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Chapter 18: Problem 98

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

Expert verified
During the electrolysis of water with a current of 2.50 A in 15.0 min, approximately 0.261 L of hydrogen gas (H₂) and 0.131 L of oxygen gas (O₂) are produced at standard temperature and pressure (STP).

Step by step solution

01

Convert time to seconds

To solve this problem, we need to have the time in seconds. So, we'll convert the 15.0 min to seconds: \[15.0 \ \text{min} \times \frac{60 \ \text{s}}{1 \ \text{min}} = 900 \ \text{s}\]
02

Calculate the charge passed during electrolysis

We need to calculate the charge passed during the electrolysis process. We can use the formula: \[Q = It\] where Q is the charge, I is the current, and t is the time. So the charge passed is: \[Q = (2.50 \ \text{A})(900 \ \text{s}) = 2250 \ \text{C}\]
03

Applying Faraday's Laws of Electrolysis

We have to determine the amount of gas formed during electrolysis using Faraday's laws. The reaction that takes place is: \[2H_2O(l) \longrightarrow 2H_2(g) + O_2(g)\] From Faraday's law, we know that the charge needed to produce one mole of a substance is given by: \[Q = nfF\] where Q is the charge, n is the number of moles, f is the number of electrons involved in the reaction, and F is Faraday's constant (\(96500\ \text{C mol}^{-1}\)). For the electrolysis of water: - 2 electrons are required to produce 1 mole of hydrogen gas (H₂) - 4 electrons are required to produce 1 mole of oxygen gas (O₂)
04

Calculate the moles of H₂ and O₂ produced

Now, we can calculate the moles of H₂ and O₂ generated during electrolysis. For H₂: \[n_{H_2} = \frac{Q}{(2 \times F)} = \frac{2250 \ \text{C}}{(2 \times 96500 \ \text{C mol}^{-1})} = 0.011672 \ \text{mol}\] For O₂: \[n_{O_2} = \frac{n_{H_2}}{2} = \frac{0.011672}{2} = 0.005836 \ \text{mol}\]
05

Calculate the volume of H₂ and O₂ produced at STP

Now that we have the moles of gases produced, we can calculate the volumes of H₂ and O₂ at STP. The volume of one mole of any gas at STP is 22.4 L/mol. So, the volumes of gases produced are: For H₂: \[V_{H_2} = n_{H_2} \times V_{molecular} = 0.011672 \ \text{mol} \times 22.4 \ \text{L/mol} \approx 0.261 \ \text{L}\] For O₂: \[V_{O_2} = n_{O_2} \times V_{molecular} = 0.005836 \ \text{mol} \times 22.4 \ \text{L/mol} \approx 0.131 \ \text{L}\] Thus, from the electrolysis of water with a current of 2.50 A in 15.0 min, approximately 0.261 L of hydrogen gas and 0.131 L of oxygen gas are produced at STP.

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