[twomblyhero]

Derive a balanced redox equation If I₂ is reduced to I- and ClO- is oxidised to ClO₃- Am I correct?

I₂(aq) +2 e-        2I-   

ClO- + O₂                ClO₃-   

I₂(aq) +ClO-  + 2H₂O                    2I- + ClO₃- + 4H+

[Hunter2]

No, no balanced equation  and there will be no oxygen be used.

PS: What was the result of your iodometry?

[twomblyhero]

This is what I have for my iodometry

The amount of thiosulphate required to reduced I₂  28.5cm³. For every 2 moles of 2S₂O₃^2- one mole of I₂ is reduced.

2S₂O₃^2-  + I₂       S₄O₆^2- + 2I^-

The moles of 2S₂O₃^2- can be calculated using the formula. The volume is divided by 1000 to give the value in dm³

volume x concentration = no. of moles
28.5cm³ x 0.1M               = 0.00285 mols
_______
   1000

Therefore, with a ratio of 2:1, the number of moles of I₂ is

0.00285
_______          = 0.001425 mols
      2

The iodine is formed, and held, by mixing KI, KIO₃ and HCl

5KI + KIO₃ + 6HCl           3H2O + 6KCl + 3I₂

The iodine is liberated from the KI

2KI             I₂ + 2e-

As it takes 2 moles of S2O3 to reduce the I2 to I-.

2S₂O₃^2-  + I₂    S₄O₆^2- + 2I-                                 
 multiply by 3 to get 3I₂

6S₂O₃^2-  + 3I₂    3S₄O₆^2- + 6I- 

Therefore, the molar ratio of S₂O₃^2-  to KI is 6:5 and the ratio of I-, IO₃ and S₂O₃^2-  is 5:1:6
And so the 5 moles iodide ions react with 1mole of iodate ions.

The concentration of iodide ions can also be determined. The volumes are divided by 1000 to give the result in dm³.





Volume of KIO₃ x concentration = volume of S₂O₃  x concentration

0.005dm³ x concentration    = 0.0285dm³ x 0.1mol dm^-3
                      Concentration  =      0.00285
                                                  _______
                                                    0.005dm

                                                  = 0.57 mol dm^-3
The ratio of KI to I₂ is 5:3 so
0.57mol dm^-3 x(5/3) = 0.95 mol dm^-3

[Hunter2]

The ratio of thiosulfate to iodide is 6:5

25.8ml of 0.1 m Thiosulfat gives 0.00258 mol  that means times 5/6 the iodide is 0,00215 mol

This was in 25 ml sample what means you will have 0.086 mol/l

Your titration value changed from 25,8 to 28,5 cm³ what was correct and your sample is now 5 ml and not 25 ml anymore.