In the electrolysis of water, 1 F of electrical energy would evolve -

Options

A. 1 mole of oxygen
B. 1 g atom of oxygen
C. 8 g of oxygen
D. 22.4 L of oxygen

Answer

To solve the question regarding the electrolysis of water and the amount of oxygen evolved from 1 Farad of electrical energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electrolysis Reaction**:
The electrolysis of water can be represented by the following reaction:
\[ 2H_2O \rightarrow 2H_2 + O_2 \]
This means that 2 moles of water produce 2 moles of hydrogen gas and 1 mole of oxygen gas.

2. **Determine the Change in Oxidation States**:
In this reaction, the oxidation state of oxygen in water (H2O) is -2, and it changes to 0 in O2. This indicates that each oxygen atom gains 2 electrons during the reaction.

3. **Calculate the Faraday Requirement**:
From the reaction, we can see that to produce 1 mole of O2, we need 4 moles of electrons (since 2 electrons are required for each of the 2 oxygen atoms). According to Faraday's laws of electrolysis:
\[ 1 \text{ mole of electrons} = 1 \text{ Faraday} \]
Therefore, to produce 1 mole of O2, we need 4 Faradays of electricity.

4. **Relate Faraday to Moles of O2**:
If 4 Faradays produce 1 mole of O2, then 1 Faraday will produce:
\[ \frac{1}{4} \text{ moles of O2} \]

5. **Convert Moles of O2 to Grams**:
The molar mass of O2 (oxygen gas) is 32 g/mol. Therefore, the mass of oxygen produced from 1 Faraday is:
\[ \frac{1}{4} \text{ moles of O2} \times 32 \text{ g/mol} = 8 \text{ grams of O2} \]

### Final Answer:
Thus, from the electrolysis of water, 1 Farad of electrical energy would evolve **8 grams of oxygen**.