Question

4. Use the information above to write net ionic equations for the two reactions that you have observed.

5. For each reaction, write the two half-reactions that make it up and calculate the overall reaction potential by adding the two half-reactions. Each overall potential should be positive in order for the reactions to occur spontaneously.


4. 2 HgCl2 (aq) + SnCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)
Hg2Cl2 (aq) + SnCl2 (aq) --> 2 Hg (s) + SnCl4 (aq)

5.
2 Hg(l) + 2 e⁻ --> Hg©ü©÷⁺ E₀ = -0.79 V
Sn©ù⁺ + 2 e⁻ --> Sn©÷⁺ E₀ = 0.15 V
Total: -0.64 V


Sn©÷⁺ + 2 e⁻ --> Sn©ù⁺ E₀ = -0.15 V
Hg©ü©÷⁺ + 2 e⁻ --> 2 Hg(l) E₀ = 0.79 V
Total: 0.64 V

Answer 1

This looks to be part of the answers for a problem by another person; however, without the same name I can't go back and look at the original post.
#4 looks ok except they are molecular equation; the problem asked for net ionic equations. Also, I don't know where the equations came from; but I don't see anything wrong with what is here. I'm assuming what went before is ok.
#5. I can't make out the funny symbols.

Answer 2

Sn2+ (aq) + 2Hg2+ (aq) + 2Cl (aq)--> Sn4+ (aq) + Hg2Cl2 (s)
Hg2Cl2 (s) + Sn2+ (aq) --> 2Hg (I) + Sn4+ (aq) + 2Cl- (aq)

5. Sn(2+) --> Sn(4+) + 2e- Eo = -0.15 V
2e+ + 2Hg(2+) + 2Cl --> Hg2Cl2 Eo = 0.85 V
Total: 0.75 V

Hg2Cl2 + 2e- --> 2Hg(2+) + 2Cl- Eo= 0.27 V
Sn(2+) + 2 e? --> Sn(4+) Eo = -0.15 V
Total: .12 V

Answer 3

Those are different from the original post, I am trying to work on it!

Answer 4

#4.Sn2+ (aq) + 2Hg2+ (aq) + 2Cl (aq)--> Sn4+ (aq) + Hg2Cl2 (s)
<b>I think 2Hg2+ should be 2HgCl2 for the complete rxn of
Sn^2+ 2HgCl2 ==> Hg2Cl2(s) + Sn^4+ + 2Cl^-
What you wrote has Sn being oxidized but Hg doesn't change.; i.e., it's one on the left and one on the right.</b>

Hg2Cl2 (s) + Sn2+ (aq) --> 2Hg (I) + Sn4+ (aq) + 2Cl- (aq)
<b> This part looks ok.</b>

Answer 5

#5. You need to redo the Hg one as a result of my comments above.
For Sn you want
Sn ^2+ ==> Sn^4+ + 2e Eo = -0.15. Written this way it is an oxidation potential.