Question

Difficulty: HardAlkali Metals: Sodium Extraction, Properties, and Compounds

During the industrial extraction of sodium metal by the electrolysis of molten sodium chloride in a Downs cell, a steady current of 9.65 A9.65\text{ A} is passed through the electrolytic cell for 50 minutes50\text{ minutes}. What mass of pure sodium metal is collected at the cathode? [Na=23\text{Na} = 23, 1 F=96,500 C mol11\text{ F} = 96,500\text{ C mol}^{-1}]

  1. 6.9 g6.9\text{ g}Answer
  2. B
    3.45 g3.45\text{ g}
  3. C
    13.8 g13.8\text{ g}
  4. D
    6.72 g6.72\text{ g}

Answer

The mass of pure sodium metal collected at the cathode is 6.9 g6.9\text{ g}.
At the cathode of the Downs cell, sodium ions undergo single-electron reduction (Na++eNa\text{Na}^+ + e^- \rightarrow \text{Na}). Passing 28,950 C28,950\text{ C} of charge transfers 0.3 mol0.3\text{ mol} of electrons. Multiplying 0.3 mol0.3\text{ mol} by the atomic mass of sodium (23 g mol123\text{ g mol}^{-1}) yields exactly 6.9 g6.9\text{ g}.

Step-by-Step Solution

1
Calculate the total electric charge (QQ) passed through the cell in seconds.
Q=I×t=9.65 A×(50×60 s)=28,950 CQ = I \times t = 9.65\text{ A} \times (50 \times 60\text{ s}) = 28,950\text{ C}.
Electric charge is determined by multiplying current in amperes by time in seconds.
2
Calculate the amount of substance (in moles) of electrons transferred.
Moles of e=QF=28,950 C96,500 C mol1=0.3 mole^- = \frac{Q}{F} = \frac{28,950\text{ C}}{96,500\text{ C mol}^{-1}} = 0.3\text{ mol}.
One Faraday (96,500 C96,500\text{ C}) corresponds to the electric charge of one mole of electrons.
3
Relate the moles of electrons to the moles of sodium metal deposited using the cathode reduction half-equation.
Cathode reaction: Na++eNa\text{Na}^+ + e^- \rightarrow \text{Na}. Thus, 1 mol1\text{ mol} of ee^- produces 1 mol1\text{ mol} of Na\text{Na}, yielding 0.3 mol0.3\text{ mol} of Na\text{Na}.
Sodium is a univalent alkali metal cation requiring 1 electron per discharged ion.
4
Calculate the mass of sodium metal deposited.
Mass of Na=0.3 mol×23 g mol1=6.9 g\text{Na} = 0.3\text{ mol} \times 23\text{ g mol}^{-1} = 6.9\text{ g}.
Mass is obtained by multiplying the number of moles by the relative atomic mass.

Key Concept

Quantitative Electrolysis of Molten Salts in Downs Cell Extraction
Estimated Time:2m 0s
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