# What quantity of electricity will liberate 0.125 mole of oxygen…

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A. 24,125 C

B. 48,250 C

C. 72,375 C

D. 96,500 C

A

## DETAILS…

According to Faraday’s law of electrolysis,

• For a mono-valent ion (eq: H, Cl¯, Na⁺, K⁺, F¯), 1 mole of its atom or molecule will be liberated by 96,500 coulombs of electricity.
• For a di-valent ion (eq: Mg²⁺, O²¯, Ca²⁺), 1 mole of it will be liberated by 2 × 96500 = 193,000 coulombs of electricity.
• For a tri-valent ion (eq: Al³), 1 mole of it will be liberated by 3 × 96500 = 289,500 coulombs of electricity.

To resolve this problem, oxygen is di-valent, it means:

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1 mole of oxygen molecule = 193000C, therefore,

0.125 mole will be = X

Cross multiplying to find X, we have:

X(1) = 0.125 × 193000

X = 24,125 coulombs

### Now for the right answer to the above question:

1. Option A is correct.
2. Option B is incorrect.
3. C is incorrect.
4. D is not the correct answer.

## KEY-POINTS…

• For a mono-valent ion, the quantity of electricity needed to liberate 1 mole of its product = 96500C
• For a di-valent ion, the quantity of electricity needed to liberate 1 mole of its product = 2 × 96500 = 193000C
• For a trivalent ion like aluminum ion, the quantity of electricity required to liberate 1 mole of its product will be = 3 × 96500C = 289500C
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