if a half equation is: Zn2+ + 2e- ----> Zn and it's E is -0.76 V would: Zn------> 2e- +Zn2+ also be -0.76V? or would it be 0.76V?
If you are talking about standard electrode potentials then they are equilibria expressed (by convention) as reductions.
Zn2+(aq) + 2e <==> Zn(s) ........... E = -0.76 V
The negative sign shows that there is a tendency for the equilibrium to move to the side of releasing electrons compared with the standard hydrogen electrode.
If you are talking about standard electrode potentials then they are equilibria expressed (by convention) as reductions.
Zn2+(aq) + 2e <==> Zn(s) ........... E = -0.76 V
The negative sign shows that there is a tendency for the equilibrium to move to the side of releasing electrons compared with the standard hydrogen electrode.
It is not a value for an equation.
If I balanced this half equation with another equation and E standard came out as positive, could I reverse the equation and e standard would be negative?
If I balanced this half equation with another equation and E standard came out as positive, could I reverse the equation and e standard would be negative?
In this case you are dealing with a difference between two Eº values and yes the positive from E(red) - E(ox) will give the direction of spontaneous change.
In this case you are dealing with a difference between two Eº values and yes the positive from E(red) - E(ox) will give the direction of spontaneous change.
Just to clarify if there was a reaction between potassium dichromate ( +1.33 V) and Chlorine ( +1.36)
How would you explain why the answer would be -0.03V? I know how to do it but don't know the reasoning behind it
Just to clarify if there was a reaction between potassium dichromate ( +1.33 V) and Chlorine ( +1.36)
How would you explain why the answer would be -0.03V? I know how to do it but don't know the reasoning behind it
As I said, in terms of standard reduction potentials, the E(cell) is given by E(red) - E(ox), where E(red) is the species being reduced and E(ox) is the species being oxidised.
Your two reagents will not react as they are both oxidising agents and can only be reduced.
As I said, in terms of standard reduction potentials, the E(cell) is given by E(red) - E(ox), where E(red) is the species being reduced and E(ox) is the species being oxidised.
Your two reagents will not react as they are both oxidising agents and can only be reduced.
According to the exam question, they do react if solid potassium dichromate is used and hot concentrated sulphuric acid is added?
An acidic solution of potassium manganate(VII) will liberate chlorine from dilute sodium chloride solution but an acidic solution of potassium dichromate(VI) will not. Solid potassium dichromate(VI) will liberate chlorine gas from concentrated hydrochloric acid
This does not say that dichromate ions are reacting with chlorine,it says that dichromate ions react with chloride ions (from hydrochloric acid,not sulfuric acid!!).
In this case the dichromate ions are being reduced and the chloride ions oxidised.
You get the value for E(cell) and find that it is negative,
However, E values only apply to standard conditions and the conditions given are anything but ...
An acidic solution of potassium manganate(VII) will liberate chlorine from dilute sodium chloride solution but an acidic solution of potassium dichromate(VI) will not. Solid potassium dichromate(VI) will liberate chlorine gas from concentrated hydrochloric acid
This does not say that dichromate ions are reacting with chlorine,it says that dichromate ions react with chloride ions (from hydrochloric acid,not sulfuric acid!!).
In this case the dichromate ions are being reduced and the chloride ions oxidised.
You get the value for E(cell) and find that it is negative,
However, E values only apply to standard conditions and the conditions given are anything but ...
Hence under these conditions there is reaction.
sorry about the confusion was tired
wait may you tell me the equation? because I thought you do the most postive minus the negative one to get the E value? so you would get +0.03? or
my equation was 2cr3+ + 3Cl2 + 7H20 ------> cr207- + 6cl- + 7H+ something like that
or has you say Er- Eox= so the reducing agent is on the left of the E cell diagram?
wait may you tell me the equation? because I thought you do the most postive minus the negative one to get the E value? so you would get +0.03? or
my equation was 2cr3+ + 3Cl2 + 7H20 ------> cr207- + 6cl- + 7H+ something like that
or has you say Er- Eox= so the reducing agent is on the left of the E cell diagram?
Hmm no. If your equation is 2Cr3++3Cl2+7H20−−−−−−>Cr2O7−+6Cl−+7H+
You need to consider the half reactions. From the equation you can see that Cr3+−−−>Cr2O7− which is an oxidation and Cl2−−−>2Cl− which is a reduction
So therefore you do ΔE∘=Echlorine−Echromium
To answer your question, you can't reverse the sign on a standard reduction potential, it's fundamentally wrong
Hmm no. If your equation is 2Cr3++3Cl2+7H20−−−−−−>Cr2O7−+6Cl−+7H+
You need to consider the half reactions. From the equation you can see that Cr3+−−−>Cr2O7− which is an oxidation and Cl2−−−>2Cl− which is a reduction
So therefore you do ΔE∘=Echlorine−Echromium
To answer your question, you can't reverse the sign on a standard reduction potential, it's fundamentally wrong
My chemistry teacher did chromium- Chlorine? so i'm confused