During the electrolysis of dilute tetraoxosulphate (VI) acid solution, 0.05 mole of electrons were passed.
During the electrolysis of dilute tetraoxosulphate (VI) acid solution, 0.05 mole of electrons were passed. What volume of gas was produced at the anode?
- A) 2.24 dm3
- B) 0.560 dm3
- C) 0.280 dm3
- D) 0.224 dm3
Correct Answer: C) 0.280 dm3
Explanation
During the electrolysis of dilute tetraoxosulphate (VI) acid solution, also known as dilute sulfuric acid (H2SO4), the products formed are hydrogen gas at the cathode and oxygen gas at the anode. The balanced equation for the reaction at the anode is:
2H2O(l) ? O2(g) + 4H+(aq) + 4e-
This equation tells us that 4 moles of electrons are required to produce 1 mole of oxygen gas. We are given that 0.05 moles of electrons were passed during the electrolysis.
To find the moles of oxygen gas produced at the anode, we can use the stoichiometric ratio:
\[\text{Moles of O}_2 = \frac{\text{Moles of electrons}}{4}\] \[\text{Moles of O}_2 = \frac{0.05}{4}\] \[\text{Moles of O}_2 = 0.0125\]
Now we can calculate the volume of gas produced at the anode using the molar volume of a gas at standard conditions (STP), which is 22.4 dm3 per mole:
\[\text{Volume of O}_2 = \text{Moles of O}_2 \times \text{Molar volume}\] \[\text{Volume of O}_2 = 0.0125 \times 22.4\] \[\text{Volume of O}_2 = 0.280\, \text{dm}^3\]Therefore, the volume of gas produced at the anode is0.280 dm3, which corresponds toOption C.