Question

How to find equivalent weight in redox reaction?
My sir gave me a formula: -
Equivalent weight,
E=
Molecular weight of compound
---------------------------------------------------
Change in oxidation number of that compound in redox reaction

Answer 1

so how to find the equivalent weight of Oxidizing Agent : -> Potassium Dichromate ?
\[\color{blue}{K_2Cr_2O_7+4H_2SO_4 \rightarrow K_2SO_4+Cr_2(SO_4)_3+4H_2O+3(O).}\]
@JFraser Sir Please help:)
very much needed.

Answer 2

Potassium Dichromate: -> \[\color{red}{K_2Cr_2O_7.}\]

Answer 3

do you know how to find the molecular wight?

Answer 4

weight*

Answer 5

yeah of course @bronzegoddess

Answer 6

It is 294

Answer 7

of k2cr2o7

Answer 8

then assign oxidation numbers to all the atoms for potassium dichromate, and do the same for the same atoms on the other side

Answer 9

the main problem is actually this:(

Answer 10

so how to do tht?

Answer 11

ON of Cr in K2Cr2O7 = +6
ON of Cr in Cr2(SO4)3 = +3
So, the change is ... ?

Answer 12

+3

Answer 13

..... I believe :
Change = new - old

Answer 14

-3

Answer 15

but answer in my book is given 294/6

Answer 16

@bronzegoddess @Callisto

Answer 17

........... because you add 3(O) at the back?!

Answer 18

i can't understand what r u saying?

Answer 19

what is the main concept?

Answer 20

wait, we started with +6 and now we have +3, so 1mole Cr gained 3 electrons, but since Cr is diatomic it gained 6electrons , isn't it? @Callisto

Answer 21

why is mole concept used here?

Answer 22

2 Cr there... Perhaps....

Answer 23

Total number of electrons transferred = 2 (3) = 6 ... Perhaps.

Answer 24

but how can we solve it by only the basis of change in oxidation number?
@bronzegoddess

Answer 25

Change in ON = 3
Since there 2 Cr there, the total electron transfer should be doubled, perhaps.

Answer 26

sorry for long reply
a net problem; so my question is that
"but in LHS also 2 cr
& in RHS also 2 cr
how can that happen which u r saying @Callisto "