Why does current flow in an electrolytic cell?

Many explanations of current flow in electrolytic cells will state that electrons flow from the external power source into the cell (via the cathode). But surely this current flow would stop if there were no reduction at the cathode (which removes electrons).


Doesn't this therefore mean that current is explained by a simultaneous combination of 'push' forces *and* 'pull' forces? You can't have current being generated in an electrolytic cell without a reaction occurring, right?


On a related note, it's also often said that the rate of electrolysis *depends* on current. But, based on my understanding, current is simply a *measure* of the rate of electrolysis: you can't have current without first having a redox reaction occurring in the electrolytic cell. Am I right in thinking this?




Let's take as an example a $1$ M $\ce{CuCl2}$ solution in water. If two platinum or charcoal electrodes are dipped into this solution, they will not react with this solution. If a small voltage (< $\pu{1.02 V}$} is applied on the electrodes, nothing happens. If now a higher continuous voltage (>$1.02$ V) is applied, electrolysis proceeds. At the cathode (negative pole), the cation $\ce{Cu^{2+}}$ is reduced into metallic copper $\ce{Cu}$. At the anode (positive pole), the ions $\ce{Cl-}$ are oxidized into some gaseous chlorine $\ce{Cl2}$ according to the equations $$\ce{Cu^{2+} + 2 e^- -> Cu}\tag {1} $$ $$\ce{2 Cl^- -> Cl2 + 2 e-}\tag{2}$$
Let's consider the situation a couple of minutes after the beginning of the electrolysis. The cathode is at least partly covered by metallic copper. The anode is at least partly covered by adsorbed bubbles of gaseous chlorine. If the outer voltage is maintained above $1.02$ Volt, the electrolysis goes on.
But if the outer voltage is < $1.02$ Volt, or even zero Volt (situation where the electrodes are just connected by a simple wire), the previous reactions will be reversed : the cell will produce current and work as a galvanic cell. The current will be reversed and go from $\ce{Cl2}$ to $\ce{Cu}$. This cell $\ce{Cu/Cl2}$ is a good example of the push-pull phenomena discussed by Michael.


Let's summarize : when two electrodes, one made of copper $\ce{Cu}$ and one containing adsorbed chlorine $\ce{Cl2}$, are related by an external wire, and are also dipped into a $\ce{CuCl2}$ solution (or even into pure water), they have a tendency to react according to the following scheme $$\ce{Cu -> Cu^{2+} + 2 e^-}\tag{3}$$ $$\ce{Cl2 + 2 e^- -> 2 Cl^-}\tag{4}$$ The electrodes both "feel like" reacting this way. They have a natural tendency to spontaneously push electrons from $\ce{Cu}$ to $\ce{Cl2}$ through the external wire. The measured voltage between the electrodes is then $\ce{1.02 V}$ (or a little less, due to Ohm's law).


If we replace the external wire by another cell $\ce{Cu/Cl2}$, both would like to send their current through the other one. But, as they are identical, nothing happens. No current is emitted.


If the external wire is removed and a "stronger" cell is connected between the two electrodes for example a cell $\ce{Zn/Cl2}$, our "weaker" $\ce{Cu/Cl2}$ cell cannot work any more according to ($3$) and ($4$). It is obliged to react in the "pull" mode. Reluctantly, the cell must accept working according to ($1$) and ($2$} which is contrary to its natural desire. As ever, the strongest imposes its will to the weaker.


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