Question

A solution is X M in Al(ClO3)3 the solution is titrated with a solution of I2. To completely react the Al(ClO3)3 in 23.24 ml, 17.52 of a 0.0457M I2 solution was needed. what is the molarity of the Al(ClO3)3 solution?

5ClO₃^-+ 3H₂O+ 3I₂ -----> 5Cl^-+ 6HIO₃

Answer 1

reduction half :-

ClO3- --------> Cl-

Cl is +5 in left and -1 in right so add 6 e- in left

ClO3- + 6e- ----------> Cl-

now balance charge by adding H+ ....-7 in left and -1 in right

ClO3- + 6e- + 6H+ --------> Cl-

now balance H and O by adding H2O

ClO3- + 6e- + 6H+ -------> Cl- + 3H2O ......(1)

this is your balanced reduction half equation ...

now oxidation half :-

I2 ---------> HIO3

balance I ...

I2 -------> 2HIO3

I is 0 in left and +5 in right and as there were two I atoms so 5 e- each are released making it a total of 10 e- so add 10e- in right.......

I2 ---------> 2HIO3 + 10e-

now balnce charge by adding H+

I2 ----------> 2HIO3 + 10e- + 10H+

now balance H and O by adding H2O

I2 + 6H2O ---------> 2HIO3 + 10e- + 10H+ .......(2)

this is your balanced oxidation half equation

multiply (1) by 5 and (2) by 3 and add

5ClO3- + 30e- + 30H+ -------> 5Cl- + 15H2O
3I2 + 18H2O ---------> 6HIO3 + 30e- + 30H+

--------------------------------------...

clearly 30 e- will cancel out

5ClO3- + 30H+ + 3I2 + 18H2O -----------> 5Cl- + 15H2O + 6HIO3 + 30H+

cancelling common terms

5ClO3- + 3I2 + 3H2O -----------> 5Cl- + 6HIO3


so 30 e- were transffered ....you should remember that no.of electrons should remain the same on both electrode ....in this way they cancel out from final equation