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Standard XII

Chemistry

introduction

MnO4- ions ar...

Question

${M n O_{4}}^{-}$ ions are reduced in acidic condition to $M n^{2 +}$ ions whereas they are reduced in neutral condition to $M n O_{2}$ . The oxidation of $25 m L$ of a solution $X$ containing $F e^{2 +}$ ions required in acidic condition, $20 m L$ of a solution $Y$ containing ${M n O_{4}}^{-}$ ions. What volume of solution $Y$ would be required to oxidise $25 m L$ of solution $X$ containing $F e^{2 +}$ ions in neutral condition?

11.4 m L

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12 m L

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33.3 m L

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35 m L

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Solution

The correct option is C $33.3 m L$
Acid medium
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
$\frac{M_{1} V_{1}}{1} = \frac{M_{2} V_{2}}{5}$
$\frac{M_{1} \times 20}{1} = \frac{M_{2} \times 25}{5}$
$M_{1} = \frac{M_{2}}{4} . . . . (i)$
Neutral medium
$M n O_{4}^{-} + 3 F e^{2 +} + 4 H^{+} \to M n O_{2} + 3 F e^{3 +} + 2 H_{2} O$
$\frac{M_{1} V_{1}}{1} = \frac{M_{2} V_{2}}{3}$
$(\frac{M_{2}}{4}) \times V_{1} = \frac{M_{2} \times 25}{3}$
$V_{1} = 33.3 m L$ .

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Similar questions

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A quantity of

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F e^{3 +}

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25.0

mL of

0.02 M

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F e^{2 +}

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F e^{3 +}

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mL of the original solution is treated with

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metal. Finally, the solution requires

40.0

mL of the same

K M n O_{4}

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The correct option is C 33.3 mLAcid mediumMnO−4+5Fe2++8H+→Mn2++5Fe3++4H2OM1V11=M2V25M1×201=M2×255M1=M24....(i)Neutral mediumMnO−4+3Fe2++4H+→MnO2+3Fe3++2H2OM1V11=M2V23(M24)×V1=M2×253V1=33.3 mL.