[kdbmvp]

Hello,

I have the following half reactions:

Fe²⁺(aq)  -> Fe³⁺(aq)+ e^- E⁰ = -0.77V
Zn²⁺ + 2e^- -> Zn(s) E⁰ = -0.76V

I.e. Fe is oxidized while Zn is reduced. According to the solution, E⁰tot = -0.01V. But I thought the answer should be -1.53V?

[Borek]

What is the difference between -20°C and -19°C? 39 degrees, or 1 degree?

[kdbmvp]

1 degree?

[Borek]

1 degree it is, no need for a question mark.

Now apply the same thinking to potentials.

[kdbmvp]

Thanks. So it's basically the potential difference between the cathode and anode?

I.e. E(total) = E(cathode) - E(anode)..?

[Borek]

Yes.

[kdbmvp]

Alright. I'm a bit confused. Because I'm using tabulated values for Standard Elecrode Potentials, where ALL half reactions are listed as reduction reactions. Is the formula for the potential difference still the same when you actually change the sign of the oxidation reaction?

[Borek]

The reaction goes both ways at exactly the same potential.

[kdbmvp]

But don't you change the sign of the potential when you turn the half-reactions around?

I find the following values for the reduction reactions:

2Fe³⁺+2e^- -> 2Fe²⁺, E⁰ = 0.77V
Zn²⁺ + 2e^- -> Zn, E⁰ = -0.76V

As Fe should be the oxidation reaction, we turn the reaction around and change sign of E⁰:

2Fe²⁺ -> 2Fe³⁺+2e^-, E⁰ = -0.77V

And E⁰tot = -1.53V ..?? That is the method that is presented in everything I find online. Find data for two reduction potentials, turn one of them around (and change sign) and then add them together.

2

[mjc123]

That is correct. The answer you gave in your first post was wrong because you were subtracting rather than adding. A common mistake. Either subtract one reduction potential from the other, or change the sign of the anode potential and add. Don't change sign and subtract!

As Fe should be the oxidation reaction

What does the fact that E°_tot is negative tell you about the reaction that occurs?

[kdbmvp]

That the reactions doesn't go in the direction we have written it..? I'm fairly sure it should be  Zn that is oxidized.