CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider the disproportionation of Iodine to iodide and iodate ions. Using Oxidation method calculate the ratio of iodate and iodide ions formed in alkaline medium : 

I2IO3+I
  1. 5:1
  2. 3:1
  3. 1:3
  4. 1:5


Solution

The correct option is D 1:5
Consider the redox reaction:
I2IO3+I
Oxidation state of I in I2=+0
Oxidation state of I in IO3=+5
Oxidation state of I in I=1
Cleary, I2 is undergoing oxidation in IO3 and in I it is undergoing reduction.
Formula used for the n-factor calculation :
nf=(|O.S.ProductO.S.Reactant|×number of atoms
using the formula of n-factor given above,
nf of IO3=5
nf of I=1
Cross multiplying these with nf of each other.
we get,
I2IO3+5I

Balancing the I  on both sides,
3I2IO3+5I
Adding the H2O to balance the oxygen,
As the LHS of the equation contains 0 oxygens while RHS contains 3 oxygens, adding 3 H2O  to LHS.
3I2+3H2OIO3+5I
adding H+ to balance hydrogen,
As LHS of the equation contains 6 hydrogens while RHS contains  0 hydrogens, so adding 6H+  to RHS.
3I2+3H2OIO3+5I+6H+
Now adding OH to both sides to combine with H+ and make it H2O,
3I2+3H2O+6OHIO3+5I+6H++OH
3I2+3H2O+6OHIO3+5I+6H2O
Eliminating  the unrequired common H2O which is present on the both sides,
3I2+6OHIO3+5I+3H2O
This is the final balanced equation.
We can also see that charge on both sides is -6. which also indicates that the equation is balanced now.
So,  the required ratio is 1:5.
 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image