The amount of electricity in Coulomb required for the oxidation of 1 mol of \( \text{H}_2\text{O} \) to \( \text{O}_2 \) is \[\_ \times 10^5 \, \text{C}.\]
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 \( 2 \times 10^5 \, \text{C} \).
