Answer to Question #129872 in Physical Chemistry for Idehenre Daniel

Question

how many grams of oxygen is liberated by the electrolysis of water after passing 0.565A for 185 seconds?

Answer 1

Oxidation at anode: 2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4e⁻

4 electrons are involved in liberation of 1 mol (32.00 g) of oxygen.

4 x F = 4 x 96,500 = 386,000 C of electricity.

We have: 0.565 A x 185 s 104.525 C of electricity

104.525 / 386,000 = 0.000271 mol

0.000271 mol x 32.00 g/mol = 0.00867 g