of electrons are required to deposit 1 mole of copper.
4. Calculate the moles of copper deposited:
- Moles of copper = (Moles of electrons) / 2 = 0.0746 mol / 2 = 0.0373 mol
5. Calculate the mass of copper deposited:
- Mass of copper = Moles of copper
RAM of copper
- Mass of copper = 0.0373 mol
64 g/mol = 2.39 g
Therefore, 2.39 grams of copper will be deposited.
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