Question

Balance this redox by half reaction method

KIO4 + KI + ? KCl + I2

Please show the steps and detail along the way

Answer 1

We will do this in a stepwise manner,

The overall equation will be,

KIO4 + KI + HCl -------> KCl + I2 + H2O

Note we have added HCl and H2O in the equation which were missing above.

In this reaction, K+ and Cl- are the spectator ions and H2O will be added later via half reactions, H+ will also be removed at this point and rewrite the equation.

IO4?(aq) + I?(aq) -----> I2(s)

We can see, oxidation state of Iodine in IO4- is getting reduced from +7 to 0 and Iodine in I- is getting oxidized from -1 to 0. So a total change of 7 e-'s.

Writing the half reactions,

IO4- ------> I [reduction)

I- -------> I [oxidation]

Balancing the half reactions,

IO4- + 7 e- + 8H+ -------> I + 4H2O

7I- ---------> 7I + 7e-

The net balanced equation will be after after adding KCl will be,

IO4- + 8H+ + 7I- ---------> 4I2 + 4H2O + KCl

Balancing the rest we get the final equation,

KIO4 + 7KI + 8HCl ------> 8KCl + 4I2 + 4H2O

We can also balance this equation by another method shown below,

Write the unbalanced equation with its ionic compounds dissociated.

K+(aq) + IO4?(aq) + K+(aq) + I?(aq) + H+(aq) + Cl?(aq) ? K+(aq) + Cl?(aq) + I2(s) + H2O(l)

Write the unbalanced net ionic equation, without the spectator ions,

K+(aq) and Cl?(aq). IO4?(aq) + I?(aq) + H+(aq) ? I2(s) + H2O(l)

The oxidation number of iodine in IO4?(aq) changes from +7 to 0, so iodine gains 7 electrons. The oxidation number of iodine in I?(aq) changes from