Inside an ideal battery, the chemical reaction is reversible (at equilibrium), thus the oxidation-reduction reaction $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$ is happening in both ways at the same rate right?

Now, for a real battery, there's resistivity inside it due to many factors, thus the chemical reaction is irreversible. My question is, during the irreversible reaction, is the reaction happening in both ways but at different rate due to resistivity inside the battery or it is just like the combustion reaction happening in one way?


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The original poster has a serious misconception by equating a reversible chemical reaction with a reaction that is in equilibrium. The electrochemical reaction that allows the battery to deliver energy to the load (such as the copper/zinc reaction mentioned) is always reversible in theory by reversing the current flow. However in addition to the electrochemical reaction, there are other chemical reactions going on inside the battery which are not reversible as well as geometrical changes that may not allow the reactants to be close enough for the electrochemical reaction to run in reverse. In batteries designed to be recharged, these geometrical and irreversible chemical changes are minimized to allow as many recharge cycles as possible. Batteries not designed to be recharged therefore usually can't be recharged fully and have very few recharge cycles not to mention a high probability of gaseous or corrosive liquid leakage.

At equilibrium, both forward and reverse reactions are going at the same rate, but this is only discernable microscopically. At a macroscopic level you see nothing happening. So a battery is never at equilibrium while you are using it since at that point the battery is completely dead. Then (or preferably sooner) it is time to recharge the battery or throw it away.

  • $\begingroup$ Hello, Than for an ideal battery how can you explain that the potential difference remains constant? And for a real battery, during discharge does the reaction happen in both ways with the first way having higher rate? or only one way? $\endgroup$ – Tonylb1 Nov 21 '15 at 7:17
  • $\begingroup$ "In reversible reactions, as the reactants react with other reactants to form products, the products are reacting with other products to form reactants." chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/… $\endgroup$ – Tonylb1 Nov 21 '15 at 7:24
  • $\begingroup$ physics.stackexchange.com/a/126977/72921 See the answer here $\endgroup$ – Tonylb1 Nov 21 '15 at 7:31

The reaction you describe is what known in chemistry as an equilibrium reaction, symbolised by the double sided arrow:

$$\mathrm{Zn}+\mathrm{Cu^{2+}}\leftrightarrow \mathrm{Zn^{2+}}+\mathrm{Cu}$$

The equilibrium constant of the reaction, $K$, which is the ratio of the products of the activities of the species on the right and the products of the activities of the species on the left is fairly high which means the reaction tends to run more or less to completion from left to right. In a cell (or battery: multiple cells in series) that results in a flow of electrons from the Zn metal to the Cu electrode, resulting in a net positive EMF.

But by reversing the current by applying the reverse voltage over the electrodes the reaction can also be made to run from right to left. The original amounts of reactants and reaction products can thus be exactly restored, assuming the total amount of charge (current times time) is the same as the charge that was discharged during 'normal' use of the cell.

Internal resistance simply arises from resistance of flow to the charge carriers (electrons and solvated ions, in this case).


When current flows it carries ions away from the battery (electrolyte) interface where the reaction can take place. That is why the reactions then take place at a different rate in the forward and reverse directions: the concentrations of ions changes.

  • $\begingroup$ this is due to resistivity? $\endgroup$ – Tonylb1 Nov 20 '15 at 22:54
  • $\begingroup$ Not so much "due to" as "gives rise to" I would say. But you might get better answers on the chemistry stack. $\endgroup$ – Floris Nov 20 '15 at 22:58

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