Is the positive electrode of a voltaic cell actually positive or just less negative? Because it seems to me that the positive electrode is just less negative relative to the negative electrode and that is why electrons flow to it. But then how do we describe the electric field generated then? Since the positive electrode isn't actually positive and doesn't exactly attract the electrons.
My understanding: In an example voltaic cell, a solid strip of zinc is placed in a $\ce{Zn(NO3)2}$ solution to form a half-cell. A solid strip of copper placed in a $\ce{Cu(NO3)2}$ solution forms a second half-cell. The strips act as electrodes, conductive surfaces through which electrons can enter or leave the half-cells. Each metal strip reaches equilibrium with its ions in solution according to these half-reactions:
$\ce{Zn(s) <=> Zn^2+ (aq) + 2 e^-}$
$\ce{Cu(s) <=> Cu^2+(aq) + 2 e^-}$
However, the position of these equilibria is not the same for both metals. Zinc has a greater tendency to ionize than copper, so the zinc half-reaction lies further to the right. As a result, the zinc electrode becomes negatively charged relative to the copper electrode.
If the two half-cells are connected by a wire running from the zinc——through a lightbulb or other electrical device——to the copper, electrons spontaneously flow from the zinc electrode (which is more negatively charged, and therefore, repels electrons) to the copper electrode. As the electrons flow away from the zinc electrode, the $\ce{Zn/Zn^2+}$ equilibrium shifts to the right (according to Le Châtelier’s principle) and oxidation occurs. As electrons flow to the copper electrode, the $\ce{Cu/Cu^2+}$ equilibrium shifts to the left, and reduction occurs. The flowing electrons constitute an electrical current that lights the bulb. The electrical current is driven by a difference in potential energy (caused by an electric field resulting from the charge difference on the two electrodes).
This question is based on readings from Chemistry: A Molecular Approach by Tro, Nivaldo J