E of the Rxn???

[G1SG2]

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I have a question that is on the back of a Gold Standard MCAT flashcard:

What is the E of the rxn if Ered = 0.34 and Eox = -0.76?

I thought I would just add them together like I always do, since I am directly given Eox and get -0.42.

However, the explanation says: E = Ered-Eox = 0.34 - (-0.76) = 1.1 V.

Can someone explain this to me (it's been a while since I did electrochem and I'm a bit rusty)? I always find the reduction and oxidation potentials and add them, and it always works. Thanks

 

[Phantastic]

The energy of an electrochemical reaction is given as:

E(rxn) = E(oxidation) + E(reduction)

Be sure that the oxidation half reaction is written as an oxidation and not a reduction. Usually, you're given a table of reduction potentials: E(red). You'll need to reverse the half rxn for your oxidation potential, by switching the sign.

As you can now see, they've used the (-) sign in their formula to account for the fact that the E(oxidation) was given as a reduction potential, and that's why they add the extra negative in.

 

[Phantastic]

It's also worth noting that:

galvanic cells will have positive E(cell)
electrolytic will have negative E(cell)

This is because galvanic is spontaneous, and electrolytic is nonspontaneous

deltaG = -nFE (note the negative sign, so +Ecell is spontaneous!)

So, as it's written, you have a net negative potential when you add them together. If this was a galvanic cell, then that alone should tell you ya added them wrong 😀

 

[G1SG2]

The energy of an electrochemical reaction is given as:

E(rxn) = E(oxidation) + E(reduction)

Be sure that the oxidation half reaction is written as an oxidation and not a reduction. Usually, you're given a table of reduction potentials: E(red). You'll need to reverse the half rxn for your oxidation potential, by switching the sign.

As you can now see, they've used the (-) sign in their formula to account for the fact that the E(oxidation) was given as a reduction potential, and that's why they add the extra negative in.

It's also worth noting that:

galvanic cells will have positive E(cell)
electrolytic will have negative E(cell)

This is because galvanic is spontaneous, and electrolytic is nonspontaneous

deltaG = -nFE (note the negative sign, so +Ecell is spontaneous!)

So, as it's written, you have a net negative potential when you add them together. If this was a galvanic cell, then that alone should tell you ya added them wrong 😀

Hey, thanks for the response. Yeah, I know the equation, but the question didn't specify what type of cell it was. I just thought it was weird that I had to reverse the Eox because it said it was the oxidation potential and I thought I was set...

 

[Phantastic]

If it doesn't specify, and there's no sign of an external power source supplying energy to the reaction, then it's safe to assume it's galvanic.