[udit]

If anyone here is familiar with redox oxidisation-reduction reactions please help me with these two questions:

1. For the following redox displacement reactions, describe what you would observe and give the net ionic equation for the reaction:

a) Bromine water is added to a solution of potassium iodide and the mixture is shaken.

b) Magnesium is added to lead nitrate solution.

I have an idea of what it would be but I am not sure, any help would be appreciated. Thanks!


And one more thing, dont redox reactions need to have an e- somewhere? like for this reaction:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g)

I did the following:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)

Cu(s) + So₄^2-(aq) + 4H⁺(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)

It's even and no electrons are needed to balance it? is this right or what?

Any help would be appreciated, I need to understand these questions.

[xiankai]

your reaction is quite wrong. are u trying to show a displacement reaction? in that case, sulphate anions do not take part in the reaction. rather, a less reactive metal cation like silver in solution is displaced.

Cu(s) + 2Ag+(aq) --> 2Ag(s) + Cu2+(aq)

electrons only appear in half equations. for example, take the above reaction, it shows reduction and oxidation taking place simultaneously. should we split them apart into reduction and oxidation seperately, we'll get:

Cu(s) --> Cu2+(aq) + 2e-

2Ag+(aq) +2e- --> 2Ag(s)

note that each reaction shows only oxidation, or just reduction. in order to balance such equations, we need to add e-. H+ and OH- are not needed because the elements are balanced, all that is left is to balance charge.

as for your questions, since they involve single displacement reactions, which means an ion is displaced by a substance from its solution. also note that the ions themselves do not change, so u cant change SO42- into SO2 and O2-.

good luck

[udit]

Ok thanks for replying. It gives the equation:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g)

And asks me to "balance the oxidation-reduction equation". So I did this:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)

Cu(s) + So₄^2-(aq) + 4H⁺(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)



But it also asks, show both oxidation and reduction half-equations. For this reaction, I put:

Oxidation: Cu(s) --> Cu2+(aq) + 2e-
Reduction: So4^2-(aq) + 4H⁺ + 2e- --> So₂(g) + 2H₂O(l)

is that basically what it is asking for?

[albicilla]

hi,

before you can start solving redox-reactions you have to understand the concept of  oxidation numbers 'cause they are necessary to keep track of the electrons.

a few guidelines:

1:the ox number of an atom in an element is 0->Cu _(s)⁰
2:(in covalent compounds with nonmetals) hydrogen is assigned a state of +1
3: oxygen usually has -2 (except in peroxides or in compounds with F)
4: the ox state of a monoatomic ion = its charge
5: in compounds, the element with greater attraction for e ^- has a negative ox number (F always -1)
6: in an electrically neutral compound the sum of the ox numbers =0

so your  reaction is correct. try to do it again, but this time with the electrons and the "half-reactions" by which i mean the oxidation reaction with Cu and a reduction reaction with SO₄^2-

 now you can start solving your half-reactions, keeping in mind that the total charge on either side of the equation must be equal (this is why you add electrons!!) in an "acidic environment" you can balance your reaction by adding H₂O  
 molecules on the side where you lack O and H⁺ on the side where you haven't got enough Hydrogen
in a "basic environment" do the same for O, but when you lack hydrogen you add H₂O on that side and OH^- on the opposite !
then you have to make sure you have an equal amount of electrons in both half-reactions by multiplying the (entire) half reaction by an integer.
afer that you can add the tho half-reactions...


but really, redox-reactions are very important, so if you really don't understand them don't hesitate and ask your teacher or buy a good textbook

[albicilla]

guess u allready knew it...good job


(damn  i'll have to learn to type faster)

hahahahaha

[udit]

no problem, your post gave some very useful stuff about oxidation numbers

[Borek]

before you can start solving redox-reactions you have to understand the concept of  oxidation numbers 'cause they are necessary to keep track of the electrons.

No. Oxidation numbers are arbitrary and are not necessary. That's and old and ongoing dispute between educators, whether to use them, or not. To balance reaction you can use half-reactions and don't bother with assigning oxidation numbers to individual atoms.

[xiankai]

hmm i never seen a reaction involving the decomposition of sulphate before... but reaction-wise i think u've balanced correctly

PS: use SO instead of So, both are different things

[albicilla]

dear Borek,

i believe you're referring to the "ion-electron" method  and yes you're right, it's easier to use in aqueous solutions; allthough it requires more "chemical insight" than the ox number -method. i therefor assumed it would be more appropriate to use the oxidation numbers.

greets

[Borek]

i believe you're referring to the "ion-electron" method  and yes you're right, it's easier to use in aqueous solutions; allthough it requires more "chemical insight" than the ox number -method. i therefor assumed it would be more appropriate to use the oxidation numbers.

At the same time oxidation numbers are in general not a thing that 'exists' - charge in the complex ions is usually not concentrated on one atom, but rather dispersed, in case of organic compounds instead of assigning oxidation numbers to covalently bonded carbon and hydrogen you may as well select them randomly.

Both methods can be applied in a mechanic way - take this, use this, add or substract to balance. No need for any understanding of what is really going on. Ox numbers are often easier, but they left wrong impression that they really exist.

But that's a philosophical discussion I just wanted to point out you were wrong stating

you have to understand the concept of  oxidation numbers

You don't have to

[sdekivit]

And one more thing, dont redox reactions need to have an e- somewhere? like for this reaction:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g)

I did the following:

Cu(s) + So₄^2-(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)

Cu(s) + So₄^2-(aq) + 4H⁺(aq) --> Cu²⁺(aq) + SO_2(g) + 2H₂O(l)

It's even and no electrons are needed to balance it? is this right or what?

Any help would be appreciated, I need to understand these questions.

it's easier to write down halfreaction. Then you'll see what actually happens in the reaction:

SO4(2-) --> SO2

SO4(2-) + 4 H(+) --> SO2 + 2 H2O

SO4(2-) + 4 H(+) + 2 e(-) --> SO2 + 2 H2O

[Bridget Jones]

Thank you - you explained it very well!