Jump to content

Changing iron(II) to iron(III) and vice versa?


Coffee Bean

Recommended Posts

Hello!

 

I'm a high school boy and redox reactions are confusing me... My lab teacher asked me to do those redox experiments and now I lost my observations paper! What a douche bag I am! How can I present it to class next Sunday :-/

So, can you guys help me? The procedure of the experiments are:

 

Changing iron(II) to iron(III)

 

1. Pour 4cm3 of iron(II) sulphate, FeSO4 into a test tube

 

2. Add potassium manganate(VII), KMnO4 until yellow solution is obtained.

i. My lab teacher said this need to be acidified, but the book didn't say so, can anyone tell me why? However, I used acidified one when doing the experiment.

ii. Can you help me with the equations here?

 

3. Divide the solution equally into 3 separate test tube.

 

4. Test with reagents: Sodium hydroxide, NaOH ; Potassium hexacyanoferrate(II), K4Fe(CN)6 and Potassium thiocyanate, KSCN

i. Can you help me with the observations and equations here for each reaction with reagents?

 

Changing iron(III) to iron(II)

 

1. Pour 4cm3 of iron(III) chloride, FeCl3 into a boiling tube.

 

2. Add half spatula of Zinc powder and warm the solution gently.

i. Can you tell me why this need to be done?

ii. Can you tell me the equations here?

 

3. Cool the mixture and filter it.

 

4. Divide the filtrate equally to 3 separate test tube.

 

5. Test with reagents: Sodium hydroxide, NaOH ; Potassium hexacyanoferrate(II), K4Fe(CN)6 and Potassium hexacyanoferrate(III), K3Fe(CN)6

i. Can you help me with the observations and equations here for each reaction with reagents?

 

I only have couple of days more to do this, please do help me...

 

Ps: Sorry if I posted this in wrong section...

Link to comment
Share on other sites

Do you know the basics of setting up two half equations and then combining them?

 

Also you mention using the acidifed one, what does this mean? You added acid to the experiment?

Edited by farmboy
Link to comment
Share on other sites

Do you know the basics of setting up two half equations and then combining them?

 

Also you mention using the acidifed one, what does this mean? You added acid to the experiment?

 

Yes, I do know how to set up two half equations and combining them.

 

My lab teacher asked me to add sulphuric acid into potassium permanganate to acidify it.

Link to comment
Share on other sites

I have to leave very shortly, so I'll just give you a few basics, though not the answers. I'd have thought you know the basics of what a redox equation is etc so I'll just give a few clues on how to start.

 

Right for the first part you have to recognise that if iron is being oxidized here then another species must be reduced in the process. Once you establish what this species is you can set up two half equations. The first will be of the form Fe2+ -> Fe3+ but you will have to add electrons to one side of that equation to show how electrons are being lost or gained. So in this instance it is.....

 

Fe2+ -> Fe3+ + e-

 

You then need to write the other half equation, this one will be more difficult but I won't tell you what it is exactly but ill give a couple of hints. There will be oxygen on one side of the reaction, which will need to be balanced on the other by adding water (I'm assuming you know about the need to balance atoms on each side of the equation). Now obviously this means that you are missing some hydrogens on the first which are then added in the form of H+. This will look something like..

 

XO4- + 5e- + 8H+ -> X2+ + 4H2O

 

So you can see that the atoms have been balanced by adding the H+ to account for the waters, this is where the acid comes in. And then I think you know what to do from there. The process is the same each time. I've got no idea what the observations would be though sorry, thats something you really need to see I think.

Edited by farmboy
Link to comment
Share on other sites

Thanks! I figured out the X will be manganese ^^

 

The observations I've been asking are simple ones... Let's say, blue precipitate formed, yellow solution turned to blue and so on...

 

Yeah I know, but I'd have to do all those equations to even guess what the change might be. Since I left secondary school I don't think I've ever had to worry about the colour change of a reaction, so it's not like I know them off the top of my head either unfortunately. I think you probably would need to have witnessed them to know exactly what happens. If you are going to have to guess then what you should do is get the right answer for every equation. Different metal ions have different colours in solution so that is what the colour will be based on, now the problem here is that you are probably going to have more than one ion in each case so that will make things more complicated. So apparently Mn2+ solutions are very pale pink or colourless and Fe3+ is yellow in solution, so that is why you get a change to yellow in that first case. You'll have to work out the rest in the same manner.

Edited by farmboy
Link to comment
Share on other sites

Just a thought.

You want to hand in a report that will get you full marks.

You won't do that unless your report says what you saw.

While you are writing those observations down, you might as well put them here too, along with any ideas you have about what they mean.

Perhaps we can help you understand them better.

Link to comment
Share on other sites

Just a thought.

You want to hand in a report that will get you full marks.

You won't do that unless your report says what you saw.

While you are writing those observations down, you might as well put them here too, along with any ideas you have about what they mean.

Perhaps we can help you understand them better.

 

No, there is no report to do. Just a presentation to help the class understand redox better. Hmm.. I can't remember much the observations as there are 6 test tubes. What do I remember is when heating zinc in iron(III) chloride, the reaction was very vigorous whoaa...

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.