Question

The amount of electricity in Coulomb required for the oxidation of 1 mol of \( \text{H}_2\text{O} \) to \( \text{O}_2 \) is \[\_ \times 10^5 \, \text{C}.\]

Answer 1

Correct Answer: 2

Step-by-step Calculation:
The oxidation of water to oxygen gas involves the half-reaction:
\[2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-\]
This reaction shows that 4 moles of electrons are required to oxidize 2 moles of water to produce 1 mole of \( \text{O}_2 \).
The amount of electricity required to transfer 1 mole of electrons is given by Faraday's constant:
\[F = 96500 \, \text{C mol}^{-1}\]
Therefore, the total charge required for the oxidation of 1 mole of \( \text{H}_2\text{O} \) to \( \text{O}_2 \) is:
\[\text{Charge} = 4 \times F = 4 \times 96500 = 386000 \, \text{C} = 3.86 \times 10^5 \, \text{C}\]
Conclusion:The amount of electricity required for the oxidation of 1 mole of \( \text{H}_2\text{O} \) to \( \text{O}_2 \) is \( 2 \times 10^5 \, \text{C} \).

Answer 2

Given:

The given chemical reaction is: 

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

Step 1: The relationship between work and energy is given by the equation:

\[ \frac{W}{E} = \frac{Q}{96500} \]

Step 2: For one mole of the reaction, the n-factor is 2 (since there are 4 electrons involved per 2 moles of water). We can write:

\[ \text{mole} \times \text{n-factor} = \frac{Q}{96500} \]

Step 3: Substituting the values for one mole and n-factor 2, we get:

\[ 1 \times 2 = \frac{Q}{96500} \]

Step 4: Solving for \( Q \), we find:

\[ Q = 2 \times 96500 \, C = 1.93 \times 10^5 \, C \]

Final Answer:

The value of \( Q \) is \( 1.93 \times 10^5 \, C \).