The ion An- is oxidised to AO-3 by MnO-4 changing to Mn2- in acid medium- Given that 2-68- 10-3 mole of An- required 1-61- 10-3 mole MnO- What is the value of n-

2.68 × 10^ 3moles of A^n+1.61 × 10^ 3moles ofMnO_4^ =53 Therefore, 5 moles of A^n+ reacts with 3 moles of MnO_4^ . Half reactions a ...

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Standard XII

Chemistry

Oxidation Number Method

The ion An+...

Question

The ion $A^{n +}$ is oxidised to $A O_{3}^{-}$ by $M n O_{4}^{-}$ changing to $M n^{2 +}$ in acid medium. Given that $2.68 \times 10^{- 3}$ mole of $A^{n +}$ required $1.61 \times 10^{- 3}$ mole $M n O^{-}$ . What is the value of n?

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Solution

$\frac{2.68 \times 10^{- 3} moles of A^{n +}}{1.61 \times 10^{- 3} moles of {M n O_{4}}^{-}} = \frac{5}{3}$
Therefore, $5$ moles of $A^{n +}$ reacts with $3$ moles of ${M n O_{4}}^{-}$ . Half reactions are:
${M n O_{4}}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O ⟶ (1)$
$A^{n +} + 3 H_{2} O - 3 e^{-} ⟶ {A O_{3}}^{-} + 6 H^{+} ⟶ (2)$
From equation $2$ , we can say that the value of $n = 2$ .

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Similar questions

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?

Q. The ion

A^{n +}

is oxidised to

{A O}_{3}^{-}

M n O_{4}^{-}

changing to

{M n}^{2 +}

in acidic solution. Given that

2.68 \times 10^{- 3}

mol pf

A^{n +}

requires

1.61 \times 10^{- 3}

mol of

M n O_{4}^{-}

. The value of

n

is :

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?

Q. The ion

A^{n +}

is oxidised to

{A O_{3}}^{-}

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

changing to

M n^{2 +}

in acid solution. Given that

2.68 \times 10^{- 3}

mol of

A^{n +}

requires

1.61 \times 10^{- 3}

mol of

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

.
What is the weight of one gram equivalent of

A C l_{n}

for the above reaction if the atomic mass of

A

97

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The ion An+ is oxidised to AO−3 by MnO−4 changing to Mn2+ in acid medium. Given that 2.68×10−3 mole of An+ required 1.61×10−3 mole MnO−. What is the value of n?

2.68×10−3moles ofAn+1.61×10−3moles ofMnO4−=53Therefore, 5 moles of An+ reacts with 3 moles of MnO4−. Half reactions are:MnO4−+8H++5e−⟶Mn2++4H2O⟶(1)An++3H2O−3e−⟶AO3−+6H+⟶(2)From equation 2, we can say that the value of n=2 .

The ion $A^{n +}$ is oxidised to $A O_{3}^{-}$ by $M n O_{4}^{-}$ changing to $M n^{2 +}$ in acid medium. Given that $2.68 \times 10^{- 3}$ mole of $A^{n +}$ required $1.61 \times 10^{- 3}$ mole $M n O^{-}$ . What is the value of n?