Electromotive force

[Cinnabans]

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Sm3+ + Rh + 6Cl- --> [RhCl6]3- + Sm
The standard reduction potentials for Sm3+ and [RhCl6]3- are -2.41 V and +0.44V respectively, calc. EMF of above rxn.

Answer: -2.85 V and the cell is electrolytic (i understand that part) but it also says the rxn would proceed spontaneously to the left, so Sm would be oxidized while [RhCl6]3- would be reduced.

Why spontaneously to the left? I thought electrolytic cells were nonspontaneous. What am I missing???

 

[PreDent2009]

Electrolytic cells are not spontaneous, but the opposite of the rxn, which proceeds to the left is spontaneous. Any reaction that is non-favorable in one direction, will be favorable in the opposite direction.

 

[hoss19]

a mathematical way to look at the rxn would be to consider the rxn in reverse (rxn'). "flip" the standard potentials. therefore, rxn'

[RhCl6]3- + Sm --> Sm3+ Rh + 6Cl-

oxidixed: Sm --> Sm3+ = - -2.41 V = +2.41 and
reduced: [RhCl6]3- --> Rh + 6Cl- = -.44V
Ecell = Eox +Ered = +2.85 V

thus in the forward direction for rxn' it is spontaneous. this means if the conditions were set up properly (ie 1/2 cells per standard galvanic cell protocol) the reaction would proceed spontaneously. the answer is not saying that the reaction proceeds spontaneously in the reverse direction within the elctrolyic cell - this is just not posible

 

[ad19]

sorry I figured out my question

 

[IWantSmile]

[RhCl6]3- + Sm --> Sm3+ Rh + 6Cl-

oxidixed: Sm --> Sm3+ = - -2.41 V = +2.41 and
reduced: [RhCl6]3- --> Rh + 6Cl- = -.44V
Ecell = Eox +Ered = +2.85 V

Sorry, I am getting confused in calculation. Would that be Ecell= Eox +Ered = +2.01V ???

 

[Sea of ASH]

n/a

 

[Sea of ASH]

Sm3+ + Rh + 6Cl- --> [RhCl6]3- + Sm
The standard reduction potentials for Sm3+ and [RhCl6]3- are -2.41 V and +0.44V respectively, calc. EMF of above rxn.

Answer: -2.85 V and the cell is electrolytic (i understand that part) but it also says the rxn would proceed spontaneously to the left, so Sm would be oxidized while [RhCl6]3- would be reduced.

Why spontaneously to the left? I thought electrolytic cells were nonspontaneous. What am I missing???

ya its non spontaneous in the forward reaction but in the reverse its spontaneous.

 

[MoooShuuu]

given values I thought that the most postitive value gets reduced and most negative is oxidized. Then use Emf= Eox+Ered.

Or do you always flip them according to the formula?😕

 

[Sea of ASH]

given values I thought that the most postitive value gets reduced and most negative is oxidized. Then use Emf= Eox+Ered.

Or do you always flip them according to the formula?😕

well it depends on what they are asking. in this case the question specifically asked what is the emf of this reaction:
Sm3+ + Rh + 6Cl- --> [RhCl6]3- + Sm

 

[MoooShuuu]

Sm3+ + Rh + 6Cl- --> [RhCl6]3- + Sm
The standard reduction potentials for Sm3+ and [RhCl6]3- are -2.41 V and +0.44V respectively, calc. EMF of above rxn.

Answer: -2.85 V

I'm very confused here. When I look at this type of problem, I see that RhCl6 has the most positive value of .44V therefore it is going to be reduced. So then Sm3+ has to be reversed to +2.41. Then add and I get 2.85. Obviously this is wrong, what should I be doing differently.

 

[chessxwizard]

You are over-thinking this one. Just look at the equation. What gets reduced? Sm! What gets oxidized? Rh!

Therefore, use the reduction potential of Sm and the oxidation potential of Rh!

-2.41 (reduction potential of Sm) + -0.44 (oxidation potential of Rh -- the reverse of the reduction potential) = -2.85 V

Since it's negative, the cell must be electrolytic in the direction written. However, since the reverse reaction has a positive potential, it will be spontaneous.

 

[MoooShuuu]

Thanks Chessxwizard...

So how do I know when to use this method vs. looking at the most positive E' and then taking the opposite of the other value?

For example:
(Equation 1) Co3+( aq ) + e ---- Co2+ (aq) E= 1.82 V

(Equation 2 ) Na+ (aq)+ e ----- Na (s) E= -2.71V

1.82 is the most positive thus will be reduced. The other Eq gets flipped, +2.71. Then add, 1.82 + 2.71= 4.53V

haha I guess I'm just confusing myself more and more...

 

[chessxwizard]

Yes, if you're given a problem like that, you have to predict which is oxidized and reduced based on the standard reduction potentials. Whichever has the higher potential will be reduced, the other will be oxidized. However, this problem tells you which is oxidized and reduced based on the chemical equation, so no guessing is involved.