Question

The quantity of electricity in Faraday, required to reduce 1 mol of \( \mathrm{Cr_2O_7^{2-}} \) to \( \mathrm{Cr^{3+}} \) is :

• A. 4F
• B. 3F
• C. 6F
• D. 5F

Hint:

Each chromium changes from +6 to +3, so two chromium atoms take 6 electrons, which is 6 F.

Answer 1

The Correct Option is C

Step 1: Write the half reaction:
In acid medium, dichromate is reduced to chromium(III).
The balanced half reaction is:
\[ \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O} \]

Step 2: Read the electrons:
One mole of dichromate takes up 6 moles of electrons.
One Faraday is the charge of one mole of electrons.

Step 3: Find the charge:
Quantity of electricity = number of moles of electrons \( \times \) 1 F.
\[ Q = 6 \times 1\,\mathrm{F} = 6\,\mathrm{F} \]

Step 4: Why the other options are wrong:
4F, 3F and 5F do not match the 6 electrons in the balanced half reaction.
3F would be correct for only one chromium atom, not for the whole ion with two chromium atoms.

Final Answer:
The charge needed is 6 F, option (C). \[ \boxed{6\,\mathrm{F}} \]