Question

2.68 × 10⁻³ moles of a solution containing an ion $A^{n+}$ require 1.61 × 10⁻³ moles of $MnO_4^-$ for the oxidation of $A^{n+}$ to $AO_3^-$ in acid medium. What is the value of n?

Answer 1

Explanation:

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Asked on November 22, 2019 by

Sparsh Paite

2.68×10

−3

mol of a solution containing an ion A

n+

required 1.6×10

−3

mol of MnO

4

−1

for oxidation of A

n+

to AO

3

−1

ion in acid medium. What is the value of n?

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ANSWER

A

n+

is oxidised to AO

3

−

Change in oxidation number =5(in AO

3

−

)−n(in A

n+

)

=5−n.....(i)

2.68×10

−3

mol of A

n+

ion react with 1.6×10

−3

mol of MnO

4

−

ions

∴1 mol of A

n+

ion will react with

2.6×10

−3

1.6×10

−3

mol of MnO

4

−

ions

2K

M

+7

nO

4

+3H

2

SO

4

→K

2

SO

4

+2

M

+2

nSO

4

+3H

2

O+5[O]

Number of equivalents of MnO

4

−

used in oxidation of A

n+

to AO

3

−

=0.597×5=2.985≈3

Thus, from equation (i), 5−n=3

n=2.

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Answer 2

Answer:

the above-mentioned post is totally correct

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