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.
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
5 answers
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
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
Those are different from the original post, I am trying to work on it!
#4.Sn2+ (aq) + 2Hg2+ (aq) + 2Cl (aq)--> Sn4+ (aq) + Hg2Cl2 (s)
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.
Hg2Cl2 (s) + Sn2+ (aq) --> 2Hg (I) + Sn4+ (aq) + 2Cl- (aq)
This part looks ok.
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.
Hg2Cl2 (s) + Sn2+ (aq) --> 2Hg (I) + Sn4+ (aq) + 2Cl- (aq)
This part looks ok.
#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.
For Sn you want
Sn ^2+ ==> Sn^4+ + 2e Eo = -0.15. Written this way it is an oxidation potential.