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 :  I2→IO−3+I−5:13:11:31:5

Solution

## The correct option is D 1:5Consider 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.

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