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Ive been struggling with this question, I cant get an intuitive hang of it. I would like an explanation which focuses on the molecular level understanding of what happens inside the electrolyte and how exactly that affects the emf in a closed circuit. I know that the electrolyte offers some resistance to the current flowing through it and im also aware of ohms law which state that current is directly proportional to the potential difference across the terminals of the cell, but you can say that im also confused as to how a decrease in current affects the potential difference because plainly its the amount of work done in bringing a test charge from one electrode to the other (strictly talking about the cell as an individual)so does that mean that this imaginary test charge lose its potential now im more confused the potential energy is gained by this test charge no so how? I hope this question makes some sense

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im also confused as to how a decrease in current affects the potential difference

You have it backwards. You can control voltage and resistance - current is the independent variable that responds to that voltage and resistance.

Electric potential energy (voltage) is just like gravitational potential energy. You can control the slope of a pipe, and the resistance of the material it's made out of, and then the flow rate of the water will respond to those variables.

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  • $\begingroup$ what do u mean by current is an independent variable? in the next para you say that the resistance and slope of a pipe is what decides the rate of the flow of water through it so it is in fact dependent on both of these factors $\endgroup$
    – sanya
    Commented Dec 23, 2022 at 6:46
  • $\begingroup$ @sanya generally speaking, potential difference and resistance are the things you have direct control over. Current you only have indirect control over (by changing voltage or resistance). I actually think I should have called current the dependent variable. $\endgroup$
    – Señor O
    Commented Dec 23, 2022 at 8:17
  • $\begingroup$ yes! that is what I was thinking $\endgroup$
    – sanya
    Commented Dec 23, 2022 at 8:30

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