A solution containing 2-68 - 10-3 mol of A n - ions requires 1-608 - 10-3 mol of MnO 4-for the complete oxidation of A n - to AO 3-in acidic medium- What is the value of n -

MnO_4^ would convert to Mn^2+ ∴ Its n factor would be 5 ∴ Equivalents of MnO_4^ =1.608× 10^ 3× 5=8.04× 10^ 3 Equivalents of A^n+=8.04× 10^ 3 n factor of A^n+ ...

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Standard XII

Chemistry

Redox Reactions

A solution co...

Question

A solution containing $2.68 \times 10^{- 3} m o l$ of $A^{n +}$ ions requires $1.608 \times 10^{- 3} m o l$ of $M n O_{4}^{-}$ for the complete oxidation of $A^{n +}$ to $A O_{3}^{-}$ in acidic medium. What is the value of $n$ ?

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Solution

$M n O_{4}^{-}$ would convert to $M n^{2 +}$
$∴$
$Its n-factor would be 5$
$∴$
$Equivalents of M n O_{4}^{-} = 1.608 \times 10^{- 3} \times 5 = 8.04 \times 10^{- 3}$
$Equivalents of A^{n +} = 8.04 \times 10^{- 3}$
$n-factor o f A^{n +} = 5 - n$
$∴ (5 - n) \times 2.68 \times 10^{- 3} = 8.04 \times 10^{- 3}$
$\Rightarrow n = 2$

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Q. A solution containing

2.68 \times 10^{- 3} mol

A^{n +}

ions requires

1.61 \times 10^{- 3} mol

M n O_{4}^{-}

for the complete oxidation of

A^{n +}

A O_{3}^{-}

in acidic medium. What is the value of

n

2.68 \times 10^{- 3} m o l

of a solution containing an ion

A^{n +}

required

1.6 \times 10^{- 3}

mol of

{M n O_{4}}^{- 1}

for oxidation of

A^{n +}

{A O_{3}}^{- 1}

ion in acid medium. What is the value of

n

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles of

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acidic medium. What is the value of

n

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acid medium. What is the value of n?

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acid medium. What is the value of n?

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A solution containing 2.68×10−3mol of An+ ions requires 1.608×10−3 mol of MnO−4 for the complete oxidation of An+ to AO−3 in acidic medium. What is the value of n ?

MnO−4 would convert to Mn2+ ∴ Its n-factor would be 5 ∴ Equivalents of MnO−4=1.608×10−3×5=8.04×10−3 Equivalents of An+=8.04×10−3 n-factor of An+=5−n ∴ (5−n)×2.68×10−3=8.04×10−3 ⇒n=2

A solution containing $2.68 \times 10^{- 3} m o l$ of $A^{n +}$ ions requires $1.608 \times 10^{- 3} m o l$ of $M n O_{4}^{-}$ for the complete oxidation of $A^{n +}$ to $A O_{3}^{-}$ in acidic medium. What is the value of $n$ ?