Metal displacement reactions (1 Viewer)

[YBK]

Could anyone please help me out with the following:

Using the activity series, write ionic equations and describe what changes (if any) would occur in the following.

Chromium is added to a solution containing Fe^(2+) ions

What I got is this:

Chromium is more reactive than iron, therefore Chromium displaces iron.

Cr(s) --> Cr^(3+)(aq) + 3e

Fe^(2+)(aq) + 2e --> Fe(s)

.: Ionic equation is:

Cr(s) + Fe^(2+)(aq) --> Fe(s) + Cr^(3+)(aq)




Does that look right? Because in the contexts textbook they have Cr^(2+) instead of 3+. Is there some kind of rule I'm not following?

Thanks!

 

[airie]

I think that's because chromium can take on the valency of 2+(for this question it does), 3+(eg. chromium sulphate), or 6+ (eg.chromium oxide)

 

[YBK]

I think that's because chromium can take on the valency of 2+(for this question it does), 3+(eg. chromium sulphate), or 6+ (eg.chromium oxide)

Thanks!

But how do we know which one it will take?

 

[tommy_1]

You recieve a standard reduction potencials sheet at the beginnign of you hsc exams telling you which valency they're expecting you to know - you dont even need to remember them - so go to the boards website and have a look at the data sheet at the front of a past hsc- if you havent printed iut off already - and see what the valency it.

 

[tommy_1]

yes the valency does change - however the one ont he data sheet is the one you'll need to know for the exam

 

[YBK]

Ah, okay, thank you. The one in that sheet is 3+, I was confused as to why it was 2+ in the answers...

 

[Riviet]

Cr(s) --> Cr^(3+)(aq) + 3e

Fe^(2+)(aq) + 2e --> Fe(s)

.: Ionic equation is:

Cr(s) + Fe^(2+)(aq) --> Fe(s) + Cr^(3+)(aq)




Does that look right? Because in the contexts textbook they have Cr^(2+) instead of 3+. Is there some kind of rule I'm not following?

Thanks!

The reason why your electrons are "supposed" to disappear in your ionic equation, is because they cancel out on both sides ie you have for example 2e on LHS and RHS, which you can subtract from both sides. If Cr were to donate 3 electrons, there would be an inequality in the number of electrons on both sides of the ionic equation when you add the two half-equations. However, this doesn't explain why Cr readily loses 2 electrons instead of 3. It just means that the 2e on both sides cancel out in your ionic equation.

 

[YBK]

Cr(s) --> Cr^(3+)(aq) + 3e (multiply by 2)
2Cr(s) --> 2Cr^(3+)(aq) + 6e

Fe^(2+)(aq) + 2e --> Fe(s) (multiply by 3)
3Fe^(2+)(aq) + 6e --> 3Fe(s)



.: Ionic equation is:
2Cr(s) + 3Fe^(2+)(aq) + ---> 2Cr^(3+)(aq) + 3Fe(s)




Now I'm sure it's right. The reason I used 3 was because it was the number on the table of standard potentials. It makes things more complicated, but should be right. I didn't know about the electron balancing thing yesterday, just did a few exercises in the context and figured it out.

 

[YBK]

The reason why your electrons are "supposed" to disappear in your ionic equation, is because they cancel out on both sides ie you have for example 2e on LHS and RHS, which you can subtract from both sides. If Cr were to donate 3 electrons, there would be an inequality in the number of electrons on both sides of the ionic equation when you add the two half-equations. However, this doesn't explain why Cr readily loses 2 electrons instead of 3. It just means that the 2e on both sides cancel out in your ionic equation.

Yeah, that's probably why they used the 2+ one

 

[airie]

Usually, during a displacement reaction like this, the valency of whatever ion that stays in the compound and not replaced would stay the same. For this question, as it was stated that the solution contains Fe^2+ ions, it means that the other ion is of the valency 2-. Therefore, chromium, in order to displace iron and form a stable compound, would need a valency of 2+

 

[YBK]

lol, we got the SAME question in our test today! :O I did it the balancing way, because we needed to find the emf.



airie, you make sense, but the problem is that we wouldn't know the cell potential of that Cr with valency 2+ since it is not on on the table of standard electrodes.

thanks!