Question 17.7: Balance the equation I2 + Cl2 + H2O→HIO3 + HCl    (unbalance...

Problem-Solving Strategy: Using Change in Oxidation Numbers to Balance Oxidation–Reduction Reactions.
1. Assign oxidation numbers:

The oxidation numbers of I_2    and    Cl_2 have changed, I_2 from 0 to +5 and Cl_2 from 0 to -1.

2. Write the oxidation and reduction steps. Balance the number of atoms and then balance the electrical charge using electrons:

I_2→2 I^{5+} + 10 e^- (oxidation) ( 10 e^- are needed to balance the +10 charge)
I_2 loses 10 electrons

Cl_2 + 2 e^-→2 Cl^-   (reduction)    (2 e^- are needed to balance the -2 charge)

Cl_2 gains 2 electrons

3. Adjust loss and gain of electrons so that they are equal. Multiply the oxidation step by 1 and the reduction step by 5:

I_2→2 I^{5+} + 10 e^-       (oxidation)

I_2 loses 10 electrons

5 Cl_2 + 10 e^-→10 Cl^-   (reduction)

5 Cl_2  gain 10 electrons

4. Transfer the coefficients from the balanced redox equations into the original equation. We need to use  1 I_2 ,  2 HIO_3  ,  5 Cl_2 ,  and     10 HCl:

I_2 + 5 Cl_2 + H_2O → 2 HIO_3 + 10 HCl      (unbalanced)

5. Balance the remaining elements, H and O:

I_2 + 5 Cl_2 + 6 H_2O → 2 HIO_3 + 10 HCl        (balanced)

6. Check: The final balanced equation contains 2 atoms of I , 10 atoms of Cl , 12 atoms of H and 6 atoms of O on each side.