Which substance acted as a catalyst?

[DeathKnight]

1st:( I'm sorry its a bit long)
1.0g of Zn in reacted with 50 cm3 of 1.0 mole/dm3 of H2SO4 to check the effect of some chemicals or elements on the reaction. 4 exp. are carried out in each exp. the time to collect 50cm3 of hydrogen was recorded. 1st exp. was done with nothing added to reaction mixture. It took 65s to collect 50cm3 of H2 and a colourless solution was obtained. In 2nd exp. CuSO4 was added as possible catalyst. The time taken was 10s and a colourless solution obtained with brown coating on Zn. In third exp. Cu(powdered) was added as possible catalyst . The time recorded was 19s and a colourless solution formed with brown solid at the bottom. In last exp. NaCl was added as possible catalyst. The time recorded was 65s and a colourless solution formed.

The question in the book is which substance acted as a catalyst?
It says the answer is: Cu and CuSO4
Why CuSo4 has been included in the ansnwer? Doesnt the defination of catalyst say it should remain chamically unchanged at the end of the reation? But here CuSO4 has been reduced to Cu.

2nd Question:

In a displacement reaction an Iron rod is dipped into aq. CuSO4. Why not the reaction stops once a layed a Cu is deposited on Iron rod but instead continues until all the CuSO4 solution is used up or whole of the iron rod dissolves away? I'm asking this because the reaction between Al and O2 Al2O3 formed stops any furthur oxidation of Al.

And thirdly: Is the following reaction a redox reaction. If it is a redox reaction which is an oxidizing agent and which one is a reducing agent? I know in the ionic equation of this reaction Zn is the reducing agent and Cu2+ is an oxidizing agent. But this really confuses me.

CuSO4(aq) + Zn(s) -------> Cu(s) + ZnSO4(aq)

Fourthly, Is it true that when an oxidizing agent is added to KI Iodine is formed which dissolves in KI to form a brown solution.

And lastly, Excess of KI(a reducing agent) is added to Fe3SO4(which i think should be an oxidizing agent). What will be the colour of the mixture after the reaction is complete? Will it be Green or brown? And what exactly will be the equation for the reaction?

Thanks in advance for any help.

 

[chem_tr]

1st:( I'm sorry its a bit long)
1.0g of Zn in reacted with 50 cm3 of 1.0 mole/dm3 of H2SO4 to check the effect of some chemicals or elements on the reaction. 4 exp. are carried out in each exp. the time to collect 50cm3 of hydrogen was recorded. 1st exp. was done with nothing added to reaction mixture. It took 65s to collect 50cm3 of H2 and a colourless solution was obtained. In 2nd exp. CuSO4 was added as possible catalyst. The time taken was 10s and a colourless solution obtained with brown coating on Zn. In third exp. Cu(powdered) was added as possible catalyst . The time recorded was 19s and a colourless solution formed with brown solid at the bottom. In last exp. NaCl was added as possible catalyst. The time recorded was 65s and a colourless solution formed.

Well, zinc and copper react with sulfuric acid to give hydrogen; with different reaction enthalpies. If copper is a catalyst to this reaction, then we can understand that it provides a lower energy state than the other compound.

• A brown coating on Zn indicates that a redox process has occurred to reduce copper while zinc has been oxidized. In control group (nothing extra added), the slowest reaction occurs.
• When you add some CuSO₄, it is converted into elemental Cu, so facilitates the oxidation of Zn. However, some Cu can also react with excess H₂SO₄ to give CuSO₄ again; this is why it is called as a catalyst.
• In the third example, you used powdered copper, which is not as good as copper ions. It cannot facilitate the oxidation of zinc, since it has already been reduced; it must have got electrons from zinc to provide a better catalyst. However, it is an indirect catalyst, since some of them is oxidized to Cu²⁺ ions by sulfuric acid, and this can do a catalytic effect on Zn.
• It is clear NaCl has no catalytic effect, for both it has no available additional oxidation states to facilitate oxidation, and its reaction time is the same as untreated (control) experiment.

The question in the book is which substance acted as a catalyst?
It says the answer is: Cu and CuSO4
Why CuSo4 has been included in the ansnwer? Doesnt the defination of catalyst say it should remain chamically unchanged at the end of the reation? But here CuSO4 has been reduced to Cu.

I disagree with the information that a catalyst is chemically unchanged. If you label a copper atom with a radioactive one and monitor its change, I doubt that it is reversibly consumed as if it were not present in the process. It can briefly be said that same kind of molecule leaves the reaction, but it doesn't imply that the atoms are the same, just a new molecule of copper sulfate is formed as a product, I think.

2nd Question:

In a displacement reaction an Iron rod is dipped into aq. CuSO4. Why not the reaction stops once a layed a Cu is deposited on Iron rod but instead continues until all the CuSO4 solution is used up or whole of the iron rod dissolves away? I'm asking this because the reaction between Al and O2 Al2O3 formed stops any furthur oxidation of Al.

It can be explained by "inactivation"; in which the formed product is inert and protects the metal being coated. In aluminum oxide, it proceeds in this way. However, I speculate that iron sulfate occurs from the redox, and it seems as reactive as the previous one; so we cannot say that an inactivation is present in this case, even a catalytic effect may be considered.

And thirdly: Is the following reaction a redox reaction. If it is a redox reaction which is an oxidizing agent and which one is a reducing agent? I know in the ionic equation of this reaction Zn is the reducing agent and Cu2+ is an oxidizing agent. But this really confuses me.

CuSO4(aq) + Zn(s) -------> Cu(s) + ZnSO4(aq)

Yes, this reaction is an obvious redox-type one. If a compound is reduced at the end, it means that the compound is an oxidizer, and vice versa. Your guess is correct; if you are confused, you can use this analogy; an oxidizer is a heavier buddy on a seesaw; it holds you up while he goes down. A reducer is the opposite one; reducer buddy is lighter than you, so you go down while he goes up.

Fourthly, Is it true that when an oxidizing agent is added to KI Iodine is formed which dissolves in KI to form a brown solution.

Yes, this is correct. Tincture of iodine (Tentur'd iode in French; I hope I wrote it correctly) is prepared upon this mechanism.

And lastly, Excess of KI(a reducing agent) is added to Fe3SO4(which i think should be an oxidizing agent). What will be the colour of the mixture after the reaction is complete? Will it be Green or brown? And what exactly will be the equation for the reaction?

In magnetite, aka $\displaystyle FeO \cdot Fe_2O_3$, the oxidizer is Fe³⁺ counterpart, as it reduced to Fe²⁺ while iodine is formed. You know the color of iodine, I think. In my opinion, colors of iron(II) and iodine will be observed together, but iodine has a sharp color, powerful enough to cover that of iron(II).

Hope that helps.

 

[DeathKnight]

Thanks a lot Chem_tr. I'll be forever in your debt. :)