Question

How many coulombs of electricity is required to produce 1g of sodium metal by reduction of sodium ion?

Hint:

Remember Faraday's first law of electrolysis: The mass of a substance deposited at an electrode is directly proportional to the quantity of electricity passed. The constant of proportionality involves the molar mass and the number of electrons in the reaction.

Answer 1

Step 1: Write the reduction half-reaction for sodium. \[ \text{Na}^+ + e^- \rightarrow \text{Na} \] This equation shows that 1 mole of electrons is required to produce 1 mole of sodium metal.
Step 2: Relate moles to mass and charge.
- Molar mass of Na = 23 g/mol.
- Charge of 1 mole of electrons = 1 Faraday (F) = 96500 Coulombs .
Step 3: Set up the proportion. From the stoichiometry, 96500 C of electricity are required to produce 23 g of Na. We need to find the charge (Q) required for 1 g of Na. \[ \frac{96500 \, \text{C}}{23 \, \text{g Na}} = \frac{Q}{1 \, \text{g Na}} \]
Step 4: Solve for Q. \[ Q = \frac{96500 \, \text{C}}{23} \approx 4195.65 \, \text{C} \] Approximately 4196 C of electricity is required.