This is was a sequence of chemical reactions lab of transforming copper. What wo
ID: 875792 • Letter: T
Question
This is was a sequence of chemical reactions lab of transforming copper.
What would be the result if a slight blue solution was poured off in the last step of adding metallic zinc (Zn) to a acidic solution to the reaction?
A. This is the anticipated result, and no yield changes are anticipated
B. If the solution decanted was even slightly blue, then Zn has been dissolved in solution and the yield of the copper would be reduced
C. If the solution decanted was even slightly blue, then not all of the [Cu(H2O)6]2+ ions have been reduced by Zn and the yield of the copper would be reduced
D. If the solution decanted was even slightly blue, this is a sign that not all of the Cu(OH)2 molecules have been transformed to CuO(s)
What would be the result if, in the by adding metallic zinc (Zn) to a acidic solution, if all of the water boiled away and the Cu was exposed to the Bunsen burner flame for a long period of time?
A. The Cu would oxidize to CuO causing the percent copper recovery to appear much too low
B. The Cu would oxidize to CuO causing the percent copper recovery to appear much too high
C. The Cu would oxidize to [Cu(H2O)6]2+ resulting in no copper recovery
D. The Cu would evaporate resulting in a very low percent recovery
Why couldn't you substitute 3M H2SO4 for concentrated HNO3 in the transformation from Cu(s) to [Cu(H2O)6]2+?
A. The necessary oxidizing agent is the NO3- ion, which is not present in H2SO4 solution
B. The necessary acid is the NO3- ion, which is not present in H2SO4 solution
C. The necessary reducing agent is the NO3- ion, which is not present in H2SO4 solution
D. The necessary base is the NO3- ion, which is not present in H2SO4 solution
Why couldn't you substitute concentrated HNO3 solution for 3M H2SO4 in a transfroming copper lab?
A. If HNO3 solution were used [Cu(H2O)6]2+ ion would not be reduced to Cu by Zn, because HNO3solution oxidizes Cu to [Cu(H2O)6]2+
B. If HNO3 solution were used in Parts IV and V, Cu(OH)2 would not be reduced to Cu by Zn, because HNO3 solution oxidizes Cu to Cu(OH)2
C.Both acids could be used
D. Neither acid should be used
Explanation / Answer
A NOTE:
There is no clarity in the questions. Still I would answer the questions based on what I could make out.
What would be the result if a slight blue solution was poured off in the last step of adding metallic zinc (Zn) to a acidic solution to the reaction? ( There is no clarity)
Answer:
C. If the solution decanted was even slightly blue, then not all of the [Cu(H2O)6]2+ ions have been reduced by Zn and the yield of the copper would be reduced
Explanation:
The equation for the reduction of Cu++ ions to metallic copper by Zn is
[Cu(H2O)6]2+ + Zn = [Zn(H2O)6]2+ + Cu
[Cu(H2O)6]2+ has blue color whereas [Zn(H2O)6]2+ is colorless. The completion of reduction [Cu(H2O)6]2+ to Cu is indicated by the blue solution becoming colorless.
Therefore if the solution decanted was even slightly blue, then not all of the [Cu(H2O)6]2+ ions have been reduced by Zn and the yield of the copper would be reduced
What would be the result if, in the by adding metallic zinc (Zn) to a acidic solution, if all of the water boiled away and the Cu was exposed to the Bunsen burner flame for a long period of time?
Answer:
B. The Cu would oxidize to CuO causing the percent copper recovery to appear much too high
Explanation:
Under the given condition Cu will be oxidized to CuO and the CuO would be formed on the surface of the metal.
1 mole of copper (63.54 g) gives 1 mole of CuO (79.54g)
Since CuO is heavier than Cu , the percent copper recovery will appear much too high
Why couldn't you substitute 3M H2SO4 for concentrated HNO3 in the transformation from Cu(s) to [Cu(H2O)6]2+?
Answer:
Explanation:
The oxidation number of Cu in Cu(s) is zero
The oxidation number of Cu in [Cu(H2O)6]2+ is +2
There is an increase in oxidation number when Cu(s) is converted into [Cu(H2O)6]2+ .This process is therefore an oxidation.
Since the transformation from Cu(s) to [Cu(H2O)6]2+ involves oxidation ,we need powerful oxidizing agent to effect this change. In acid medium NO3- ion is a powerful oxidizing agent.
The necessary oxidizing agent NO3- ion, is not present in H2SO4 solution.
Hence we cannot substitute 3M H2SO4 for concentrated HNO3 in the transformation from Cu(s) to [Cu(H2O)6]2+?
Why couldn't you substitute concentrated HNO3 solution for 3M H2SO4 in a transforming copper lab?
A. If HNO3 solution were used [Cu(H2O)6]2+ ion would not be reduced to Cu by Zn, because HNO3solution oxidizes Cu to [Cu(H2O)6]2+
B. If HNO3 solution were used in Parts IV and V, Cu(OH)2 would not be reduced to Cu by Zn, because HNO3 solution oxidizes Cu to Cu(OH)2
C.Both acids could be used
D. Neither acid should be used
There is no clarity in the question.For instance what do Parts IV and V.refer to ?
Answer:
A. If HNO3 solution were used [Cu(H2O)6]2+ ion would not be reduced to Cu by Zn, because HNO3solution oxidizes Cu to [Cu(H2O)6]2+
Explanation;
The conversion of [Cu(H2O)6]2+ ion into Cu(s) involves reduction ( Note that the oxidation number of copper decreases from +2 to 0 and decrease in oxidation number is reduction)
This process should therfore be carried out in the absence of oxidizing agents. Since HNO3 is a powerful oxidizing agent [Cu(H2O)6]2+ ion would not be reduced to Cu by Zn
Cu formed will be oxididized to Cu ++ ions which will exist as hydrated ions [Cu(H2O)6]2+
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