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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

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    $\begingroup$ As all electrodes are positive (see absolute potential of SHE), negative electrodes are less positive than positive ones. $\endgroup$
    – Poutnik
    Commented May 5 at 4:59

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Is the positive electrode of 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.

This is a good question because most general chemistry textbooks never explicitly talk about it. They rarely connect or bring electrostatic phenomena to cells.

When we say positive electrode in potentiometry (no current flowing), it actually means an electrostatically positive electrode. All it means, in other words, that there is a disbalance of positive and negative charges on the electrode in an "absolute" sense. You do recall the definitions of positive and negative charges, right? A glass rod is positive by definition when it is rubbed with silk. In modern words, the glass rod is stripped of electrons. In the same way, a positive sign designated electrode means that it is stripped of electrons.

If we had a charge sensor with the positive electrode or by contact we say that the electrode is electrostatically positive. Search charge sensors on YouTube. Vernier also makes them.

positive electrode is just less negative relative to the negative electrode

Positive electrode means it is stripped of electrons, and there is a net electrostatic positive charge. At an atomic level, the number of protons is in a very very slight excess than the number of electrons in that material.

At the time of posting, there are two downvotes and two upvotes. The net result is zero! If there is a slight disbalance, one of the mathematical signs will prevail. These mathematical signs have nothing to do with electrostatic positive or negative.

The OP further asked in the comments...

how is the copper electrode the +ve one here in the example I gave please? Going by the stripped of electrodes defn, wouldn't we call the zinc electrode the +ve one then? Since that's the one that loses electrons. But in actuality, the zinc electrode is the -ve. But going back to the copper, how is the number of protons in excess of its electrons?

The excellent questions you are asking have been asked for more than a century by brilliant minds (two Nobel prize winners, Nernst and Marcus among others). Forget about the copper electrode and think about the zinc rod immersed in a solution of $\ce{Zn^{2+}}$ ions (zinc sulfate). The bulk solution is completely electrically neutral. All the magic (Nobel prize-winning work of electron transfer studies) is occurring at the interface of the solution and the metal rod. Suppose a very very small amount of zinc dissolves from the rod, becoming $\ce{Zn^{2+}}$. Now there is an equilibrium for this process but microscopically nothing is static, some zinc ions are becoming zinc metal as well but there is a net disbalance. Slightly more zinc has dissolved. Where can the electron go? It resides on the surface of the metal, the metal becomes negatively charged- there is an excess of electrons in the metal. Zn is electrostatically negatively charged.

I cannot find the calculation (I hope someone else can), it takes a very small number of electrons disbalance for a metal to develop a huge potential difference.

Now why is copper electrode positive-- the same process is occurring. Cu rod is dipped in copper sulfate, some copper is attempting to dissolve but slightly more copper ions deposit by stripping electrons from copper. The copper rod is short of electrons and it is electrostatically positive.

Why do Cu and Zn behave differently in terms of redox processes? Because they are different elements, their electronic energy levels are different.

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  • $\begingroup$ So how is the copper electrode the +ve one here in the example I gave please? Going by the stripped of electrodes defn, wouldn't we call the zinc electrode the +ve one then? Since that's the one that loses electrons. But in actuality the zinc electrode is the -ve. But going back to the copper, how is the number of protons in excess of its electrons? $\endgroup$
    – funso
    Commented May 4 at 21:33
  • $\begingroup$ @funso I expanded the answer. $\endgroup$
    – AChem
    Commented May 4 at 22:39
  • $\begingroup$ thank you! I didn't think about how the Cu would be undergoing a similar process as the zinc but the opposite with more Cu ions being deposited on it and stripping the Cu of it electrons. But this is affected by what the other electrode its connected to right? because with a copper and silver galvanic cell the copper becomes the negative electrode. So the equilibrium, which determines it charge, is dependent on the other electrode? in copper/zinc more copper ions are deposited and copper is +ve, but in copper/silver its the opposite. $\endgroup$
    – funso
    Commented May 5 at 0:47
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    $\begingroup$ Right, earlier we were talking about isolated electrodes. When we connect two electrodes, the Fermi levels of the two metals determines which one will acquire a net positive charge and which one will be net negative. $\endgroup$
    – AChem
    Commented May 5 at 1:34
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    $\begingroup$ In each case, the sign is "absolute" electrostatic sign whether we are talking about isolated electrodes or we are talking about connected electrodes. Measuring the single electrode potential is an ill-posed problem. $\endgroup$
    – AChem
    Commented May 5 at 1:35

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