Question

What is the quantity of current required to liberate 16g of O\(_2\) (g) during electrolysis of water? (Given 1F = 96500C)

• A. \(4.825 \times 10^4 \, C\)
• B. \(9.65 \times 10^4 \, C\)
• C. \(2.895 \times 10^5 \, C\)
• D. \(4.825 \times 10^5 \, C\)
• E. \(1.93 \times 10^5 \, C\)

Hint:

In electrolysis, use the mole ratio of the reaction to calculate the amount of charge needed for a given mass of a substance. Remember that 1 mole of electrons corresponds to 96500 C.

Answer 1

Answer: (E)

In the electrolysis of water, the reaction is: \[ 2 {H}_2{O} (l) \rightarrow 2 {H}_2 (g) + {O}_2 (g) \] From the reaction, 1 mole of O\(_2\) is produced by the passage of 4 moles of electrons.
The molar mass of O\(_2\) is 32 g. Therefore, the moles of O\(_2\) in 16 g are:
\[ {Moles of O}_2 = \frac{16}{32} = 0.5 \, {mol} \] The charge required to produce 1 mole of O\(_2\) is 4 moles of electrons, which corresponds to: \[ {Charge for 1 mole of O}_2 = 4 \times 96500 \, C = 386000 \, C \] For 0.5 moles of O\(_2\), the charge required is: \[ {Charge for 0.5 moles of O}_2 = \frac{386000}{2} = 193000 \, C \] Thus, the total charge required to liberate 16g of O\(_2\) is \(1.93 \times 10^5 \, C\).