It is found experimentally that the potential of an electrochemical cell with applied current density varies as follows:

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To produce the passage of current through a cell it is necessary to apply a potential difference that exceeds the equilibrium f.e.m by at least the overpotential of the cell. Furthermore, the potential of the cell will decrease as it generates current, because under such conditions it no longer works reversibly and therefore does less work than the maximum. When the current is high and close to the limiting value of one of the electrodes, a sharp drop in the work potential is observed.

I know this is what is observed, but I can't explain why. Why does this happen?

  • $\begingroup$ What is the meaning of "potencial de salida" and "potential de la pila" ? Other question ? Are you speaking of an electrolysis or of a cell producing the currant ? $\endgroup$ – Maurice Feb 28 at 18:12
  • $\begingroup$ The "potencial de la pila" is the voltage measured during the experiment with a multimeter. The "potencial de salida" is simply the product of the current and the voltage. I am referring to the case of a fuel cell, such as a methanol cell. @Maurice $\endgroup$ – aprendiendo-a-programar Feb 28 at 18:31
  • $\begingroup$ The voltage that you measure is equal to the redox potential $E$ if the current is zero. If the current, the measured voltage is equal to ${E - rI}$ where $r$ is the internal resistance. In principle this internal resistance should be a constant. But in your reported data, it is obviously not the case. The internal resistance is about $\pu{1 ohm}$ in the beginning, and decreases when the current is increasing. This is surprising, and I don't know its reason. $\endgroup$ – Maurice Feb 28 at 20:25

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