During electrolysis of an aqueous NaOH solution in the anode, 2.8 L of Oxygen are released (under conditions considered normal). How much hydrogen is free in the cathode?
H2O? Twice the O2, is H2. 5.6L
To determine the amount of hydrogen gas released at the cathode during the electrolysis of an aqueous NaOH solution, we need to consider the stoichiometry of the reaction and the molar volume of gases at standard temperature and pressure (STP).
First, let's write the balanced equation for the electrolysis of water:
2H2O(l) → 2H2(g) + O2(g)
From this equation, we can see that for every 2 moles of water (2H2O) electrolyzed, we get 2 moles of hydrogen gas (2H2) released.
Given that 2.8 L of oxygen gas (O2) is released, we need to use the concept of stoichiometry to find the amount of hydrogen gas (H2) released. This can be calculated using the molar volume of gases at STP, which is equal to 22.4 L/mol.
Since the stoichiometric ratio of oxygen to hydrogen is 1:2, we can use the following equation to find the moles of hydrogen gas released:
Moles of H2 = (2.8 L of O2 / 22.4 L/mol) * (2 mol of H2 / 1 mol of O2)
Let's calculate the moles of hydrogen gas:
Moles of H2 = (2.8 / 22.4) * 2 = 0.25 mol
Therefore, the amount of hydrogen gas released at the cathode during the electrolysis of the NaOH solution is 0.25 moles.