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

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 $2 \times 10^5 \, \text{C}$.