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In the electrolysis of water shown below, a current of 2 amps is applied to 180 mL of H2O(l) for 6 hours and 42 minutes. How many grams of H2(g) are formed? (Faraday’s constant = 96,500 C/mol)2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)Question 27 Answer Choices   A.  0.25 g   B.  0.5 g   C.  5.0 g   D.  10.0 g

Question

In the electrolysis of water shown below, a current of 2 amps is applied to 180 mL of H2O(l) for 6 hours and 42 minutes. How many grams of H2(g) are formed? (Faraday’s constant = 96,500 C/mol)2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)Question 27 Answer Choices   A.  0.25 g   B.  0.5 g   C.  5.0 g   D.  10.0 g

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Solution

First, we need to convert the time to seconds because the current is in amperes (which is coulombs/second).

6 hours and 42 minutes is equal to 24120 seconds (6 hours * 60 minutes/hour * 60 seconds/minute + 42 minutes * 60 seconds/minute).

Next, we calculate the total charge passed through the solution using Ohm's law (Q = It), where I is the current and t is the time.

So, Q = 2 A * 24120 s = 48240 C.

The number of moles of electrons (n) can be calculated using Faraday's law (Q = nF), where F is Faraday's constant.

So, n = Q / F = 48240 C / 96500 C/mol = 0.5 mol.

From the balanced chemical equation, we can see that 2 moles of water produce 1 mole of hydrogen gas. Therefore, 0.5 moles of electrons will produce 0.5/2 = 0.25 moles of hydrogen gas.

Finally, we convert moles of hydrogen gas to grams using the molar mass of hydrogen gas (H2), which is approximately 2 g/mol.

So, mass = 0.25 mol * 2 g/mol = 0.5 g.

Therefore, the answer is B. 0.5 g.

This problem has been solved

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