In the following homework problem, the answer says:

"$E ^{o}_{cell}$ is negative, which makes the $\ce{Fe}$ and $\ce{Ga}$ half-reactions non-spontaneous."

But I found $E ^{o}_{cell}$ to be positive - NOT negative:

$$E ^{o}_{cell} = E ^{o}_{cathode} -E ^{o}_{anode} = E ^{o}_{Fe} - E ^{o}_{Ga} = -0.409\,\mathrm{V} - (-0.56\,\mathrm{V}) = +0.151\,\mathrm{V}$$

which means that these half-reactions would actually be spontaneous.

Am I solving this problem wrong or did the writer of this question make a mistake?

After all, $\ce{Fe}$ should be the cathode since it has a more positive voltage. Right?

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  • $\begingroup$ It says nonspontaneous, so aren't you looking for the combination with the negative voltage? $\endgroup$
    – Zhe
    Nov 29 '16 at 3:09
  • $\begingroup$ Yes. But they're saying that Fe and Ga are the combination with the negative voltage. However, after subtracting them, I ended up with a positive voltage. So is their answer wrong? $\endgroup$
    – Czar Great
    Nov 29 '16 at 3:12
  • $\begingroup$ I'm not seeing that. They're saying that the gallium is reduced and the iron is oxidized, so you should reversing the iron equation and its potential. The sum looks negative to me. $\endgroup$
    – Zhe
    Nov 29 '16 at 3:26
  • $\begingroup$ Even if you reverse the iron's equation, you would still end up with a positive cell potential: 0.409−(−0.56)=+0.969Volts. So I don't think you're right... $\endgroup$
    – Czar Great
    Nov 29 '16 at 3:30
  • $\begingroup$ I'm still confused by what you're trying to do. The gallium reduction potential is $-0.509\, \mathrm{V}$. The iron oxidation potential is $+0.409\,\mathrm{V}$. The sum is negative. $\endgroup$
    – Zhe
    Nov 29 '16 at 3:32

You need two half reactions for the total redox reaction: one reduction and one oxidation. The convention is that we provide all potentials as standard reduction potentials. If you need an oxidation potential, you're going to need to reverse the reduction half reaction and negate the sign standard potential.

This means that the total potential of the cell is the raw sum of the oxidation and reduction potentials. In your specific example, you want: $$\ce{Ga^{3+} + 3e- -> Ga}, E_{red}^{\varnothing} = -0.560\,\mathrm{V}$$ $$\ce{Fe -> Fe^{2+} + 2e-}, E_{ox}^{\varnothing} = +0.409\,\mathrm{V}$$

The EMF of the cell is the sum:

$$E_{cell}^{\varnothing} = E_{red}^{\varnothing} + E_{ox}^{\varnothing}$$

And this value is negative.

Now, as a shortcut, you can identify which of the two reduction half reactions needs to be reversed to make an oxidation, and you can add that value--negated--to the potential for the reduction half reaction, as above.

What your teachers or book meant was the following. I've changed the names of the variables so that they don't conflict with the standard notation (which is why we're talking past each other).

$$E_{cell}^{\varnothing} = E_{red,1}^{\varnothing} - E_{red,2}^{\varnothing}$$

where $E_{red,1}^{\varnothing}$ is the standard reduction potential for the reduction half reaction and $E_{red,2}^{\varnothing}$ is the standard reduction potential for the reduction that corresponds to the oxidation half-reaction. This is hellishly confusing, which why I think you are confused. The key here is that

$$E_{\mathrm{cathode}} = E_{red,1}^{\varnothing} = E_{red}^{\varnothing}$$


$$E_{\mathrm{anode}} = E_{red,2}^{\varnothing} = -E_{ox}^{\varnothing}$$

Note also that ordinarily, for a spontaneous reaction, iron would constitute the cathode of the cell because it would be the reduction. However, in this case, because the reaction is non-spontaneous, the iron half reaction is actually going to be an oxidation, so that means it's actually going to be the anode. This is the other factor you're confused about. You're trying to apply shortcuts for a spontaneous cell to a non-spontaneous one. Try to understand what the underlying principle is.

  • $\begingroup$ Thank you for the detailed answer. This makes much more sense now. I was taught the shortcut before the underlying principle which was what led to this confusion. $\endgroup$
    – Czar Great
    Nov 29 '16 at 4:19

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