[lonewanderer]

In my solutions manual for General Chemistry, it clearly states that E° gets reversed when reaction is reversed.

Yet, later in, in a problem, you need to take a reduction half reaction and its E value from the appendix and REVERSE it and use it as an oxidation half reaction. In the appendix, the reduction half reaction E value is given as -.25, yet, in the problem, the solution manual reverses the reaction to use it as an oxidation BUT DOES NOT reverse the sign, and uses E = -.25 for the oxidation half reaction.

This seems like a contradiction to me. Does the reversing rule of E° only apply to E°cells or does it apply to all E° with all full reactions and half reactions?//???

[lonewanderer]

Nevermind, I found the solution

If you are using the convention:

E°cell = E°cathode(reduction) - E°anode(oxidation), then you should NOT change the sign of E° when reversing the reaction.

If you are using:

E°cell = E°(reduction) + E°(oxidation) , then you should change the sign of E° when reversing the reaction.

This is what I understand from what I searched

[Loyal]

Nevermind, I found the solution

If you are using the convention:

E°cell = E°cathode(reduction) - E°anode(oxidation), then you should NOT change the sign of E° when reversing the reaction.

If you are using:

E°cell = E°(reduction) + E°(oxidation) , then you should change the sign of E° when reversing the reaction.

This is what I understand from what I searched

They are two ways to write the same equation.  The difference is how you set up your data.   

The first equation is if you set up your equation as strictly reduction or oxidation potentials.   The second equation is if you convert one to oxidation potentials and leave the other as a reduction potential.  So in the end it is the same deal, but just two different ways to write it.

Think about it this way.  For the first equation I will use Silver and Hydrogen.

H⁺ + e^-    1/2H₂
Ag⁺ + e^-  Ag

Let's say we want to first have Silver Metal oxidize to silver ion using acid.

E⁰_cell1 = E⁰_Hydrogen - E⁰_Silver

This reaction is not very favorable since Silver is around 0.799 V while Hydrogen is designated 0.

So if we want to see what the reaction between silver ions and hydrogen is like we would set it up like this.

E⁰_cell2 =  E⁰_Silver - E⁰_Hydrogen

Which
E⁰_cell2 = -E⁰_cell1