Change in standard reduction potential during redox

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rolark

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Okay, I'm stuck and have searched around the forums - can't find the answer I need. Hopefully someone here can help.

I need to utilize the following equation with a particular redox pair:

∆G = -👎(F)(∆E)

I'm wondering about ∆E, the change in standard reduction potential. To get that value, I know have to find the difference in standard reduction potential between the oxidized and reduced species (the former is 0.03 volts and the latter is -0.22 volts).

Here's my question: should I subtract the oxidized species from the reduced species or vice-versa? And why? The values are pretty substantially different (one leads to a positive ∆G, the other negative).

Thanks much!
 
Okay, I'm stuck and have searched around the forums - can't find the answer I need. Hopefully someone here can help.

I need to utilize the following equation with a particular redox pair:

∆G = -👎(F)(∆E)

I'm wondering about ∆E, the change in standard reduction potential. To get that value, I know have to find the difference in standard reduction potential between the oxidized and reduced species (the former is 0.03 volts and the latter is -0.22 volts).

Here's my question: should I subtract the oxidized species from the reduced species or vice-versa? And why? The values are pretty substantially different (one leads to a positive ∆G, the other negative).

Thanks much!

Hmm ... I think the main idea is that if the cell is Galvanic/Voltaic delta E = positive but if it is Electrolytic delta E = negative. Once you know this, and I am not sure whether they told you this in the question or not, - they should have though - you can easily figure out whether to subtract or add to get the desired value (sign).
 
Okay, I'm stuck and have searched around the forums - can't find the answer I need. Hopefully someone here can help.

I need to utilize the following equation with a particular redox pair:

∆G = -👎(F)(∆E)

I'm wondering about ∆E, the change in standard reduction potential. To get that value, I know have to find the difference in standard reduction potential between the oxidized and reduced species (the former is 0.03 volts and the latter is -0.22 volts).

Here's my question: should I subtract the oxidized species from the reduced species or vice-versa? And why? The values are pretty substantially different (one leads to a positive ∆G, the other negative).

Thanks much!

You gotta see if the equation given to you is spontaneous or not, that way you can do the subtraction correctly to end up with either a -G (spontaneous) or +G (nonsp.)
 
Okay, I just must be missing something here. If anyone has the Berkeley Review books and can look at their bio 2 book, this question is in the ETC section of metabolism. I'd appreciate it if someone could help me understand why the answer has a positive change in free energy when the question says free energy should decrease in the ETC.
 
Okay, I just must be missing something here. If anyone has the Berkeley Review books and can look at their bio 2 book, this question is in the ETC section of metabolism. I'd appreciate it if someone could help me understand why the answer has a positive change in free energy when the question says free energy should decrease in the ETC.

In this question, 2 electrons are involved, so n = 2.

In the reaction given, FAD is reduced to FADH2 (because it is gaining more bonds to Hydrogen). If FAD is reduced, then succinate has to oxidize to fumarate.
If something is being reduced, then electrons appear on the reactant side in the half reaction. So FAD + 2e --> FADH2 is correctly written and has an E of -0.22 Volts (since it is correctly written, you don't have to switch the sign).
For fumarate half raction, the given half reaction reduces fumarate to succinate. But we want the opposite; succinate --> fumarate + 2e
So you switch the sign and it now becomes -0.03
When you add both of the half reactions, total E becomes -0.25
And finally just plug it in from here:
G = -(2)(-0.25)(F)
G = +(0.50)(F) ... so the answer is B
 
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