GCHEM: reduction potentials

[PreDental88]

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I don't know why I'm having such a hard time with calculating the standard reduction potential.

I understand that the species with a higher reduction potential is more likely to be reduced, and the species with a lower reduction potential is more likely to be oxidized.

For instance, why is this not the case in this scenario:

EG: Given that the standard reduction potentials for Sm3+ and [RhCl6]3- are -2.41V and +0.44 V, respectively. Calculate the EMF of the following reaction:

Sm3+ + Rh + 6Cl- --> [Rhcl6]3- + Sm

It says that Sm3+ is being reduced. This confused me because Sm3+ has the lower reduction potential. So this made me think that the species with the higher magnitude reduction potential was reduced, but another problem proved my theory wrong.

If anyone can help clarify this concept, I would greatly appreciate it!

Thanks

 

[UndergradGuy7]

In the reaction
Sm3+ + Rh + 6Cl- --> [Rhcl6]3- + Sm

Sm3+ is being reduced.

What you are thinking is that the molecule with the more positive V will be reduced and the other one will be oxidized. Like in Sm3+ and [RhCl6]3- are -2.41V and +0.44 V, respectively. If the reaction was spontaneous and was written in the spontaneous way then the molecule with the more positive V would be reduced, which would be [RhCl6]3- because it is 0.44V. Sm3+ would be oxidized because it is -2.41V.

In the problem you posted though. They gave you an equation. From the equation you have to tell which is being oxidized and reduced, so you don't look at which is bigger V or not. The problem did not say the reaction is spontaneous, so the molecule with the most positive V does not have to be reduced. If it is non-spontaneous and we add electricity to it, then the molecule with the smaller V can be reduced.

 

[PreDental88]

In the reaction
Sm3+ + Rh + 6Cl- --> [Rhcl6]3- + Sm

Sm3+ is being reduced.

What you are thinking is that the molecule with the more positive V will be reduced and the other one will be oxidized. Like in Sm3+ and [RhCl6]3- are -2.41V and +0.44 V, respectively. If the reaction was spontaneous and was written in the spontaneous way then the molecule with the more positive V would be reduced, which would be [RhCl6]3- because it is 0.44V. Sm3+ would be oxidized because it is -2.41V.

In the problem you posted though. They gave you an equation. From the equation you have to tell which is being oxidized and reduced, so you don't look at which is bigger V or not. The problem did not say the reaction is spontaneous, so the molecule with the most positive V does not have to be reduced. If it is non-spontaneous and we add electricity to it, then the molecule with the smaller V can be reduced.

That makes perfect sense, you are a lifesaver!

 

[rnbaxter]

So EMF should be -2.41-.44 right, therefore nonspontaneous?

 

[UndergradGuy7]

So EMF should be -2.41-.44 right, therefore nonspontaneous?

Yes for the equation he wrote. It is non-spontaneous because E is negative.