Chemistry, asked by kritikabehera61, 1 year ago

The equivalent weight of iodine in the reaction 2Nas2o3+I2=2NaI+Na2s4o6

kritikabehera61: yeah...

kritikabehera61: I m bit afraid of that book

drnmishrap5oyxk: I've posted the answer below though, and yeah, if you do have it, read the chapter thoroughly.

drnmishrap5oyxk: Which standard are you in?

kritikabehera61: I m a dropper

drnmishrap5oyxk: Ok, so I understand it's a bit tough for you to understand these concepts,
But don't fret, the book's amazing. I did the whole chapter of redox from it in 9th, Try it out, solve all of it's questions, keep in touch for solutions of the whole lot if you want.

kritikabehera61: thanks so much.... I will try my best to solve them

drnmishrap5oyxk: You'd surely be able to do it, or if you don't want to, just read it's theory,

kritikabehera61: 2 days before I was full of determination that anyhow I will solve that book.... BT my bad luck.... it was hard

drnmishrap5oyxk: Best of luck though!

Answers

Answered by drnmishrap5oyxk

SO here, the average oxidation number for each atom of I in the I2 molecule is 0 and for the atom of I in NaI is -1.

Thus the change in oxidation number for each atom of I is 1, but as their are 2 I atoms involved, the n factor comes out to be 2.

Now we know that, Equivalent Weight = Molecular Weight/ n factor.

There fore the Equivalent weight of I2 in this reaction = Mw/2 = 127 x 2 / 2 = 127

Inform if wrong, approve if correct.

Answered by sandavada

Answer:

Oxidation state of I changes from 0 to −1.

Therefore, change in oxidation state is 1.

Change, per mole of iodine molecule is 1×2 (as we have two atoms of I) = 2.

So, Eq. Wt is =

molecular weight / 2

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