If equal volumes of 1$M$ ${\rm{KMn}}{{\rm{O}}_4}$ and $1M{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ solutions are allowed to oxidise ${{\rm{F}}^{2 + }}$ to ${{\rm{F}}^{3 + }}$ in acidic medium volume of oxidant required for one mole of ${{\rm{F}}^{2 + }}$ will be :

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} > {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} < {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} = {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

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NEET UG CHEMISTRY MCQ: REDOX REACTIONS

1 Hard

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Analyzing +4 | -1

If equal volumes of 1

M

\mathrm{KMn}{\mathrm{O}_4}

and

1M{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

solutions are allowed to oxidise

{\mathrm{F}^{2 + }}

{\mathrm{F}^{3 + }}

in acidic medium volume of oxidant required for one mole of

{\mathrm{F}^{2 + }}

will be :

View answer and explanation

Correct option: A

{\mathrm{V}_{\mathrm{KMn}{\mathrm{O}_4}}} > {\mathrm{V}_{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}}}

Option A is correct because KMnO4 requires 5 moles of Fe2+ per mole of KMnO4, while K2Cr2O7 requires 6 moles of Fe2+ per mole of K2Cr2O7, thus KMnO4 will have a larger volume.

NEET UG CHEMISTRY MCQ: REDOX REACTIONS

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