Expansion Work Done in Complete Electrolysis of Water

To find the expansion work done during the complete electrolysis of water, we proceed with the following steps:

Step 1: Write the balanced chemical equation for the electrolysis of water The electrolysis of liquid water into hydrogen and oxygen gas is represented by: $\text{H}_2\text{O}(l) \longrightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g)$

Step 2: Determine the number of moles of water electrolyzed The molar mass of water ($\text{H}_2\text{O}$) is: $M_{\text{H}_2\text{O}} = 2 \times 1 + 16 = 18\text{ g mol}^{-1}$

The number of moles of $\text{H}_2\text{O}$ in $144\text{ g}$ is: $n_{\text{H}_2\text{O}} = \frac{144\text{ g}}{18\text{ g mol}^{-1}} = 8\text{ mol}$

Step 3: Calculate the moles of gaseous products formed From stoichiometry:

• Moles of $\text{H}_2(g)$ produced $= 8\text{ mol}$
• Moles of $\text{O}_2(g)$ produced $= \frac{1}{2} \times 8 = 4\text{ mol}$

Thus, the change in the number of moles of gas is: $\Delta n_g = n_{\text{gas, products}} - n_{\text{gas, reactants}} = (8 + 4) - 0 = 12\text{ mol}$

Step 4: Calculate the work done Assuming ideal gas behavior and that the molar volume of liquid water is negligible compared to the gaseous products: $w = -P\Delta V = -\Delta n_g R T$

Given:

• $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$
• $T = 300\text{ K}$

Substituting the values: $w = -12\text{ mol} \times 8.3\text{ J K}^{-1}\text{ mol}^{-1} \times 300\text{ K}$ $w = -29880\text{ J} = -29.88\text{ kJ}$

The magnitude of the expansion work done is $29.88\text{ kJ}$ (or $-29.88\text{ kJ}$ by IUPAC thermodynamic sign convention).

Final Answer: The expansion work done is $-29.88$ (or $29.88$).