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
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.
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
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.
#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.
To write the net ionic equations for the reactions, we need to identify the spectator ions and remove them from the equation. The net ionic equations include only the ions or molecules that participate in the actual reaction.
For the first reaction:
2 HgCl2 (aq) + SnCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)
Step 1: Write the balanced overall equation by balancing the number of atoms on each side:
2 HgCl2 (aq) + SnCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)
Step 2: Identify the ions that are spectator ions. In this case, the spectator ions are Cl- because they appear on both sides of the equation in the same form.
Step 3: Remove the spectator ions from the equation:
2 HgCl2 (aq) + SnCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)
The net ionic equation for the first reaction is:
2 Hg2+ (aq) + Sn2+ (aq) --> Sn4+ (aq) + Hg2Cl2 (s)
Now let's move on to the second reaction:
Hg2Cl2 (aq) + SnCl2 (aq) --> 2 Hg (s) + SnCl4 (aq)
Step 1: Write the balanced overall equation by balancing the number of atoms on each side:
Hg2Cl2 (aq) + SnCl2 (aq) --> 2 Hg (s) + SnCl4 (aq)
Step 2: Identify the spectator ions. In this case, the spectator ions are Cl- because they appear on both sides of the equation in the same form.
Step 3: Remove the spectator ions from the equation:
Hg2Cl2 (aq) + SnCl2 (aq) --> 2 Hg (s) + SnCl4 (aq)
The net ionic equation for the second reaction is:
Hg2Cl2 (aq) + Sn2+ (aq) --> 2 Hg (s) + Sn4+ (aq)
Now, let's move on to calculating the overall reaction potential by adding the two half-reactions.
For the first reaction:
2 Hg(l) + 2 e- --> Hg2+ E₀ = -0.79 V
Sn2+ + 2 e- --> Sn4+ E₀ = 0.15 V
To calculate the overall reaction potential, we need to add the two half-reactions together:
2 Hg(l) + 2 e- + Sn2+ --> Hg2+ + Sn4+
E₀total = -0.79 V + 0.15 V = -0.64 V
For the second reaction:
Sn4+ + 2 e- --> Sn2+ E₀ = -0.15 V
Hg2+ + 2 e- --> 2 Hg(l) E₀ = 0.79 V
To calculate the overall reaction potential, we need to add the two half-reactions together:
Sn4+ + 2 e- + Hg2+ --> Sn2+ + 2 Hg(l)
E₀total = -0.15 V + 0.79 V = 0.64 V
Therefore, the overall reaction potential for the first reaction is -0.64 V, and for the second reaction is 0.64 V.