Calculate the volume of oxygen produced at room temperature when a concentrated H2SO4 is electrolysed for 30.0 min using a current.

Steps to Solve

1. Calculate Total Charge (Coulombs)

Given current ($I = 0.5 , A$) and time ($t = 30 , min = 30 \times 60 , s = 1800 , s$), we find the total charge produced during electrolysis: $$ Q = I \times t = 0.5 , A \times 1800 , s = 900 , C $$

2. Determine Moles of Electrons

Using Faraday's constant ($F = 96500 , C/mol$), we can determine the number of moles of electrons ($n_e$) produced: $$ n_e = \frac{Q}{F} = \frac{900 , C}{96500 , C/mol} \approx 0.00932 , mol $$

3. Calculate Moles of Oxygen Produced

From the half-reaction: $$ 4 OH^- \rightarrow 2 O_2 + 4 H_2O + 4 e^- $$

It shows that 4 moles of electrons produce 1 mole of $O_2$. Hence, the moles of $O_2$ produced ($n_{O_2}$) is: $$ n_{O_2} = \frac{n_e}{4} = \frac{0.00932 , mol}{4} \approx 0.00233 , mol $$

4. Convert Moles of Oxygen to Volume

Using the ideal gas law, at room temperature, 1 mole of gas occupies approximately $24.0 , dm^3$. So, the volume of $O_2$ produced ($V_{O_2}$) can be calculated as: $$ V_{O_2} = n_{O_2} \times 24.0 , dm^3/mol = 0.00233 , mol \times 24.0 , dm^3/mol \approx 0.056 , dm^3 $$