Chem6 Flashcards

1
Q

Why are graphite anodes replaced in aluminum extraction?

A

O₂ reacts with them: 2C(s) + ³⁄₂O₂(g) → CO(g) + CO₂(g), causing decay.

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2
Q

What two advantages do modern fluoride salt mixtures have over cryolite?

A
  1. Lower melting point (saves energy).
  2. Lower density (eases Al separation).
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3
Q

At which electrode does metal deposition always occur?

A

Cathode (reduction site).

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4
Q

What gases form when graphite reacts with oxygen?

A

CO and CO₂.

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5
Q

Why is lowering the melting point crucial in aluminum extraction?

A

Reduces energy costs (from 2045°C → 950°C).

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6
Q

How does lower density of the modern mixture help?

A

Molten aluminum sinks more easily for collection.

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7
Q

What is the key difference between cryolite and the modern fluoride mixture?

A

The modern mix has Na/Ca/Al fluorides (not just NaAlF₆).

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8
Q

What happens to aluminum ions (Al³⁺) during electrolysis?

A

They gain electrons at the cathode: Al³⁺ + 3e⁻ → Al(l).

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9
Q

What type of reaction occurs at the anode in all three experiments?

A

Oxidation (loss of electrons).

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10
Q

What is the primary energy cost in aluminum extraction?

A

Electricity for electrolysis (hence the need for lower MP mixtures).

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11
Q

Why is fluorspar (CaF₂) added to the aluminum extraction mix?

A

Lowers melting point further, saving energy.

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12
Q

What is the industrial name for the aluminum extraction process?

A

Hall-Héroult process.

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