How many Faradays are required to reduce 1 mol of \(Cr_2O_7^{2-}\) to \(Cr^{3+}\) in acidic medium ?
Show Hint
- 2
- 3
- 6
- 4
The Correct Option is C
Solution and Explanation
The number of Faradays equals the number of moles of electrons transferred. One Faraday is the charge of one mole of electrons.
Step 1:
In \(Cr_2O_7^{2-}\), each Cr is in the +6 oxidation state and is reduced to \(Cr^{3+}\) (+3). The change per Cr atom is \(6 - 3 = 3\) electrons.
Step 2:
There are 2 Cr atoms per dichromate ion, so total electrons gained = \(2 \times 3 = 6\). The half reaction is \(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\).
Answer: Option (C) 6 Faradays.
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