The Student Room Group

Scroll to see replies

Reply 1
Second question attached...

well all I know is this - you definitely need an oxidising agent from 2+ to 6+, but acid or alkali and why??
Reply 2
OK... first one:
1. Silver chloride ppts out (and it's white...)
2&3. Barium sulphate ppts out
4. Barium nitrate and sodium chloride remain in sln.

As far as the second question is concerned, my transition metal chem is very patchy - I ignore d-block questions wherever possible. A little unfortunate, seeing as next term, I have, I think, 12 lectures on d and f block chem... argh!
Reply 3
cpchem
OK... first one:
1. Silver chloride ppts out (and it's white...)
2&3. Barium sulphate ppts out
4. Barium nitrate and sodium chloride remain in sln.

As far as the second question is concerned, my transition metal chem is very patchy - I ignore d-block questions wherever possible. A little unfortunate, seeing as next term, I have, I think, 12 lectures on d and f block chem... argh!

First q: of course... I don't know why I imagined barium nitrate to be insoluble. I mean hey nitrate and sulphate is almost the same right?

For the second q: adding acid won't do much to the complex, so that's why I guess the alkaline solution, which apparently is correct...
The answer is alkaline conditions because the Mn(VI) state will disproportionate in acid conditions to Mn(II) and Mn(VII)
Reply 5
charco
The answer is alkaline conditions because the Mn(VI) state will disproportionate in acid conditions to Mn(II) and Mn(VII)

How am I supposed to know that??
Reply 6
How am I supposed to know that??

By looking at one of those famed Frost diagrams...
Reply 7
cpchem
By looking at one of those famed Frost diagrams...

Then I'm not supposed to know that.
Reply 8
I would have said alkali for the second one. This is my (A-level) reasoning:

This is the half equation I came up with:

[Mn(H20)6]2+ <-----> MnO42- + 4e- + 2H2O + 8H+

I know the waters should cancel but anyway..

If you look at it in terms of Le Chatilier's principle, adding acid shifts the equilibrium to the left and alkali shifts it to the right, therefore you would use an oxidising agent and alkali.

Somebody please shout if the above is complete and utter rubbish!
Manganese 2+ can be oxidised to both Mn(VII) and Mn(VI)

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

Mn2+ + 4H2O --> MnO4(2-) + 8H+ + 4e

Each of these half equations has a redox potential that must be considered against other half equation redox potentials when looking for probable reactions.

5MnO4(2-) + 8H+ --> Mn2+ + 4MnO4(-) + 4H2O

also has a redox potential which makes that process favourable in the direction shown in the presence of a suitable oxidising agent.

In answer to the plea for clemency - it's unfortunate but in order to know inorganic TM reactions you simply have to learn them.
Reply 10
charco
Manganese 2+ can be oxidised to both Mn(VII) and Mn(VI)

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

Mn2+ + 4H2O --> MnO4(2-) + 8H+ + 4e

Each of these half equations has a redox potential that must be considered against other half equation redox potentials when looking for probable reactions.

5MnO4(2-) + 8H+ --> Mn2+ + 4MnO4(-) + 4H2O

also has a redox potential which makes that process favourable in the direction shown in the presence of a suitable oxidising agent.

In answer to the plea for clemency - it's unfortunate but in order to know inorganic TM reactions you simply have to learn them.

Wait... how do you know the favourable acid/base conditions, without knowing what the oxidising agent is...

I think the most probable explanation, as Kyle_S-C pointed out to me in a PM, is that alkaline conditions help remove the hydrogens in the water ligands, as MnO4- has none.
Please note that in my previous posts I showed the disproportionation reactions producing Mn2+. The extent of the disproportionation depends of the severity of the conditions and although the equations previously posted are not incorrect, it is generally accepted that these reactions produce the Mn(IV) state in the form of MnO2 along with the Mn(VII) in the form of MnO4(-). This is corrected below:
------------------------------------------------------------------------------------------------------------

The half equation for oxidation of Mn2+ in alkaline conditions is:

Mn2+ + 4H2O --> MnO4(2-) + 8H+ + 4e

The half equation for oxidation of Mn2+ in acid conditions is:

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

so your argument could be equally applied to both reactions.

