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Electrochemistry

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(edited 6 years ago)
Reply 1
Avatar for Yme2day
Yme2day
12
•Vanadium species at the end of the species= V^3+
Since the electropotential value is less than the Magnate one( -0.26 in comparison to +1.52) Meaning Vanadium will undergo oxidation. Therefore from V2+ to V3+

•Oxidation state of Vanadium in final species= (|||)
Since itll be 3+

• half equation of forming V3+ = V2+ --> V3+ + e-
You've got to oxidise V2+
Original post by Qer


And what are your thoughts so far?
Reply 3
Avatar for Yme2day
Yme2day
12
If you have the markscheme let me know if I'm right btw :smile:
Reply 4
Avatar for Yme2day
Yme2day
12
Original post by charco
And what are your thoughts so far?


Do you think my answers are correct?
Original post by Yme2day
Do you think my answers are correct?


No, they are not.
Reply 6
Avatar for Qer
Qer
OP
16
Original post by Yme2day
•Vanadium species at the end of the species= V^3+
Since the electropotential value is less than the Magnate one( -0.26 in comparison to +1.52) Meaning Vanadium will undergo oxidation. Therefore from V2+ to V3+

•Oxidation state of Vanadium in final species= (|||)
Since itll be 3+

• half equation of forming V3+ = V2+ --> V3+ + e-
You've got to oxidise V2+



You have done exactly the same thing that I did but that's not the right answer.
Original post by Qer
You have done exactly the same thing that I did but that's not the right answer.


With redox reactions you can decide whether or not a reaction proceeds by looking at the cell potential of the proposed reaction.

The cell potential is obtained using the formula E(cell) = E(red) - E(ox). If the value of E(cell) is positive, the reaction is spontaneous. If it is between 0 and +0.3 it is likely to arrive at equilibrium. If it is greater than +0.3V then the reaction is likely to go to completion.

As the manganate(VII) ion is the oxidising agent it is the species that forms the reduced half of the equation. i.e. E(red)

Now you use the values for the vanadium half-equations to see which ones will be oxidised by the manganate(VII) ion (in acidic solution)
Reply 8
Avatar for Qer
Qer
OP
16
Original post by charco
With redox reactions you can decide whether or not a reaction proceeds by looking at the cell potential of the proposed reaction.

The cell potential is obtained using the formula E(cell) = E(red) - E(ox). If the value of E(cell) is positive, the reaction is spontaneous. If it is between 0 and +0.3 it is likely to arrive at equilibrium. If it is greater than +0.3V then the reaction is likely to go to completion.

As the manganate(VII) ion is the oxidising agent it is the species that forms the reduced half of the equation. i.e. E(red)

Now you use the values for the vanadium half-equations to see which ones will be oxidised by the manganate(VII) ion (in acidic solution)


1.52-0.34=1.18
1.52-1.00=0.52
These both reaction is spontaneous then
Original post by Qer
1.52-0.34=1.18
1.52-1.00=0.52
These both reaction is spontaneous then


yes
Reply 10
Avatar for KESbian
KESbian
9
VO2^+ Will be the Vanadium species present, with oxidation state +5.
This is because MnO4- is strong enough an oxidising agent to oxidise Vanadium to its +5 oxidation state. We can tell this by looking at the E(red) values for the V species. None of the Vanadium E(red) values are more positive than E(red) for MnO4-.
Original post by charco
With redox reactions you can decide whether or not a reaction proceeds by looking at the cell potential of the proposed reaction.

The cell potential is obtained using the formula E(cell) = E(red) - E(ox). If the value of E(cell) is positive, the reaction is spontaneous. If it is between 0 and +0.3 it is likely to arrive at equilibrium. If it is greater than +0.3V then the reaction is likely to go to completion.

As the manganate(VII) ion is the oxidising agent it is the species that forms the reduced half of the equation. i.e. E(red)

Now you use the values for the vanadium half-equations to see which ones will be oxidised by the manganate(VII) ion (in acidic solution)


I don't quite understand... So if the Electropotential value is E of red - E of ox then that means:

1.52 - -0.26 = 1.52 + 0.26 = +1.78 V.

So this is greater than +0.3V meaning that the reaction will go to completion....i don't understand how this helps us get the answer...
Original post by Yme2day
I don't quite understand... So if the Electropotential value is E of red - E of ox then that means:

1.52 - -0.26 = 1.52 + 0.26 = +1.78 V.

So this is greater than +0.3V meaning that the reaction will go to completion....i don't understand how this helps us get the answer...


You will see from the table that manganate(VII) ions have a more positive potential than all of the vanadium half-equations. It can consequently oxidise all of the vanadium species up to V(V)
Original post by charco
You will see from the table that manganate(VII) ions have a more positive potential than all of the vanadium half-equations. It can consequently oxidise all of the vanadium species up to V(V)


I think I understand but arent these separate half equations? Or are the successive equations of the same original vanadium that started out at V2+?
Original post by Yme2day
I think I understand but arent these separate half equations? Or are the successive equations of the same original vanadium that started out at V2+?


They are all separate half-equations,

but if the first vanadium half-equation makes a compound that can start another half-equation then that's what happens.
Original post by charco
They are all separate half-equations,

but if the first vanadium half-equation makes a compound that can start another half-equation then that's what happens.


Aha so this means that hypothetically, for example, if there was another half equation of Vanadium with an electropotential value of +1.41 V for instance then the Magnate will still cause it to be oxidised and the final product of that half equation containing Vanadium would be the answer instead?
Original post by Yme2day
Aha so this means that hypothetically, for example, if there was another half equation of Vanadium with an electropotential value of +1.41 V for instance then the Magnate will still cause it to be oxidised and the final product of that half equation containing Vanadium would be the answer instead?


yes
Original post by charco
yes


You've been great help, thanks for the clarification!

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