6N.11

[Maya Gollamudi 1G]

How do you solve problem 6N.11:
Use data from Appendix 2B to calculate the solubility product of Hg2Cl2 & compare this number with the value listed in Table 6I.1 and comment on any difference

[JasonLiu_2J]

The solubility product is essentially the equilibrium constant of the solubility expression for Hg2Cl2. When Hg2Cl2 dissolves in solution, you get the equation Hg2Cl2(s) --> Hg2^2+(aq) + 2Cl-(aq). Looking at appendix 2B, you find a reduction half reaction with the form Hg2Cl2(s) --> 2Hg(l) + 2Cl-(aq). Afterwards, you want to find an oxidation half reaction that, when paired with the reduction half reaction, gives you the solubility expression. Looking again at the appendix, you will find the reduction reaction Hg2^2+(aq) +2e- --> 2Hg(l). Converting this to the oxidation half reaction will give you 2Hg(l) --> Hg2^2+(aq) +2e-. Combining the two half reactions will now give you the solubility expression you want. From here, you can find the overall cell potential of the solubility expression using the cell potentials from the half reactions, and finally use the equation nFE=RTln(K) to solve for K, the solubility product. Hope this helps!

[owenbella1]

To solve:
Derive: Hg2Cl2(s) --> Hg2^2+(aq) + 2Cl-(aq). Then from appendix 2B, find a reduction half reaction Hg2Cl2(s) --> 2Hg(l) + 2Cl-(aq) and Hg2^2+(aq) +2e- --> 2Hg(l).
The convert to oxidation: 2Hg(l) --> Hg2^2+(aq) +2e-. Now combine the oxidation and reduction: Hg2Cl2(s) --> Hg2^2+(aq) + 2Cl-(aq). This verifies that 2 e- are transferred, so n=2.
Now find the Standard potential for the reaction from the values derived from the appendix: .27 - .79 = -.52
Then use nFE=RTln(K) to solve for K giving 2.85 x 10^-18 which is pretty close to the answer in the book!