The amount of electricity in Coulomb required for the oxidation of 1 mol of $$H_2O$$ to $$O_2$$ is ______ $$\times 10^5$$ C.

Find the charge required for oxidation of 1 mol $$H_2O$$ to $$O_2$$.

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

This shows that 2 mol of water loses 4 mol of electrons to produce 1 mol of $$O_2$$.

From the balanced equation: 2 mol $$H_2O$$ gives 4 mol $$e^-$$.

Therefore: 1 mol $$H_2O$$ gives 2 mol $$e^-$$.

$$ Q = n_e \times F = 2 \times 96500 = 193000 \text{ C} = 1.93 \times 10^5 \text{ C} \approx 2 \times 10^5 \text{ C} $$

The answer is 2 $$\times 10^5$$ C.