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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?

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    $\begingroup$ 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? $\endgroup$ – M. Farooq Aug 1 '20 at 14:57
  • $\begingroup$ Oh yes. I'll edit it @M.Farooq. Thanks for pointing it out :) $\endgroup$ – sai-kartik Aug 1 '20 at 15:00
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    $\begingroup$ 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. $\endgroup$ – Zhe Aug 1 '20 at 17:29
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    $\begingroup$ Be prepared that any task description can contain an error. If you underestimate your knowledge, overtrusting autority, you may not notice it. $\endgroup$ – Poutnik Aug 2 '20 at 8:12
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    $\begingroup$ 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. $\endgroup$ – Zhe Aug 2 '20 at 12:55

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