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Chemistry A level

Can someone explain this to me on how to work out the process? Thanks

99D68537-2905-4784-A4E0-D9EA095F266F.jpg.jpeg
Original post by AllGrade999999
Can someone explain this to me on how to work out the process? Thanks

99D68537-2905-4784-A4E0-D9EA095F266F.jpg.jpeg

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)
Reply 2
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)


Ok thanks so how would I write this first equation?
Original post by AllGrade999999
Ok thanks so how would I write this first equation?

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 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.
Reply 4
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 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.


So if it is negative, it doesn’t work

What is it, if it’s the other way round?
Original post by AllGrade999999
So if it is negative, it doesn’t work

What is it, if it’s the other way round?

Just reverse the equation and change the sign...
Reply 6
Original post by charco
Just reverse the equation and change the sign...


I’m confused

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