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Oxidising Agents

Hi

How can you tell what the strength of an oxidising agent will be? For example, between KBrO or KBrO₃ how could you tell which would be the stronger oxidising agent? Many thanks for any help!

Reply 1

shengoc

12

hidden088

Hi

How can you tell what the strength of an oxidising agent will be? For example, between KBrO or KBrO₃ how could you tell which would be the stronger oxidising agent? Many thanks for any help!

BrO3- is a stronger oxidising agent than BrO-, as the first one has bromine in oxidation state +5 and the second one has Br in oxidation state +1.

These are quite unstable oxidation state for Br, hence it would want to get reduced. However, saying that, the abilities as an oxidising/reducing agent can be affected by pH, so the strength might vary in acidic/basic conditions.

Look at KMnO4 very powerful oxidising agent, Mn is in +7 - it wants to get reduced to Mn2+, its most stable oxidation state in acidic medium - cf Frost diagram.

OP

Ok thanks very much!

my penis is stronger :yep:

Reply 4

Waterstorm

16

The higher oxidation states are oxidising agents, and lower oxidation states are reducing agents.

Reply 5

hidden088

OP

2

Ok thanks, I have the following question from my college teacher:

Write the half-reactions and the balanced equation for the cell reaction for the following galvanic cell:

(a) Cu (s)│Cu2+ (aq)║Cu+ (aq)│Cu (s)

How will I go about this? Never seen anything like this before :s-smilie:

Don't even know what a galvanic cell is :P

Reply 6

icedsarcasm

6

You can use standard electrode potentials to work out how strong an oxidising/reducing agent something is. The higher the number, the stronger an oxidising agent it is - the lower, the stronger a reducing agent.

hidden088

Ok thanks, I have the following question from my college teacher:

Write the half-reactions and the balanced equation for the cell reaction for the following galvanic cell:

(a) Cu (s)│Cu2+ (aq)║Cu+ (aq)│Cu (s)

How will I go about this? Never seen anything like this before :s-smilie:

Don't even know what a galvanic cell is :P

Standard electrode potentials are:
Cu2+ (aq)/ Cu (s) = +0.34V
Cu+ (aq) / Cu (s) = +0.52V

In the cell diagram, they've been arranged into reduced / oxidised // oxidised / reduced, but it's important not to pay attention to that order when you're working out equations. Use the ones in the data you're given, which is always oxidised / reduced.

Now put the most negative one on the top.

<--------------
Cu2+ (aq)/ Cu (s)
Cu+ (aq) / Cu (s)
--------------->

The arrows show the 'anti clockwise rule' which is how you work out the direction of the reaction.

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

There are two electrons in the first equation and only one in the second, so to balance it you have to times everything in the second equation by two:

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

And then combine the half equations to get the full balanced equation:

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

Reply 7

shengoc

12

hidden088

Ok thanks, I have the following question from my college teacher:

Write the half-reactions and the balanced equation for the cell reaction for the following galvanic cell:

(a) Cu (s)│Cu2+ (aq)║Cu+ (aq)│Cu (s)

How will I go about this? Never seen anything like this before :s-smilie:

Don't even know what a galvanic cell is :P

Hi there, dont even worry about a galvanic cell, the name "cell" is the keyword here, just do with what you would do with conventional cell reaction.

From formal cell reaction like the one above, always, and i mean always write down the half cell equations, meaning the one with electrons involved.

(1) Cu2+ (aq) + 2e ----> Cu (s) E = +0.34V
(2) Cu+ (aq) + e -----> Cu (s) E = +0.52V

Then from the cell above, it is always oxidation on the LHS and reduction on the RHS,

Therefore, in this case, Cu+ is reduced to Cu, so
E(cell) = RHE - LHE = +0.52 - (+0.34) = +0.18 V

Don't worry about the number of electrons, the E(cell) value is independent of the number of electrons - it would be more important if you go into electrochemistry in detail.

And one more thing, don't worry about changing the sign when doing this arithmetic, it is always RHE - LHE reduction potential values.

However, for the complete balanced equation, the number of electrons matter as you would need to subtract the correct multiples of electrons, so that they dont appear in your balanced eqn, in this case, it would be
(2) x 2 - (1) in order to get rid of the number of electrons, and also making sure that Cu+ is reduced to Cu in that process, you get

2Cu+(aq) -----> Cu(s) + Cu2+(aq) E(cell) = +0.18 V

Eqn is balanced, and the E(cell) value you calculated is for this forward process, ie reduction of Cu+ to Cu, but of course this is a disproportionation reaction.

Reply 8

hidden088

OP

2

Ok thanks for the nice explanation!

I also have this:

Ag (s),AgI (s)│I^- (aq)║Cl^- (aq)│AgCl (s),Ag (s)

If I understand the convention correctly, this means

The cathode is Ag, AgI + e^- -> I^- + Ag

Then at the anode is again Ag, Cl^- + Ag -> AgCl + e^-

So the overall reaction would be:

AgI + Cl^- -> AgCl + I^-?

Reply 9

hidden088

OP

2

I also thought that the anode/oxidation always is on the LHS, in this case it doesn't appear to be....

Reply 10

shengoc

12

hidden088

I also thought that the anode/oxidation always is on the LHS, in this case it doesn't appear to be....

I always try and write out the half cell equation using the standard reduction potential style, ie reduction, and then the E values next to it.

Then, line them up so that they fit the reaction that you have, ie oxidation on the left and reduction on the right.

Ag (s),AgI (s)│I- (aq)║Cl- (aq)│AgCl (s),Ag (s)

So AgCl(s) + e- -----> Ag(s) + Cl-(aq) (ie Ag+ is reduced)
and also, AgI(s) + e- ----> Ag(s) + I-(aq) (This is the standard reduction potential)

These are lined up these ways as they would appear in your data book, or equivalent.

So we want the overall reaction with AgCl being reduced to Ag, so it is eqn 1 - eqn 2(same number of electrons - less worry).

Hence, AgCl(s) + I-(aq) ----> AgI(s) + Cl-(aq)

This is the reaction as given by the formal cell above, and i know you must be thinking that wait, hang on, this reaction doesn't happen. And yes it doesn't because E(cell) would be negative and delta G = -nFE where E = E(cell), if E is negative, delta G is positive, so this reaction is not spontaneous.

However, you are required to write it in this form as it is a convention, RHE - LHE.

Hope that helps you understand better.