Voltaic cell equilibrium confusion

Question

In a voltaic cell, taking one with $\ce{Zn}$ and $\ce{Cu }$ electrodes for an example, an EMF is formed because the equilibria at the $\ce{Zn}$ electrode is shifted more towards the $\ce{Zn^2+ + 2e-}$ side than the copper electrode’s equilibrium is shifted towards the $\ce{Cu^2+ + 2e-}$ side. Electrons, I understand, then flow from the $\ce{Zn}$ electrode to the $\ce{Cu}$ electrode and redox occurs.

The problem I have is that, if electrons migrate from the $\ce{Zn}$ electrode in this way, surely this will mess up the equilibrium at the $\ce{Zn}$ electrode and, even though it will attempt to shift back to counteract the change, the overall result will be a loss of delocalised electrons to participate in the $\ce{Zn}$ equilibrium. This would imply a decreased emf towards the copper electrode and hence an increasingly more positive $\pu{E_{cell}}$ value over time (assuming the $\ce{Zn}$ electrode to be on the left of the setup).

Am I correct or have I gone wrong and, if so, where is my mistake?

Answer 1

Consider electrodes as electron pumps that try to maintain their equilibrium potential, according to their electrochemical state.

When $\ce{Zn}$ and $\ce{Cu}$ electrodes (of e.g. the Daniell cell) are connected by external circuit, they mutually try to force their potential to each other, causing the current flowing across this circuit.

The potential of $\ce{Zn}$ electrode (anode) becomes higher then the equilibrium one and the oxidation of $\ce{Zn}$ occurs, partially ( because of limiting reaction kinetics) acting against increased potential, releasing electrons to the circuit and positive ions to the electrolyte.

The potential of $\ce{Cu}$ electrode (cathode) becomes lower then the equilibrium one and the reduction of $\ce{Cu^2+}$ occurs, partially acting against decreased potential, withdrawing electrons from the circuit and positive ions from the electrolyte.

The cell voltage over time decreases because the cell open voltage over time decreases, because the electrode open potentials converge to each other. Another drop of cell voltage under load is due increasing of internal resistence over time, mostly due depletion of electroactive components.

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