KCET · Chemistry · Electrochemistry
How many Coulombs of electricity are required for the oxidation of one mol of water to
dioxygen?
- A \( 9.65 \times 10^{4} C \)
- B \( 1.93 \times 10^{4} C \)
- C \( 1.93 \times 10^{5} \mathrm{C} \)
- D \( 19.3 \times 10^{5} \mathrm{C} \)
Answer & Solution
Correct Answer
(C) \( 1.93 \times 10^{5} \mathrm{C} \)
Step-by-step Solution
Detailed explanation
For the oxidation of one mole of water to dioxygen,
\( \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_{2}+2 e^{-} \)
\( 2 \mathrm{~F} \) Coulombs of electricity is required.
Therefore, electricity in \( \mathrm{C} \) is
\[
\begin{array}{l}
=2 \times F \\
=2 \times 96500 \\
=193000 C \\
=1.93 \times 10^{5} C
\end{array}
\]
\( \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_{2}+2 e^{-} \)
\( 2 \mathrm{~F} \) Coulombs of electricity is required.
Therefore, electricity in \( \mathrm{C} \) is
\[
\begin{array}{l}
=2 \times F \\
=2 \times 96500 \\
=193000 C \\
=1.93 \times 10^{5} C
\end{array}
\]
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