Charging an electrochemical capacitor

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Charging a symmetrical electrochemical capacitor to 1V results in one electrode reaching +0.5V and the other -0.5V, with both starting at 0V. When measuring the voltage across each electrode, they move in opposite directions to these potentials. The 0V point is identified as the equilibrium state where no current flows after discharging the device. During constant current charge-discharge testing, the voltage increases from 0V to the set limit and then decreases upon discharge. Understanding these voltage dynamics is crucial for effective capacitor operation and testing.
nubie43
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I have a very basic question confusing me about charging electrochemical capacitors -

When you charge a symmetrical electrochemical capacitor to 1V, because one electrode becomes positively charged and the other negatively charged, the electrode potential across one will be +0.5V, and the electrode potential across the other will be -0.5V?

So if you were to measure the CV across each electrode they would both start at 0V, and travel in opposite directions to + and - 0.5V?
 
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You mention a 0v point. Do you have an explanation at which point in the capacitor exactly where this is?
 
Im not really sure about that either. When testing a cell for example by constant current charge discharge, you apply the current, and the voltage will increases to your set limit from 0V then, goes back on discharge.

I suppose 0V would be when the system is at equilibrium, where no current flows after you discharge the device?
 
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