Question

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

Short answer

The volumes of H₂(g) and O₂(g) produced at STP from the electrolysis of water by a current of 2.50 A in 15.0 minutes are approximately 0.261 L and 0.1305 L, respectively.

Step-by-step solution

Calculate the electric charge passed

Electric charge, Q, is calculated using the formula: Q = I × t, where Q is the electric charge, I is the current (in Amperes, A), and t is the time in seconds (s). We are given: Current, I = 2.50 A Time, t = 15.0 minutes = 15.0 × 60 seconds = 900 s Now let's calculate the charge passed: Q = I × t Q = 2.50 A × 900 s Q = 2250 Coulombs (C)

Calculate moles of electrons transferred

Using Faraday's Law, we can calculate the moles of electrons transferred: \(n_e = \frac{Q}{F_e}\), where \(n_e\) is the moles of electrons, Q is the electric charge, and \(F_e\) is the Faraday constant (96485 C/mol). \(n_e = \frac{2250 C}{96485 C/mol}\) \(n_e = 0.0233 \, mol\)

Calculate moles of H₂ and O₂ produced

For the electrolysis of water, the balanced chemical equation is: 2H₂O (l) → 2H₂ (g) + O₂ (g) For every 4 moles of electrons, 2 moles of H₂ and 1 mole of O₂ are produced. Moles of H₂ produced: \(n_{H_2} = \frac{1}{2} n_e\) Moles of O₂ produced: \(n_{O_2} = \frac{1}{4} n_e\) \(n_{H_2}=\frac{1}{2} \times 0.0233\, mol = 0.01165 \, mol \) \(n_{O_2}=\frac{1}{4} \times 0.0233\, mol = 0.005825 \, mol \)

Convert moles to volume at STP

The molar volume of any gas at Standard Temperature and Pressure (STP) is 22.4 L/mol. To find the volumes of H₂ and O₂ produced, we can use the formula: Volume = Moles × Molar Volume Volume of H₂ produced: \(V_{H_2} = n_{H_2} \times 22.4 \, L/mol\) Volume of O₂ produced: \(V_{O_2} = n_{O_2} \times 22.4 \, L/mol\) \(V_{H_2} = 0.01165 \, mol \times 22.4 \, L/mol = 0.261 \, L\) \(V_{O_2} = 0.005825 \, mol \times 22.4 \, L/mol = 0.1305 \, L\) The volumes of H₂(g) and O₂(g) produced at STP from the electrolysis of water by a current of 2.50 A in 15.0 minutes are approximately 0.261 L and 0.1305 L, respectively.