Question

If electrolysis of water is carried out for a time duration of 2 hours, how much electric current in amperes would be required to liberate 100 ml of O\(_2\) gas measured under standard conditions of temperature and pressure?

• A. 0.1723 A
• B. 4.178 A
• C. 0.8616 A
• D. 0.2393 A

Hint:

In electrolysis problems, remember to use Faraday’s laws of electrolysis, and always ensure to convert volumes of gases into moles using the molar volume at STP.

Answer 1

The Correct Option is D

The reaction for the electrolysis of water is: \[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \] From this reaction, it is clear that 1 mole of O\(_2\) is released for every 4 moles of electrons (since 4 moles of electrons are required to produce 2 moles of H\(_2\) and 1 mole of O\(_2\)). Step 1: Calculate moles of O\(_2\) liberated The volume of O\(_2\) is given as 100 ml at standard conditions (STP). At STP, 1 mole of any gas occupies 22.4 liters (or 22400 ml). Therefore, the moles of O\(_2\) liberated are: \[ \text{moles of O}_2 = \frac{\text{volume of O}_2}{\text{molar volume at STP}} = \frac{100 \, \text{ml}}{22400 \, \text{ml/mol}} = 4.464 \times 10^{-3} \, \text{mol} \] Step 2: Calculate the total charge required From the reaction, we know that for every 1 mole of O\(_2\), 4 moles of electrons are involve(D) The total moles of electrons required to liberate 4.464 × 10\(^{-3}\) mol of O\(_2\) is: \[ \text{moles of electrons} = 4 \times 4.464 \times 10^{-3} = 1.786 \times 10^{-2} \, \text{mol} \] The charge required to produce 1 mole of electrons is 96500 C (Faraday constant). Therefore, the total charge required is: \[ Q = 1.786 \times 10^{-2} \, \text{mol} \times 96500 \, \text{C/mol} = 1729.39 \, \text{C} \] Step 3: Calculate the current The total charge is released in 2 hours (or 7200 seconds). Using the relation \( Q = I \times t \), where \( I \) is the current and \( t \) is the time, we can calculate the current: \[ I = \frac{Q}{t} = \frac{1729.39 \, \text{C}}{7200 \, \text{s}} = 0.2393 \, \text{A} \] Thus, the required current is 0.2393 (A)