KCET · Chemistry · Electrochemistry
How many coulombs are required to oxidise 0.1 mole of \(\mathrm{H}_2 \mathrm{O}\) to oxygen?
- A \(1.93 \times 10^5 \mathrm{C}\)
- B \(1.93 \times 10^4 \mathrm{C}\)
- C \(3.86 \times 10^4 \mathrm{C}\)
- D \(9.65 \times 10^3 \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(1.93 \times 10^4 \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The reaction involved is a follows
\(\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2+2 e^{-}\)
So, from the equation it is clear that 1 mole of \(\mathrm{H}_2 \mathrm{O}\) required 2 Faradays \(=2 \times 96500 \mathrm{C}\)
So, 0.1 mole of \(\mathrm{H}_2 \mathrm{O}\) will require \(=\frac{2 \times 96500 \times 0.1}{1}\)
\(\begin{aligned} & =19300 \mathrm{C} \\ & =1.93 \times 10^4 \mathrm{C}\end{aligned}\)
\(\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2+2 e^{-}\)
So, from the equation it is clear that 1 mole of \(\mathrm{H}_2 \mathrm{O}\) required 2 Faradays \(=2 \times 96500 \mathrm{C}\)
So, 0.1 mole of \(\mathrm{H}_2 \mathrm{O}\) will require \(=\frac{2 \times 96500 \times 0.1}{1}\)
\(\begin{aligned} & =19300 \mathrm{C} \\ & =1.93 \times 10^4 \mathrm{C}\end{aligned}\)
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