Questions:
1. In each test tube, a precipitate formed upon DROPWISE addition of the second reagent (steps 5 and 6). What are these precipitates?
(a) First test tube with HgCl©ü :
(b) Second test tube with SnCl©ü :
2. In each test tube, which reagent was limiting and which was in excess in this first part of the experiment?
3. In the second part of the experiment, the excess reagent in each test tube becomes the limiting reagent upon addition of a large quantity of the other reagent. Explain what happens in each test tube. (Hint: in one of the test tubes a second reaction with the products of the first is involved.)
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.
Answers:
1. (a) First test tube with HgCl2: Formed a white percipitate of Hg2Cl2 or mercurous chloride.
(b) Second test tube with SnCl2: Formed a brown percipitate of Hg or Mercury.
2. In test tube one, the limiting reageant would be the SnCl2 and the excess reagant would be the HgCl2. In the second test tube, the limiting reagant would be HgCl2, and the excess reagant would be the SnCl2.
3. After adding 5mL of SnCl2 to the first test tube, HgCl2 then became the limitting reagant and the white percipitate changed to brown in color and formed Hg. After adding 5mL of HgCl2 to test tube number two, SnCl2 became the limitting reagant, however, the percipitate stayed brown in color and Hg and Hg2Cl2 were both present.
I really need help!! I have no idea how to do number 4 or 5, and I'm not sure if number 3 is explained in depth enough?
I am wondering what you dropped into the test tubes.
Test tube one had 5mL of HgCl2, and for the first reaction I added 5 drops of SnCl2. I did the exact oppisite for test tube two. Then I added 5mL of SnCl2 to test tube one for the second reaction, and 5mL of HgCl2 to test tube two.
No problem, I'm here to help! Let's break down each question and explain how to solve them:
4. To write net ionic equations, we need to first write the balanced chemical equations for the reactions observed. From your answers, we know that in the first test tube, HgCl2 reacts with SnCl2, and in the second test tube, SnCl2 reacts with HgCl2. Here are the balanced chemical equations:
(a) First test tube: HgCl2 + SnCl2 -> Hg2Cl2 + SnCl4
(b) Second test tube: SnCl2 + HgCl2 -> SnCl4 + Hg
Now, let's write the net ionic equations:
(a) First test tube net ionic equation: Hg2+ + 2Cl- + Sn2+ + 4Cl- -> Hg2Cl2 + Sn4+
(b) Second test tube net ionic equation: Sn2+ + 4Cl- + Hg2+ + 2Cl- -> Sn4+ + Hg
In these net ionic equations, we've only included the ions that are involved in the reaction. Hg2Cl2 is a solid precipitate, so it is not included as ions.
5. To write the two half-reactions and calculate the overall reaction potential, we need to identify the oxidation and reduction half-reactions for each reaction.
(a) First test tube half-reactions:
Oxidation half-reaction: Sn2+ -> Sn4+ + 2e-
Reduction half-reaction: 2Hg2+ + 4e- -> 2Hg
Now, let's calculate the overall reaction potential by adding the two half-reactions together:
Overall reaction potential = reduction potential + oxidation potential
Please provide the reduction potentials and oxidation potentials for the half-reactions, so I can calculate the overall reaction potential for you.
I hope this explanation helps you with questions 4 and 5. If you have any further questions or need more clarification, feel free to ask!