-----------------------------------------------------------------------------------------------------------

These are redox half-equations and so only apply in the presence of an oxidising agent - which one, doesn't matter providing that it's strong enough.

However, the Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation - it is a reaction):

3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O

The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:

3MnO4(2-) + 2H2O --> MnO2 + 2MnO4(-) + 4OH-

This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur, as the MnO4(-) ion is a powerful oxidising agent and can drive the reaction backwards.

However, what we are really comparing is the Gibbs free energy of the two disproportionation equations. In the case of acid conditions:

3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O

The Gibbs Free energy makes the forward reaction feasible. In the case of alkaline conditions, however, Gibbs Free energy makes the backward reaction equally feasible:

3MnO4(2-) + 2H2O <-- MnO2 + 2MnO4(-) + 4OH-

As stated in a previous post, this can be predicted with electrode potentials. If you break down each disporportionation into two half equations and compare the redox potentials using E = E(red) - E(ox).

OK, let's do it!
------------------------------------------------------
Acid conditions:

3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O

This disproportionation can be though of as two half equations:
MnO4(2-) + 4H+ + 2e --> MnO2 + 2H2O ........... Eº = +2.26V
MnO4(2-) --> MnO4(-) + 1e ............................Eº = +0.56V

Calculating Eº = E(red) - E(ox) = +2.26V - +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means tha the reaction is spontaneous as shown) so the forward reaction proceeds.

Alkaline conditions:

3MnO4(2-) + 2H2O --> MnO2 + 2MnO4(-) + 4OH-

This disproportionation can be though of as two half equations:
MnO4(2-) + 2H2O + 2e --> MnO2 + 4OH- ........... Eº = +0.67V
MnO4(2-) --> MnO4(-) + 1e..............................Eº = +0.56V

Calculating Eº = E(red) - E(ox) = 0.67 - 0.56 = 0.11V

Although this is a positive value, it is very small (under 0.3V) indicating that an equilibrium will be established (under standard conditions) and that the reaction will proceed in the forward direction by only an immeasurable amount, and as the equation is in equilibrium, by Le Chatelier, addition of base will drive the equilibrium in the reverse direction.
Reply 12
I forgot how amazingly dull inorganic chemistry is. This is why inorganic chemists decide to claim the majority of the periodic table - you can put all of the elements in a sack, and call it "inorganic", and it's still nowhere near as interesting as what an organic chemist can do with a just a few p-block elements...
and that's without studying diborane and the boranes...
Reply 14
Diborane is quite good, I'll give you that. We're forever being told about who first explained the bonding in it, which is fair enough, since it is quite impressive... what with him only being a second year and everything.
Reply 15
Doesn't inorganic chemistry contain most of solid state chemistry which is pretty interesting? That said, I didn't really enjoy the inorganic we covered in chemistry.
Reply 16
Solid state chemistry is perhaps the most boring thing we've looked at so far...
Reply 17
cpchem
Solid state chemistry is perhaps the most boring thing we've looked at so far...

Solid state chemistry already sounds boring. Learn to love the most "boring" thing ever - it makes life so much easier.
Reply 18
Well materials science seems largely about solid state chemistry (given that it's impossible to find books on materials science but easy to find them about solid state chemistry). It covers superconductivity...I can't see how it's that bad...
Reply 19
---- ANOTHER QUESTION ----

CsCl has a lower lattice enthalpy than NaCl, because, in simple terms, Na+ is more charge dense than Cs+.

But I could imagine the reaction between Cs and Cl to be more exothermic than Na and Cl - so doesn't that mean the bond between Cs and Cl should be stronger than Na and Cl?

Indeed CsCl is considered to be more ionic than NaCl, as electrons are transferred from one atom to another very distinctly. But still, the lattice enthalpy of NaCl is still larger than that of CsCl?

I'm SO confused: lattice enthalpy, bond strength, exothermicity of the reaction, the ionicitiy of the bond.