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If polarisation means: "any mechanical side-effect by which isolating barriers develop at the interface between electrode and electrolyte."

And if concentration polarisation is "changes in the electrolyte concentration due to the passage of current through the electrode/solution interface. Here polarization is understood as the shift of the Electrochemical potential difference across the cell from its equilibrium value. The cause of the changes in concentration (emergence of concentration gradients in the solution adjacent to the electrode surface) is the difference in the rate of electrochemical reaction at the electrode and the rate of ion migration in the solution from/to the surface."

Thus, should concentration polarisation cause a decrease in the performance/efficiency of a galvanic cell over time and would this be shown with a decrease of voltage/current through the cell? Or does voltage and current remain the same throughout a galvanic cell reaction?

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There are two different factors that cause a battery to have a lower energy output.

First regardless of the "state" of the battery, there is the open cell voltage and the obtained voltage under load. State meaning here how much of the battery is used up. For the open cell voltage the voltage is measured with a high impedance voltmeter so that essentially no current (meaning a very very small current) is being drawn. Next the battery is put under some load typical for its use and the voltage is measured again while under the load. This second load voltage measurement will always be lower that the first measurement for the open cell voltage. This voltage drop is caused by the diffusion limits of the chemicals in the battery to and from the electrodes.

This behavior can be essentially modeled by assuming that the battery has an internal resistance. So as the current draw increases the voltage drop internal to the battery increases and the output voltage drops.

Now the second problem is related to the Nernst equation. Let's assume that the battery is used up in 10% increments in a series of tests. At the start of each test the voltage and current is measured. Of course during using each 10% of the battery the reactants in the vicinity of the electrodes are used up and both the voltage and current out for a given load drops.

Now if the battery is left to equilateral for some time then the reactants redistribute and the battery regains some power. However the battery will not recovery to the level at the start of the previous load test. Essentially the bulk concentration of the reactants is dropping by about 10% after each test. So even after a recovery period both the voltage and current that the battery can supply for a constant load drops.

Now with all of this being said there are many types of batteries. Some types produce a more constant power output than others.

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  • $\begingroup$ So in an experiment with a simple Cu/Zn electrochemical cell, if you attached a load (i.e. 3 ohms) into the circuit and left the cell to operate over a period of time, would the cell decrease in voltage and current (as the electrodes are being used up)? And would a larger load drain the "bulk concentration of the reactants" quicker and thus also reduce the voltage and current quicker? $\endgroup$ – Sam Sep 27 '16 at 5:16
  • $\begingroup$ Yes and yes. A larger load of course means less resistance in the load so that more current flows from the battery. $\endgroup$ – MaxW Sep 27 '16 at 5:38
  • $\begingroup$ What do you mean by "less resistance in the load"? $\endgroup$ – Sam Sep 30 '16 at 1:38
  • $\begingroup$ The load has a resistance. Since V=IR, in order to increase I you have to decrease R. $\endgroup$ – MaxW Sep 30 '16 at 4:28
  • $\begingroup$ Ah okay that makes sense, thanks for the help. $\endgroup$ – Sam Sep 30 '16 at 11:11

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