Question

How many coulombs are required for the oxidation of \(1\) mole of \(H_2O\) to \(O_2\)?

• A. \(1.93 \times 10^5\;C\)
• B. \(9.65 \times 10^4\;C\)
• C. \(3.86 \times 10^5\;C\)
• D. \(4.825 \times 10^5\;C\)

Hint:

Always balance electrochemical half-reactions first. Then calculate: \[ \text{Charge} = (\text{moles of electrons}) \times 96500 \] Remember: \[ 1F = 96500\;C \]

Answer 1

The Correct Option is A

Concept: The quantity of electricity required in electrochemical reactions is calculated using Faraday's law: \[ Q = nF \] where:
• \(Q\) = charge in coulombs
• \(n\) = number of moles of electrons transferred
• \(F\) = Faraday constant \[ 1F = 96500\;C \] During oxidation of water, electrons are released.

Step 1: Writing oxidation half-reaction of water.
Oxidation of water is: \[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \] From the equation: \[ 2\;\text{moles of } H_2O \text{ release } 4\;\text{moles of electrons} \] Therefore: \[ 1\;\text{mole of } H_2O \text{ releases } 2\;\text{moles of electrons} \]

Step 2: Calculating charge required.
Using: \[ Q=nF \] Here: \[ n=2 \] Thus, \[ Q=2 \times 96500 \] \[ Q=193000\;C \] \[ Q=1.93 \times 10^5\;C \] Therefore, the required charge is: \[ 1.93 \times 10^5\;C \]