Question

How many Coulombs are required to oxidise 0.1 mole of H₂O to oxygen?

• A. 1.93×10⁵ C
• B. 1.93×10⁴ C
• C. 3.86×10⁴ C
• D. 9.65×10³ C

Answer 1

The Correct Option is B

The balanced half-reaction for the oxidation of water to oxygen is:

$$2 \, \text{H}_2\text{O} \rightarrow O_2 + 4 \, \text{H}^+ + 4e^-$$

This equation shows that for 2 moles of water (H₂O), 4 moles of electrons (4e⁻) are required.

For 1 mole of water, 2 moles of electrons are needed.

Now, to calculate the total charge (in Coulombs) required:

• The charge required to transfer 1 mole of electrons is 1 Faraday = 96500 C.
• Since we need 2 moles of electrons to oxidize 1 mole of H₂O, the charge required is:

  Charge required = 2 × 96500 C = 193000 C = 1.93 × 10⁵ C

For 0.1 mole of water, the charge required will be:

Charge for 0.1 mole = 0.1 × 1.93 × 10⁵ C = 1.93 × 10⁴ C

Thus, the number of Coulombs required to oxidize 0.1 mole of H₂O to O₂ is 1.93 × 10⁴ C, making option (B) the correct answer.

Answer 2

The oxidation of water to oxygen occurs via electrolysis, where each water molecule loses 4 electrons to produce oxygen. - 1 mole of water = 6.022 × 10²³ molecules.
The total charge required to oxidize 1 mole of H₂O can be calculated using Faraday's law:
$Q = n \times F$,
where: 
$n$ is the number of moles of electrons and $F$ is Faraday’s constant (96,485 C/mol). Since 1 mole of water requires 4 moles of electrons: $$Q = 0.1 \times 4 \times 96,485 = 1.93 \times 10⁴ \, C$$