HELPPPPPPPPPPP!!! THERMODYNAMICS/CHEMISTRY SIMPLEE!!!!!!! 3. By reference to the

Question

HELPPPPPPPPPPP!!! THERMODYNAMICS/CHEMISTRY SIMPLEE!!!!!!!

3.         By reference to the EH-pH diagram for the uranium/water system based on unit activity for all the dissolved uranium species at 25oC, answer the following questions.

(a)        Write the balanced chemical equations for lines 4, 8, and 9.

(b)        Recommend a reagent and a pH range suitable for the dissolution of UO2(s) to UO22+.

(c)        What would you expect to happen if the pH of the solution in part (b) was changed to 10? Give reasons for your answer

Explanation / Answer

a) 4. Because is a continuos oblicuous line we know that this an equilibrium that depends on the pH and voltage of the solution and that occurs between two solid species.

UO2(s) + H2O(l) ========= UO3(s)  +H2(g)

8. When we have an oblicuos line we know that we have dependency from the pH and V. When this line is strong and discontinuous this means that we have the same species but this can affect the water descomposition to hydrogen and oxygen gaseous.

UO2(s) + H2O(l) ========= UO2(s)  +H2(g) + O2(g)

9. For this occurs the same that in the line 8 but with UO3

UO3(s) + H2O(l) ========= UO3(s)  +H2(g) + O2(g)

b) For this reaction to occur we would have to be over the line 2 with the equilibrium:

UO2(s) ========= UO2+2(ac)

This would have to be in a determined Voltage around 0,2V and in the range of pH of more than 2 and less than 4,5.

We could use sulfate, because the pH of this anoin is acid so it would favor the solubility of the UO2.

UO2(s) + H2SO4 ========= UO2+2(ac) + HSO4- + H+

c) We would have a change of the specie  UO2+2(ac) to  UO3(s), the SO4- ion won´t affect because it will be only in this form.

UO2+2(ac) +OH- +H2O ======== UO3(s) + 1/2H2O +H2(g)

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