Do galvanic cells depend on distance?

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Discussion Overview

The discussion revolves around whether the distance between electrodes in galvanic cells affects their voltage output. Participants explore various aspects of this question, including theoretical implications and practical considerations.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that the voltage remains constant regardless of the distance between electrodes, arguing that only the electric field weakens and current decreases with increased distance.
  • Another participant questions the relevance of electrostatic repulsion between electrodes in a liquid electrolyte, indicating uncertainty about its impact on the system.
  • A later contribution states that increasing the surface area of electrodes does not change the voltage of a galvanic cell.

Areas of Agreement / Disagreement

Participants express differing views on the influence of distance and surface area on galvanic cell voltage, indicating that multiple competing perspectives remain unresolved.

Contextual Notes

Some assumptions about the behavior of electric fields and potentials in galvanic cells are not fully explored, and the discussion lacks clarity on the specific conditions under which the claims are made.

Biker
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I have seen some sites say that it doesn't and other says the opposite. So let's see
h2vcudiag.gif

Source of the pic: Chemguide

Lets look at the hydrogen (Oxidation) half cell. If we assume that it contributes x volts until the electron reachs the copper, then no matter how big is the distance the voltage is going to be the same ( only the Electric field will be weak and the current will decrease)

However if we look at both cells together and try to move the copper away from the hydrogen and look at the contribution of the copper electrode, First opposing the current, by exerting a force in the other direction. So when you move it away, you are making the potential of the electrons in the platinum decrease.
So if we assume that the voltage at the copper electrode ( without the contribution of hydrogen) is y and the voltage at the hydrogen is z at the initial state then you move it away z will decrease so the difference will increase thus opposing the current moreIs this a explanation true?
 
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Any answer?
 
Biker said:
However if we look at both cells together and try to move the copper away from the hydrogen and look at the contribution of the copper electrode, First opposing the current, by exerting a force in the other direction. So when you move it away, you are making the potential of the electrons in the platinum decrease.
These are electrodes immersed in liquid electrolyte, and you're envisioning a tiny electrostatic repulsion between the plates due to the charges on each? I don't think that comes into the picture, but we'll see if someone can set you right.
 
NascentOxygen said:
These are electrodes immersed in liquid electrolyte, and you're envisioning a tiny electrostatic repulsion between the plates due to the charges on each? I don't think that comes into the picture, but we'll see if someone can set you right.
Nvm, Anyone who is reading this thread. I already figured it out

Thanks NascentOxygen
 
Last edited:
Even if you make the surface area of each electrode 100 times greater, a galvanic cell's voltage doesn't change.
 

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