Question

This is a two part solution. Please include step by step solution that way I can study from this. Thank you.

Equation (24.2): 3I^-(aq) + H2O2(aq) + 2H3O^+(aq) ---> I3^-(aq) + 4H2O(l)

Equation (24.11): 2 S2O3^2-(aq) + I3^-(aq) ---> 3 I^-(aq) + S4O6^2-(aq)

Part 1:

Add equations 24.2 and 24.11 together.What happens to I^-1 and I₃^-1?

What should happen to their concentrations while S₂O₃^-2 is present?

Inspect equations 24.2 and 24.11 again.

What will happen to the concentration of I₃^-1 when S₂O₃^-2 is gone?

Part B:

When would the solution turn blue if S₂O₃^-2 was never added to the solution?

When would the solution turn blue if starch was never added to the solution?

Explain your answers.

Answer 1

3I^-(aq) + H2O2(aq) + 2H3O^+(aq) ---> I3^-(aq) + 4H2O(l)

2 S2O3^2-(aq) + I3^-(aq) ---> 3 I^-(aq) + S4O6^2-(aq)

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3I^-(aq) + H2O2(aq) + 2H3O^+(aq) +2 S2O3^2-(aq) ------>  3 I^-(aq) + S4O6^2-(aq)+ 4H2O(l)

I- is retained as a product if the reactions are added as redox reactions takes place twice when once it is reduced and then oxidized back.I^3- is an intermediate in the rxn

With S2O3^2- present the concentration of I^3- reduces as it is oxidised by the former.While I^- concentration increases as it is the oxidised product

If S2O3^2-  is all used up then the concentration of I^3-  will increase in the solution

Part B

The solution would turn blue at the moment H2O2 is added to I^- solution,first drop ,(the solution containing starch ) as I3^-  reacts with starch to give blue complex.

It would never turn blue  as I3^-  reacts with starch to give blue complex.