standard electrode potential questions

[nicz888]

---

Members do not see this ad.

Here is the complete question,
Based on the following standard electrode potentials for the two half rxns, what is the standard electrode potential for the rxn

i have a question on standard electrode potentials for the two half rxn.
from cliffy DAT book.

here is my take on those
for Q2,
Ag gain electron so, it is the E(reduction), change sign for I since it is the E(oxidation).
so 0.7-0.5 = 0.2

which is the right answer

but for Q1

Sn gain electron so, it is the E(reduction), change the sign for I since it is the E(oxidation).
so i did -0.2-0.5 = -0.7

but the answer said it is +0.7

is my reasoning wrong?

 

[jinlighter]

Ill give it a shot:

From the cell diagrams, I don't see an ionic medium ( ex. (xM of SO4-) or something) for either one of these cells. Therefore, I have to assume it is in an aqueous medium. If that is true, then I can not go by what the cell diagram says is a reduction or an oxidation ( cath or anode) because of water's interference with these reaction states. As such, I would have to look at the actual reduction potentials to determine what is reduced and what is oxidized. From my experience, the reduction potential with the largest number is the cathode(or reduced species) and the smaller is the anode( oxidated species).

This way when you calculate Eocell via EoCell= EoCath+EoAnode( switching the sign on EoAnode ofcourse), for Q1 and Q2 you come up with:
Q1=0.5+(0.2)=0.7
and
Q2= 0.7+(-0.5)=0.2

I hope this helps.

 

[UCB05]

i have a question on standard electrode potentials for the two half rxn.
from cliffy DAT book.

here is my take on those
for Q2,
Ag gain electron so, it is the E(reduction), change sign for I since it is the E(oxidation).
so 0.7-0.5 = 0.2

which is the right answer

but for Q1

Sn gain electron so, it is the E(reduction), change the sign for I since it is the E(oxidation).
so i did -0.2-0.5 = -0.7

but the answer said it is +0.7

is my reasoning wrong?

Given their representation of the cell in the convention (Anode||Cathode) it should be -0.7V.

 

[nicz888]

Ill give it a shot:

From the cell diagrams, I don't see an ionic medium ( ex. (xM of SO4-) or something) for either one of these cells. Therefore, I have to assume it is in an aqueous medium. If that is true, then I can not go by what the cell diagram says is a reduction or an oxidation ( cath or anode) because of water's interference with these reaction states. As such, I would have to look at the actual reduction potentials to determine what is reduced and what is oxidized. From my experience, the reduction potential with the largest number is the cathode(or reduced species) and the smaller is the anode( oxidated species).

This way when you calculate Eocell via EoCell= EoCath+EoAnode( switching the sign on EoAnode ofcourse), for Q1 and Q2 you come up with:
Q1=0.5+(0.2)=0.7
and
Q2= 0.7+(-0.5)=0.2

I hope this helps.

but for Q1 Sn = -0.2 is the E(cath) so i would not switch the sign right? i would only switch E(anode) which in Q1 is 0.5. so...

i am confuse

 

[GiTsticker]

So assuming these are galvanic cells, the first step is to determine which half reaction has the highest positive reduction potential. This one will stay positive. (Unless they are both negative, then you choose the smallest negative one). Switch the other one and add them together.
So for Q2 .5 is greater than -.2 so you switch the -.2 to .2 and add it to .5 to get .7

 

[nze82]

So assuming these are galvanic cells, the first step is to determine which half reaction has the highest positive reduction potential. This one will stay positive. (Unless they are both negative, then you choose the smallest negative one). Switch the other one and add them together.
So for Q2 .5 is greater than -.2 so you switch the -.2 to .2 and add it to .5 to get .7

I don't agree!
E = +0.5v is the potential for the reduction reaction. However, when you look at the cell, I2 is designated as the anode, so it must be oxidized (You can't ignore how the cell is setup and start choosing your own anodes and cathodes). Therefore, you have to flip the reaction associated with this part of the cell, which will in turn flip the sign of E and change it to E = -0.5.

 

[Shinpe]

I don't agree!
E = +0.5v is the potential for the reduction reaction. However, when you look at the cell, I2 is designated as the anode, so it must be oxidized (You can't ignore how the cell is setup and start choosing your own anodes and cathodes). Therefore, you have to flip the reaction associated with this part of the cell, which will in turn flip the sign of E and change it to E = -0.5.

I somewhat agree. If the actual question was available, it would be more clear. The way it is represented, it cannot be a galvonic cell (where E would be +0.7). Right now it's an electrolytic cell with (E=-0.7), so unless the question told you to find E if those reactions were to be used in a galvonic cell, then the answer should be -0.7.

 

[GiTsticker]

That's right, we need to know if it's galvanic or electrolytic. The problem probably says electrolytic and that's why the given answer is +.7.
You are right nze lol I ignored the cell setup. lol good thing I don't have to know it anymore...EVER

 

[Shinpe]

That's right, we need to know if it's galvanic or electrolytic. The problem probably says electrolytic and that's why the given answer is +.7.
You are right nze lol I ignored the cell setup. lol good thing I don't have to know it anymore...EVER

Wait, so you're saying that an E = -0.7 for galvonic means +0.7 for electrolytic?
Does anyone know if that's right or wrong?
I thought you'd just say that the E is -0.7, and the fact that it is negative shows that it doesn't go in a galvonic cell or it is electrolytic.
I personally don't think saying it's -0.7 for galvonic and +0.7 for electrolytic is right. But I'm not always right either lol

 

[nze82]

Wait, so you're saying that an E = -0.7 for galvonic means +0.7 for electrolytic?
Does anyone know if that's right or wrong?
I thought you'd just say that the E is -0.7, and the fact that it is negative shows that it doesn't go in a galvonic cell or it is electrolytic.
I personally don't think saying it's -0.7 for galvonic and +0.7 for electrolytic is right. But I'm not always right either lol

When E>0, then the reaction occurs spontaneously, and the cell is an GALVANIC cell. When E<0, then the reaction doesn't occur spontaneously, so you need a battery to run the reaction, and that defines a ELECTROLYTIC cell.

 

[nicz888]

see my original post, i edited it.

 

[UCB05]

When E>0, then the reaction occurs spontaneously, and the cell is an electrolytic cell. When E<0, then the reaction doesn't occur spontaneously, so you need a battery to run the reaction, and that defines a galvanic cell.

errr, isnt that the other way around? A galvanic cell is the type with E>0 and spontaneous. An Electrolytic cell requires electrical input.

 

[GiTsticker]

Ya E>0 is galvanic.
If the question doesn't specify spontaneous/nonspontaneous galvanic/electrolitic or 0>E/>0 then there isn't enough info. The given answer though, implies nonspontaneous/electrolytic/E<0. This is because reduction is occuring at the annode (electrolytic).

 

[nze82]

errr, isnt that the other way around? A galvanic cell is the type with E>0 and spontaneous. An Electrolytic cell requires electrical input.

That could very well be the case, when I start posting stuff after 12:00AM.