MHT CET · Chemistry · Electrochemistry
How much charge in coulombs is required for the reduction of one mole of \(\mathrm{Al}^{3+}\) to Al?
- A \(1.930 \times 10^{4} \mathrm{C}\)
- B \(2 \cdot 895 \times 10^{5} \mathrm{C}\)
- C \(2 \cdot 895 \times 10^{4} \mathrm{C}\)
- D \(1.930 \times 10^{5} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(2 \cdot 895 \times 10^{5} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Al}^{3 *}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}\)
1 mole of electron \(=1 \mathrm{~F}=96500\) coulombs
\(\therefore 3 \mathrm{e}^{-}=3 \mathrm{~F}=3 \times 96500\) \(=2,89,500=2.895 \times 10^{5}\) coulombs
1 mole of electron \(=1 \mathrm{~F}=96500\) coulombs
\(\therefore 3 \mathrm{e}^{-}=3 \mathrm{~F}=3 \times 96500\) \(=2,89,500=2.895 \times 10^{5}\) coulombs
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