Number of moles of acidified potassium dichromate required to liberate 6 moles of iodine from an aqueous solution of iodate ions

I came across this question in a recent exam I had taken:

How many moles of acidified $$\ce{K2Cr2O7}$$ is required to liberate 6 moles of $$\ce{I2}$$ from an aqueous solution of $$\ce{I-}$$ ?

So this is how I approached it (by writing the corresponding equations):

\require{cancel} \begin{align} \ce{ Cr2O7^2- + 14 H+ + \cancel{6e^-} &-> 2Cr^3+ + 7H2O} \\ \ce{6 I- &-> 3I2 + \cancel{6e^-}}\\ \hline \ce{Cr2O7^2- + 6I- + 14H+&-> 2Cr^3+ + 3 I2 + 7H2O} \end{align}

Clearly, every mole of $$\ce{K2Cr2O7}$$ liberates $$3$$ moles of $$\ce{I2}$$ from an aqueous solution containing $$\ce{I-}$$.

So, I concluded that $$2$$ moles of $$\ce{K2Cr2O7}$$ are required to liberate 6 moles of $$\ce{I2}$$.
But the answer key given to me says that the correct answer for this question is $$1$$. Am I going wrong somewhere?

• Yes you are right, it should be 2 mol dichromate leading to 6 mol iodine. Can you balance the equation first (hand written), iodide ion is missing on the left hand side? Aug 1 '20 at 14:57
• Oh yes. I'll edit it @M.Farooq. Thanks for pointing it out :) Aug 1 '20 at 15:00
• Your answer is right, but only for the reaction you have written. There could be other reduction half-reactions that you might select, though based on the statement of the question, you have selected the correct oxidation half-reaction.
– Zhe
Aug 1 '20 at 17:29
• Be prepared that any task description can contain an error. If you underestimate your knowledge, overtrusting autority, you may not notice it. Aug 2 '20 at 8:12
• Try consulting a table of half reactions. Judging by the expected answer, you're looking to reduce the dichromate down to elemental chromium, which, in all honesty, seems like a worse choice than what you've selected.
– Zhe
Aug 2 '20 at 12:55