Seperation and Qualitative Analysis of Cations Post Lab Please help me finish th

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Seperation and Qualitative Analysis of Cations Post Lab

Please help me finish the questions that have not be crossed out, I only have until the ned of the night and my final grade depends on it. PLEASE I WILL LOVE YOU

Here is the lab procedure:

I know its a lot to digest, but it would mean the WORLD to me if you could help me finish these questions.

3. Write half-reactions that show how H202 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations ed in your testing. formed, and then dissolved as complexes. these situations for which to write a b lect three 5. When Min ions are separated from the mixture, they go through a series of oxidizing and reducing steps. Write the reaction equations that describe this process. NH3 solution is used, in three separate insta s to separate two i each of these cases, one ion forms a precipitate and the other ion remains in so n as a complex. Describe each of these instances using reaction equations 7. Als and 21 ions are amphoteric. Describe how the amphoteric nature of these ions is shown in your testing.

Explanation / Answer

Frm the test tube 2, u have Mn2+.

when u add H2O2 u wl get Mn2+ oxidation to Mn4+

Mn2+ + H2O2 + 2 OH- = MnO2 + 2 H2O

again it will oxidised to Mn7+

Mn2+ + 4H2O --> MnO4- + 8H+ + 5e-

The available elctrons reduce the NaBiO3

NaBiO3 + 6H+ + 2e- --> Bi3+ + Na+ + 3H2O

The last step is to put them together. We need to eliminate the electrons from each side, so we multiply the first equation by 2 and the second by 5. This will give 10 electrons on each side so they will cancel out and we will get:

2Mn2+ + 5NaBiO3 + 14H+ --> 2MnO4- + 5Bi3+ + 5Na+ + 7H2O

MnO4- is pink colour, so if supernatent is pinl colour then Mn2+ is presen.

b)  an amphoteric compound is a molecule or ion that can react both as an acid as well as a base

Al3+ reacts with acid like HCl and base like NaOH

Al(OH)3 + HCl ----> AlCl3 + H2O

Al(OH)3 + NaOH ----> Na[Al(OH4)]

like wise Zn2+ also behaves

Zn(OH)2 + HCl ----> ZnCl2 + H2O

Zn(OH)3 + NaOH ----> Na4[Zn(OH3)]

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