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Aluminum metal is obtained commercially by electrolyzing Al2O3 mixed with cryolite (Na3AlF6).Electrochemistry Chemistry Question

Problem Context

Aluminum metal is obtained commercially by electrolyzing Al2O3 mixed with cryolite (Na3AlF6).

a.

Explain why electrolysis is used rather than heating the Al2O3 either directly or in the presence of C (as is done to extract Fe or Zn from their ores).

Model Answer

Al-O bonds are too strong to be broken by simple heating of the oxide, even in the presence of carbon. The heat of formation of Al2O3 is much more negative than that of CO2, so the reaction: 2Al2O3 + 3C → 4Al + 3CO2 is endothermic.

b.

State the purpose of the Na3AlF6.

Model Answer

Na3AlF6 is added to lower the melting point of Al2O3. Lowering the melting point also lowers the amount of energy needed to carry out the process.

c.

Write the two half-reactions that occur during electrolysis and indicate which of the two occurs at the cathode

Model Answer

Al3+ + 3e– → Al (cathode) and 2O2– → O2 + 4e–

d.

How many moles of electrons must pass through the cell to produce 5.00 kg of Al? (Assume 100% efficiency.)

Model Answer

5.00 × 103 g Al × (1 mol / 26.98 g) = 185.33 mol Al and 185.3 mol Al × (3 mol e– / 1 mol Al) = 556. mol e–

e.

Determine the current required (in amperes) if the aluminum in d. is produced in 10.0 hours.

Model Answer

We need to determine the charge (in coulomb) and time (in seconds) to calculate current.
556 mol e– × (96500 C / 1 mol e–) = 5.365 × 107 C.
and 10.0 hours × (60 min / 1 hour) × (60 sec / 1 min) = 3.6 × 104 sec.
so,

f.

Calculate the volume of gas formed in the process in d. at 25 ˚C and 720 mmHg.

Model Answer

556 mol e– × (1 mol O2 / 4 mol e–) = 139 mol O2
Now use Ideal Gas Law, V = nRT / P
= (139 mol O2)(0.0821 L atm mol-1 K-1)(298 K) / (720 mmHg × (1 atm / 760 mmHg))
= 3590 L

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