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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?

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    $\begingroup$ Electrons, I understand, then flow from the copper electrode to the Zn electrode and redox occurs. Exactly the opposite. $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 9:30
  • $\begingroup$ Apologies, that was a typing error which I would fix if I could make edits. It is not what my question is about, though. $\endgroup$
    – P0W8J6
    Commented Jan 24, 2022 at 9:33
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    $\begingroup$ What do you mean by you cannot edit? Anybody who is registered can edit their posts. AFAIK, guest users would not have 129 reputation points. // See also policy-on-(Am I right?)-questions $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 9:38
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    $\begingroup$ You must have noticed in real life that voltage of cells under load decreases, immediately and also in time. $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 9:44
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    $\begingroup$ Does the question manifest research effort ? Is the question clear and useful? Compare the answers with the mouse hint at the downvoting icon. $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 12:49

1 Answer 1

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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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  • $\begingroup$ Thanks for taking the time to respond, Poutnik, but my question remains unanswered: does the potential difference between the electrodes decrease progressively over time because the equilibrium at the Zn doesn’t reform electrons at the same rate that they are drawn towards and used up in redox at the Cu electrode? $\endgroup$
    – P0W8J6
    Commented Jan 24, 2022 at 13:32
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    $\begingroup$ No, it does not. If it did, current would stop. $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 13:44
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    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Jan 24, 2022 at 14:10

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