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 can help you with questions 4 and 5.
4. To write net ionic equations, we need to identify the ions involved in the reactions and write the balanced equations for the precipitation reactions.
(a) First test tube with HgCl2:
The net ionic equation can be written as:
Hg2²⁺(aq) + 2Cl⁻(aq) → Hg2Cl2(s)
(b) Second test tube with SnCl2:
The net ionic equation can be written as:
Sn²⁺(aq) + 2Cl⁻(aq) → SnCl2(s)
In both cases, the precipitates formed are indicated by "(s)" in the equations.
5. To write the half-reactions and calculate the overall reaction potential, we need to balance the oxidation and reduction half-reactions.
(a) First test tube with HgCl2:
Half-reactions:
Sn²⁺(aq) → Sn(s) + 2e⁻ (oxidation)
Hg2²⁺(aq) + 2e⁻ → Hg2Cl2(s) (reduction)
The overall reaction potential can be calculated by adding the reduction potential (E°) of the reduction half-reaction with the oxidation potential (E°) of the oxidation half-reaction.
(b) Second test tube with SnCl2:
Half-reactions:
Hg²⁺(aq) + 2e⁻ → Hg(s) (reduction)
Sn²⁺(aq) → SnCl2(s) + 2e⁻ (oxidation)
Again, calculate the overall reaction potential by adding the reduction potential (E°) of the reduction half-reaction with the oxidation potential (E°) of the oxidation half-reaction.
To determine if the reactions occur spontaneously, the overall reaction potential needs to be positive. If it is negative, the reaction will not occur spontaneously.