Back
to top
Chemistry A level
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
Can someone explain this to me on how to work out the process? Thanks
![Name: 99D68537-2905-4784-A4E0-D9EA095F266F.jpg.jpeg
Views: 13
Size: 36.3 KB]()
0
reply
Report
#2
(Original post by AllGrade999999)
Can someone explain this to me on how to work out the process? Thanks
![Name: 99D68537-2905-4784-A4E0-D9EA095F266F.jpg.jpeg
Views: 13
Size: 36.3 KB]()
Can someone explain this to me on how to work out the process? Thanks
Hence, the values are referred to as standard reduction potentials.
The equilibria of course can go either way, but only if there is an appropriate species that can either receive or donate the electrons involved.
The spontaneity of a possible (hypothetical) reaction depends on the cell potential once a reducing agent is put together with an oxidising agent.
The reducing agents appear on the right hand side and the oxidising agents on the left hand side.
The cell potential is calculated as E(cell) = E(reduced species) - E(oxidised species)
If E(cell) is positive the hypothetical reaction can happen (is spontaneous), although it is possible that an equilibrium may result
If E(cell) is positive and greater than 0.3V the reaction is likely to go all the way
If E(cell) is negative then the proposed reaction is not possible (under standard conditions)
0
reply
(Original post by charco)
By convention, all of the equilibria are shown as reductions.
Hence, the values are referred to as standard reduction potentials.
The equilibria of course can go either way, but only if there is an appropriate species that can either receive or donate the electrons involved.
The spontaneity of a possible (hypothetical) reaction depends on the cell potential once a reducing agent is put together with an oxidising agent.
The reducing agents appear on the right hand side and the oxidising agents on the left hand side.
The cell potential is calculated as E(cell) = E(reduced species) - E(oxidised species)
If E(cell) is positive the hypothetical reaction can happen (is spontaneous), although it is possible that an equilibrium may result
If E(cell) is positive and greater than 0.3V the reaction is likely to go all the way
If E(cell) is negative then the proposed reaction is not possible (under standard conditions)
By convention, all of the equilibria are shown as reductions.
Hence, the values are referred to as standard reduction potentials.
The equilibria of course can go either way, but only if there is an appropriate species that can either receive or donate the electrons involved.
The spontaneity of a possible (hypothetical) reaction depends on the cell potential once a reducing agent is put together with an oxidising agent.
The reducing agents appear on the right hand side and the oxidising agents on the left hand side.
The cell potential is calculated as E(cell) = E(reduced species) - E(oxidised species)
If E(cell) is positive the hypothetical reaction can happen (is spontaneous), although it is possible that an equilibrium may result
If E(cell) is positive and greater than 0.3V the reaction is likely to go all the way
If E(cell) is negative then the proposed reaction is not possible (under standard conditions)
0
reply
Report
#4
(Original post by AllGrade999999)
Ok thanks so how would I write this first equation?
Ok thanks so how would I write this first equation?
example:
Zn2+ + 2Cr2+ ==> Zn + 2Cr3+
For this equation to "work" the zinc gets reduced and the chromium(II) gets oxidised. So fill in the Eº values for the equation:
E(cell) = E(red) - E(ox) = -0.76 - -0.41 = -0.33V
The answer is negative, so the equation as written is not feasible (non-spontaneous).
It would, however work the other way around.
0
reply
(Original post by charco)
Choose 1 reducing agent (RHS) and one oxidising agent (LHS) and write a hypothetical equation:
example:
Zn2+ + 2Cr2+ ==> Zn + 2Cr3+
For this equation to "work" the zinc gets reduced and the chromium(II) gets oxidised. So fill in the Eº values for the equation:
E(cell) = E(red) - E(ox) = -0.76 - -0.41 = -0.33V
The answer is negative, so the equation as written is not feasible (non-spontaneous).
It would, however work the other way around.
Choose 1 reducing agent (RHS) and one oxidising agent (LHS) and write a hypothetical equation:
example:
Zn2+ + 2Cr2+ ==> Zn + 2Cr3+
For this equation to "work" the zinc gets reduced and the chromium(II) gets oxidised. So fill in the Eº values for the equation:
E(cell) = E(red) - E(ox) = -0.76 - -0.41 = -0.33V
The answer is negative, so the equation as written is not feasible (non-spontaneous).
It would, however work the other way around.
What is it, if it’s the other way round?
0
reply
Report
#6
(Original post by AllGrade999999)
So if it is negative, it doesn’t work
What is it, if it’s the other way round?
So if it is negative, it doesn’t work
What is it, if it’s the other way round?
0
reply
(Original post by charco)
Just reverse the equation and change the Eº sign...
Just reverse the equation and change the Eº sign...
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Have your say on 2021 assessmentOfqual and DfE want students' thoughts
- Who has received uni offers?
- Starting my new job - AMA!
- Win a year of Headspace
- Private candidates support thread