[xstrae]

Hi,

1.44g of FeC₂O₄ was dissolved in sulphuric acid(dil.) and the solution was diluted to 100cm3. Which of the following solution of KMnO4 will be able to oxidise it completely? (multiple options may be correct)

(A) 0.01M , 400cm3
(B) 0.02M , 200cm3
(C) 0.01M , 600cm3
(D) 0.05N , 600cm3

My work:
For complete oxidation, Milli equivalents of FeC₂O₄ = Milli equivalents of KMnO4

Milli equivalents of FeC₂O₄ = wt/eq.wt x 1000
                                                                    = 1.44/72 x 1000  = 20
(  eq wt = m.w / n factor = 144/2  )

In acidic medium, Mn⁺⁷ ---->  Mn⁺²
therefore, n factor of KMnO4 = 5

N = n x M
Ml. eq of KMnO4 = N x V

Putting in the values, I got (A) and (B) as the correct answer. My book, however, lists that (C) and (D) are the correct options. Can someone please tell me where I am going wrong? Thank you.

[Borek]

How does the oxidation of FeC₂O₄ looks like?

[Albert]

Moreover, why did you solve it using equivalents? It's easier to do it through traditional moles, isn't it?

[xstrae]

Fe²⁺ --- > Fe⁺³
so 'n' factor is 1 right? Am I doing anything else wrong? Because now, none of the options seem to match. thanks.

Moreover, why did you solve it using equivalents? It's easier to do it through traditional moles, isn't it?

I find equivalents much easier. Maybe its because I have had a lot of practice in that.

[Borek]

What about oxalate?

[xstrae]

ohh sorry.

C₂³ ----> 2C⁺⁴  + 2e-

So 'n' factor is 3 right? Hence, (C) and (D). I understand now. Thanks a lot for your *delete me*