  Question

During the disproportionation of Iodine to iodide and iodate ions, the ratio of iodate and iodide ions formed in alkaline medium is:  I2→IO−3+I−

A
1 : 5  B
5 : 1  C
3 : 1  D
1 : 3  Solution

The correct option is A 1 : 5Steps for Balancing redox reactions: Identify the oxidation and reduction halves. Find the oxidising and reducing agent. Find the n-factor of  oxidising and reducing agents. Cross multiply the oxidising and reducing agent with the n-factor of each other. Balance the atoms other than oxygen and hydrogen. Balance oxygen atoms. Balance hydrogen atoms. For basic medium: If x oxygens are less on one side than the other side add  x H2O units to that side. As soon as we add x H2O units for balancing oxygen, we add generally 2x (or whatever suitable) H+ ions on the opposite side to balance hydrogen.Then after that we add equal  (equal to number of  H+ added) number of OH− ions to the both sides and combine H+ and OH− ions to form H2O. Then we cancel H2O which is common to both sides and which can be eliminated thus finally OH− is left on one side and the equation is balanced. Then we can verify whether the equation is balanced or not by checking the balance of charge on both sides. formula used for the n-factor calculation, nf=(|O.S.Product−O.S.Reactant|×number of atoms Taking the given equation and following the above mentioned steps: 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, I is undergoing oxidation in IO−3 and   in I− it is undergoing reduction. 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.  Chemistry

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