Number of coulombs required to oxidise one mole of $H_{2}O$ to $O_{2}$ are:

Question

Accepted Answer

To determine the number of coulombs required to oxidize one mole of $ H_2O $ to $ O_2 $, we need to consider the balanced chemical equation for the oxidation process:

$$ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- $$

This equation shows that 4 moles of electrons are required to oxidize 2 moles of water to produce 1 mole of oxygen gas. Therefore, to oxidize 1 mole of $ H_2O $, 2 moles of electrons are needed.

The charge of one mole of electrons is given by Faraday's constant, which is approximately 96500 coulombs per mole of electrons. Therefore, the total charge required to oxidize 1 mole of $ H_2O $ is:

$$ 2 \text{ moles of electrons} \times 96500 \text{ C/mole} = 193000 \text{ C} $$

This value can be expressed in scientific notation as $ 1.93 \times 10^5 $ C, which corresponds to Option 1.

**Explanation for why each of the other options is incorrect:**

- **Option 2: 96500 C** - This option represents the charge required for 1 mole of electrons. However, since 2 moles of electrons are needed to oxidize 1 mole of $ H_2O $, this option underestimates the required charge by a factor of 2.

- **Option 3: $\frac{96500}{2}C$** - This option suggests that only half a mole of electrons is needed, which is incorrect. The balanced equation clearly shows that 2 moles of electrons are required for the oxidation of 1 mole of $ H_2O $.

- **Option 4: $19.3 \times 10^5 C$** - This option overestimates the required charge by a factor of 10. The correct calculation shows that $ 1.93 \times 10^5 $ C is needed, not $ 19.3 \times 10^5 $ C.

In conclusion, Option 1 is the correct answer because it accurately reflects the charge required to oxidize one mole of $ H_2O $ to $ O_2 $ based on the stoichiometry of the reaction and Faraday's constant.