Positive electrode of an electrochemical cell? I had the following problem:



Relevant information from Data Book:


  $\ce{Ag+ + e- &lt;=&gt; Ag}$
  
  $E^0 = +0.80$
  
  and 
  
  $\ce{Fe^3+ + e^- &lt;=&gt; Fe^2+}$
  
  $E^0 = +0.77$


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?
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