-factor of when it is oxidized to and is: | Redox Reactions and Volumetric Analysis — Chem-Mantra

Question

$n$-factor of $CrI_3$ when it is oxidized to $CrO_4^{2-}$ and $IO_4^-$ is:

Answer: C

💡 Solution & Explanation

$Cr^{3+} \rightarrow Cr^{+6}$ ($3 e^-$). <br>$3I^{-} \rightarrow 3I^{+7}$ ($3 \times 8 = 24 e^-$). <br>Total electrons lost = $3 + 24 = 27$.

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