The correct option is B 1:5
Consider the redox reaction:
I2→IO−3+I−
Oxidation state of I in I2=+0
Oxidation state of I in IO−3=+5
Oxidation state of I in I−−=−1
Cleary, I2 is undergoing oxidation in IO−3 and in I− it is undergoing reduction.
Formula used for the n-factor calculation :
nf=(|O.S.Product−O.S.Reactant|×number of atoms
using the formula of n-factor given above,
nf of IO−3=5
nf of I−=1
Cross multiplying these with nf of each other.
we get,
I2→IO−3+5I−
Balancing the I on both sides,
3I2→IO−3+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+3H2O→IO−3+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+3H2O→IO−3+5I−+6H+
Now adding OH− to both sides to combine with H+ and make it H2O,
3I2+3H2O+6OH−→IO−3+5I−+6H++OH−
3I2+3H2O+6OH−→IO−3+5I−+6H2O
Eliminating the unrequired common H2O which is present on the both sides,
3I2+6OH−→IO−3+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.