Does the negative terminal of a battery have greater electron concentration then the positive terminal?

Else how would charge start flowing once a conducting wire is connected?


Starting with a cell with zero net charge the electrochemical reactions within the cell make it energetically favourable to move charges from one end of the cell, through the cell, to the other end of the cell.
The cell is moving charges like a pump moves water.

The end at which electrons accumulate is called the negative terminal and the end from which electrons have moved away is called the positive terminal.

Whilst this movement of charges is going on the charges on the terminals set up an electric field within the cell in opposition to the movement produced by the electrochemical reactions.

Eventually the forces due to the electrochemical reactions trying to move the charges are balanced by the forces due to the electric field set up within the cell due to the redistribution of charges and then there is no net migration of charges within the cell.

The cell in this condition has a "surplus" of electrons at the negative end and a "deficit" of electrons at the positive end.

When a conducting metal wire is connected across the terminals of the cell, electrons move from the negative terminal, through the wire towards the positive terminal whilst all the time within the cell the electrochemical reactions are causing electrons to move from the positive terminal to the negative terminal trying to maintain the imbalance of charges on the two terminals.

  • $\begingroup$ 1. From the second last para of your answer I am concluding that the electron concentration at the negative terminal is indeed higher than that at the positive terminal. Please confirm if that is correct. 2. Would it create an electric field near the terminals of the battery in the air (even though magnitude may be negligible)? $\endgroup$ Sep 7 '17 at 1:39
  • $\begingroup$ @user2555452 The electron concentration would be higher at the negative terminal and there would be ab electric field in the air which would not be strong enough to cause ionisation of the air. The shuttling ball experiment shows the existence of such a field although a much higher voltage source than one cell is required. m.youtube.com/watch?v=2Rh8fJnvisA with an explanation here practicalphysics.org/moving-charges-are-electric-current.html $\endgroup$
    – Farcher
    Sep 7 '17 at 4:19

In a word, No.

The negative terminal of a battery is what it is because the chemicals at that end are set up so that electrons are more easily "unleashed" than at the positive end.

The topic of "how easily are electrons unleashed?" is a different topic than "how many electrons are there per volume?" (i.e. what is the concentration?)

Now, let's address the other part of your question: "Else, how would the charge start flowing once a conducting wire is connected."

You are correct that say, 500 electrons squashed into a smaller area will have more electrostatic repulsion than 500 electrons scattered out across a larger area. But this isn't the main source of EMF in a battery. The main source of EMF in a battery comes from the fact that one of the half-reactions is looking to donate electrons, and the other half-reaction is looking to accept electrons.

I suppose other readers are wondering "how do the electrons in one reaction know that they are needed at the other terminal?" While this is a rather interesting question indeed, we must understand that it is in the same pile of paradoxes as "how does the marble rolling around in the bowl know that it must obey the laws of potential energy" and "why don't magnets ever get confused about the opposites attract rule?" -- I don't think science really has answers to these questions, interesting as they may be.

  • $\begingroup$ I don't think science really has answers to these questions Yes it does. A marble doesn't have to "know" anything, a magnet cannot get "confused", and this is not a proper answer to the question. There is no "paradox", the laws of electromagnetism still apply and the current will answer to a difference of potential... so a difference in $e^-$ concentration must be there. $\endgroup$
    – MrBrushy
    Jun 26 '17 at 9:23

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