How many moles of dichromate ion will be required to oxidize one mole of \( Fe^{2+} \) to \( Fe^{3+} \) ions in acidic solution?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1/6
In acidic solution the dichromate ion is a strong oxidising agent. Each chromium goes from +6 to +3, so one dichromate ion (two Cr atoms) accepts 6 electrons, while each \( \ce{Fe^{2+}} \) ion gives only one:
\( \displaystyle \ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O} \)
\( \displaystyle \ce{Fe^2+ -> Fe^3+ + e-} \)
Multiplying the iron half-reaction by 6, so that the electrons given equal the electrons taken, and adding:
\( \displaystyle \ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 6Fe^3+ + 7H2O} \)
So 1 mol of dichromate oxidises 6 mol of \( \ce{Fe^{2+}} \), and for 1 mol of \( \ce{Fe^{2+}} \):
\( \displaystyle n(\ce{Cr2O7^2-}) = \frac{1}{6}\ \mathrm{mol} \)
The answer is 1/6 mol of dichromate ion.