I had the following problem:

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Relevant information from Data Book:

$\ce{Ag+ + e- <=> Ag}$

$E^0 = +0.80$


$\ce{Fe^3+ + e^- <=> Fe^2+}$

$E^0 = +0.77$

This means $E_\mathrm{cell}$ is +0.03. (right)

And (I thought) that the left ($\ce{Fe}$) half cell has the positive electrode/anode as the ions lose electrons to the anode. I thought that the half equation with the more positive $E^0$ value went forwards.

However this is wrong. Where did I go wrong in my reasoning?

  • $\begingroup$ Your reasoning was correct, except for the part that Fe is positive electrode. As $Fe^{2+}$ lose electron, it produce electron, so that the electrode becomes negative. Ag electrode becomes negative for the same reason. $\endgroup$ – Huy Ngo Feb 7 '17 at 15:07

As the standard reduction potential of Ag electrode is more than that of Fe electrode, Ag electrode will act as cathode (where reduction or gain of electrons takes place). The Ag electrode will be a positive electrode because it is electron deficient and thats why reduction is taking place at this electrode.

The Fe electrode will act as anode (where oxidation or loss of electrons takes place) and it will be the negative electrode because there is excess of electrons on this electrode.

Ecell = (Ereduction)cathode - (Ereduction)anode

      = +0.80 - (+0.77)

      = 0.03V
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  • $\begingroup$ So the Ag electrode is positive because the electrons flow through the circuit to it and the Fe electrode is negative because electrons flow away from it? $\endgroup$ – Mirte Feb 8 '17 at 5:14
  • $\begingroup$ Yes thats the reason. At Fe electrode Fe+3 loses a electron so there is excess of electrons here and at the Ag electrode Ag+ from the AgSo4 solution gets the electron and gets reduced to Ag and deposits itself of the Ag electrode. $\endgroup$ – Mitchell Feb 8 '17 at 5:18

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