My Chemistry Thread

[alchemy]

^Bump for the previous question.

Also, I'm slightly confused on determining states (particularly in redox reactions).
For example, MnO₄^-(aq) +4H⁺+2e^- ---> MnO₂(s) + 2H₂O. How do we know MnO₄ is aqueous on the left hand side, but MnO₂ is a solid?

EDIT: I should've clarified my question. Say we were asked to write the half equation for the reduction of MnO₄^- to MnO₂, how would we know what states the reactants and products are in?

[Yacoubb]

^Bump for the previous question.

Also, I'm slightly confused on determining states (particularly in redox reactions).
For example, MnO₄^-(aq) +4H⁺+2e^- ---> MnO₂(s) + 2H₂O. How do we know MnO₄ is aqueous on the left hand side, but MnO₂ is a solid?

EDIT: I should've clarified my question. Say we were asked to write the half equation for the reduction of MnO₄^- to MnO₂, how would we know what states the reactants and products are in?

MnO₄^2- is a polyatomic ion, meaning that it in a solution, the permanganate ions would be aqueous. MnO2 is a solid; we know that basically <anything> oxide (except Na, K or any other element in group 1) is insoluble. Thus, it wont be aqueous, we know it cannot be liquid, so it must be solid.

I'll give you a little tip for solubility:

Nitrates:
Any compound containing a nitrate ion is soluble.

Sulphate:
Any compound containing a sulphate ion is soluble, EXCEPT for (Calcium Sulphate, Barium Sulphate and Lead Sulphate).

Carbonates, Hydroxides, Oxides, Phosphates and Sulfides
Excluding any compounds containing any of the above anions and an cation from group 1 (or NH4+) (e.g. KOH, Na3PO4), every compound that contains one of the above anions is insoluble in water.

Halides - e.g. chloride, iodides, etc.
EXCEPT for Silver <halide>, Lead <halide> or Mercury <halide>, all compounds that contain a halide anion is soluble in water.

[alchemy]

OMG thanks Yacoubb! I completely neglected my solubility rules when doing redox questions, but will keep in mind to remember them from now on.

[Yacoubb]

OMG thanks Yacoubb! I completely neglected my solubility rules when doing redox questions, but will keep in mind to remember them from now on.

No worries. That chart is literally my best friend. Note that when you do organic chem questions, states are not important.

Remember your famous 5 soluble ions! Na+, NH4+, K+, CH3COO- and NO3-

Good luck

[alchemy]

I think I'm making a silly mistake on the following question:

Balance the following redox reaction by separating them into two half-equations, balancing each equation and then combining the pair into a balanced complete redox reaction.
SO₂^–(aq) + MnO₄^–(aq) + H⁺(aq) ——> SO₃^2–(aq) + Mn²⁺(aq).

So my working out was:
SO₂^–(aq) + H₂O --> SO₃^2- +2H⁺ + 1e^-
MnO₄ +8H⁺+5e^---> Mn²⁺ + 4H₂O
Multiply the first equation by 5 to get: 5SO₂^–(aq) + 5H₂O --> 5SO₃^2- +10H⁺ + 5e^-
Combine both equations together to get: 5SO₂^–(aq) + H₂O + MnO₄ --> 5SO₃^2- + Mn²⁺ + 2H⁺

But the answer is: 5SO₃^2- + 2MnO₄^- + 6H⁺ --> 5SO₄^2- +2Mn²⁺ +3H₂O

What have I done wrong?

[alchemy]

What have I done wrong?

Wait, I think I might've figured out my mistake. The first line of working out is meant to have 2e^- on the right hand side, correct?
Only problem is I don't understand why things like I^- result from I₂ + 2e^-. If I₂ has an oxidation number of 0, then shouldn't just one electron be added to it to it to make it -1. As in 0+(-1)=-1. 

[Rishi97]

Wait, I think I might've figured out my mistake. The first line of working out is meant to have 2e^- on the right hand side, correct?
Only problem is I don't understand why things like I^- result from I₂ + 2e^-. If I₂ has an oxidation number of 0, then shouldn't just one electron be added to it to it to make it -1. As in 0+(-1)=-1. 

You haven't balance this properly.The charge on the LHS is +3 but the RHS is +2
MnO4 +8H++5e---> Mn2+ + 4H2O

[alchemy]

You haven't balance this properly.The charge on the LHS is +3 but the RHS is +2
MnO4^- +8H++5e---> Mn2+ + 4H2O

The charge on the LHS is +7, isn't it, because -1+8=7? So that +7 - 5 (because of the 5 electrons added) give +2 on the RHS.

[Rishi97]

The charge on the LHS is +7, isn't it, because -1+8=7? So that +7 - 5 (because of the 5 electrons added) give +2 on the RHS.

No because there are 8H + so a charge of 8, then you minus the 5 electrons. So 8-5=3

[alchemy]

No because there are 8H + so a charge of 8, then you minus the 5 electrons. So 8-5=3

Oh right, thanks Rishi! I never knew that all this time

Are you or anyone else able to explain this for me please:

Is the first line of working out is meant to have 2e^- on the right hand side?
Only problem is I don't understand why things like 2I^- result from I₂ + 2e^-. If I₂ has an oxidation number of 0, then shouldn't just one electron be added to it to it to make it -1. As in 0+(-1)=-1. 

[lzxnl]

Two iodine atoms. Each one gets an electron

[Rishi97]

If you are still a bit confused, give me the original question and I'll work it out and try to show you how it all works out

[alchemy]

Lol, I just figured out that the book answers for this exercise are almost entirely incorrect. Seems like they fixed it up on the 'enhanced' book...

I'm also somewhat puzzled with this question as I end up with unusually large numbers when trying to balance the redox equations:

NO(g) + Cr₂O₇^2-(aq) --> NO₃^-(aq) + Cr³⁺(aq)

Interestingly enough, the 'enhanced' book has removed this question and added a somewhat similar one, but with water on the right hand side as a product too. I was wondering if this question is still doable?

[lzxnl]

Lol, I just figured out that the book answers for this exercise are almost entirely incorrect. Seems like they fixed it up on the 'enhanced' book...

I'm also somewhat puzzled with this question as I end up with unusually large numbers when trying to balance the redox equations:

NO(g) + Cr₂O₇^2-(aq) --> NO₃^-(aq) + Cr³⁺(aq)

Interestingly enough, the 'enhanced' book has removed this question and added a somewhat similar one, but with water on the right hand side as a product too. I was wondering if this question is still doable?

Preliminary check: oxidation numbers? NO goes from +2 to +5 in NO₃^-, while Cr₂O[/sub]7[/sub]^2- is +6 and Cr³⁺ is +3
Hmm. The numbers should be fine

NO + 2H₂O => NO₃^- + 4H⁺ + 3e^-
Cr₂O₇^2- + 14H⁺ + 6e^- => 2Cr³⁺ + 7H₂O

Overall I get 2NO + Cr₂O₇^2- + 6H⁺ => 2NO₃^- + 2Cr³⁺ + 3H₂O

The numbers aren't too bad.

[alchemy]

Probably a stupid question, but why is the anode negatively charged, if that's where oxidisation (loss of electrons) is occurring?