Getting really confused with this redox electrochemical question Watch

Stormragexox
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Hey guys i an so lost my teacher told me that the electrode potential is negative for the third reaction but I am really confused to why?
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Stormragexox
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Stormragexox
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(Original post by Stormragexox)
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hassan1236945
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By third reaction do you mean the one where fe3+ is getting reduced?
(Original post by Stormragexox)
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BobbJo
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Zn + Fe2+ -> Zn2+ + Fe
Ecell = -0.44 + 0.76 = +0.32 V > 0 hence feasible

Zn + 2 Fe3+ -> Zn2+ + 2Fe2+
Ecell = 0.77 - (0.76) = +1.53V > 0 hence feasible

The electrode potential for Fe3+/Fe2+ is positive and equal to + 0.77 V, as given in the question

but your second line should be
Fe -> Fe2+ + 2e- E = + 0.44 V
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Stormragexox
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thx for replying XD My doubt when I was trying to attempt this question was why is Fe being oxidised to fe2+ initially I thought that Zn acts as a reducing agent therefore that would be the obly reaction where the electrode potential sign would be reversed and all the other reactions would have the same electrode potential with the same sign
(Original post by BobbJo)
Zn + Fe2+ -> Zn2+ + Fe
Ecell = -0.44 + 0.76 = +0.32 V > 0 hence feasible

Zn + 2 Fe3+ -> Zn2+ + 2Fe2+
Ecell = 0.77 - (0.76) = +1.53V > 0 hence feasible

The electrode potential for Fe3+/Fe2+ is positive and equal to + 0.77 V, as given in the question

but your second line should be
Fe -> Fe2+ + 2e- E = + 0.44 V
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BobbJo
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(Original post by Stormragexox)
thx for replying XD My doubt when I was trying to attempt this question was why is Fe being oxidised to fe2+ initially I thought that Zn acts as a reducing agent therefore that would be the obly reaction where the electrode potential sign would be reversed and all the other reactions would have the same electrode potential with the same sign
The question is asking for the reaction of Zn with Fe2+ and with Fe3+.

Zn does not react with Fe.
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Stormragexox
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(Original post by BobbJo)
The question is asking for the reaction of Zn with Fe2+ and with Fe3+.

Zn does not react with Fe.
So Fe2+ reacts with 2e- to form Fe then so we don't have to change the electrode potential sign of the reaction? Because initially I thought Fe3+ is being reduced and then fe2+ is being reduced( as the question states fe as to be a product and Zn is being oxidised to zn2+so that is the only reaction where the sign changes
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BobbJo
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(Original post by Stormragexox)
because initially I thought Fe3+ is being reduced and then fe2+ is being reduced( as the question states fe as to be a product and Zn is being oxidised to zn2+so that is the only reaction where the sign changes
Yes, you thought correctly. Fe3+ reacts with Zn to form Fe2+ and Zn2+. Fe2+ reacts with Zn to form Fe and Zn2+.

For Zn -> Zn2+ + 2e- E = +0.76 V, you reverse the sign here as the equation is reversed

So Fe2+ reacts with 2e- to form Fe then so we don't have to change the electrode potential sign of the reaction?
Fe2+ + 2e- -> Fe E = -0.44 V, you don't change the sign of the electrode potential
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Stormragexox
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(Original post by BobbJo)
Zn + Fe2+ -> Zn2+ + Fe
Ecell = -0.44 + 0.76 = +0.32 V > 0 hence feasible

Zn + 2 Fe3+ -> Zn2+ + 2Fe2+
Ecell = 0.77 - (0.76) = +1.53V > 0 hence feasible

The electrode potential for Fe3+/Fe2+ is positive and equal to + 0.77 V, as given in the question

but your second line should be
Fe -> Fe2+ + 2e- E = + 0.44 V

(Original post by BobbJo)
Yes, you thought correctly. Fe3+ reacts with Zn to form Fe2+ and Zn2+. Fe2+ reacts with Zn to form Fe and Zn2+.

