WBJEE · Chemistry · Redox Reactions
The equivalent weight of \(\mathrm{KIO}_3\) in the given reaction is (\(\mathrm{M}=\) molecular mass) :
\(2 \mathrm{Cr}(\mathrm{OH})_3+4 \mathrm{OH}^{-}+\mathrm{KIO}_3 \rightarrow 2 \mathrm{CrO}_4{ }^{2-}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{KI}\)
- A \(\mathrm{M}\)
- B \(\mathrm{M} / 2\)
- C \(\mathrm{M} / 6\)
- D \(\mathrm{M} / 8\)
Answer & Solution
Correct Answer
(C) \(\mathrm{M} / 6\)
Step-by-step Solution
Detailed explanation
Hint : \(\mathrm{Cr}(\mathrm{OH})_3+4 \overline{\mathrm{O}} \mathrm{H}+\mathrm{K~} \mathrm{I}^{+5} \mathrm{O}_3 \longrightarrow 2 \mathrm{CrO}_4^{-2}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{K~} \mathrm{I}^{-1}\) Change in oxidation state of iodine \(=6\) \(\therefore\) Equivalent…
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