For Zn -> Zn2+ + 2e- E = +0.76 V, you reverse the sign here as the equation is reversed


Fe2+ + 2e- -> Fe E = -0.44 V, you don't change the sign of the electrode potential
So the ecell value should be calculated separately for each reaction is it possible to calculate the overall ecell value as well because thats what I originally attempted to do but apparently it was wrong
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BobbJo
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(Original post by Stormragexox)
So the ecell value should be calculated separately for each reaction is it possible to calculate the overall ecell value as well because thats what I originally attempted to do but apparently it was wrong
You can't.

Fe3+ + 3e- <-> Fe E = not given in the question = -0.04 V
Zn2+ + 2e- <-> Zn E = -0.76 V
Ecell = -0.04 - (-0.76) = +0.72 V
It is impossible to calculate this since E(Fe3+/Fe) is not given.
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Stormragexox
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(Original post by BobbJo)
You can't.

Fe3+ + 3e- <-> Fe E = not given in the question = -0.04 V
Zn2+ + 2e- <-> Zn E = -0.76 V
Ecell = -0.04 - (-0.76) = +0.72 V
It is impossible to calculate this since E(Fe3+/Fe) is not given.
That makes sense now thank you so much
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Stormragexox
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(Original post by BobbJo)
You can't.

Fe3+ + 3e- <-> Fe E = not given in the question = -0.04 V
Zn2+ + 2e- <-> Zn E = -0.76 V
Ecell = -0.04 - (-0.76) = +0.72 V
It is impossible to calculate this since E(Fe3+/Fe) is not given.
Hey there XD I have come across another question recently and only got one of the reactions correct
https://www.ocr.org.uk/Images/175443...d-elements.pdf
For question 7 do all the reactions have to be feasible?
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BobbJo
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(Original post by Stormragexox)
Hey there XD I have come across another question recently and only got one of the reactions correct
https://www.ocr.org.uk/Images/175443...d-elements.pdf
For question 7 do all the reactions have to be feasible?
Part (c)(i)? It says to predict " three reactions that might be feasible". You have to give 3 reactions for which Ecell > 0
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Stormragexox
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(Original post by BobbJo)
Part (c)(i)? It says to predict " three reactions that might be feasible". You have to give 3 reactions for which Ecell > 0
Thank you for replying so if that were the case I only figured out the feasible reaction between 4 and 5 I am struggling to figure out anyother feasible reactions
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BobbJo
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3: Al3+ + 3e- -> Al E = - 1.66V
4: Fe3+ + e- -> Fe2+ E = +0.77V
5: I2+ 2e- -> 2I- E = +0.54V

give any reactions for which Ecell is positive

such as
2Fe3+ + 2I- -> 2Fe2+ + I2
3Fe3+ + Al -> 3Fe2+ + Al3+
2Al + 3I2 -> 2Al3+ + 6I-
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Stormragexox
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(Original post by BobbJo)
3: Al3+ + 3e- -> Al E = - 1.66V
4: Fe3+ + e- -> Fe2+ E = +0.77V
5: I2+ 2e- -> 2I- E = +0.54V

give any reactions for which Ecell is positive

such as
2Fe3+ + 2I- -> 2Fe2+ + I2
3Fe3+ + Al -> 3Fe2+ + Al3+
2Al + 3I2 -> 2Al3+ + 6I-
But i thought 3 and 4 won't be feasible as the ecell is -ve?
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BobbJo
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(Original post by Stormragexox)
But i thought 3 and 4 won't be feasible as the ecell is -ve?
What you are referring to by 3 and 4?
3: Al3+ + 3e- -> Al E = - 1.66V
4: Fe3+ + e- -> Fe2+ E = +0.77V
This?
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Stormragexox
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(Original post by BobbJo)
What you are referring to by 3 and 4?
3: Al3+ + 3e- -> Al E = - 1.66V
4: Fe3+ + e- -> Fe2+ E = +0.77V
This?
yes
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BobbJo
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(Original post by Stormragexox)
yes
You need to use 2 equations from 3,4,5 to write an equation for the overall reaction whose Ecell is positive.
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