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I've been thinking about batteries a whole lot, and I've uncovered why the battery symbol looks like a pair of cap-like things in circuit diagrams: it's modeled as two cap-like things using an equivalent circuit diagram:enter image description here

We see here that the two sides have a similar assigned voltage. The actual voltage is dependent on the chemical reaction. Here's the problem. I'm having quite an issue looking at the right side, the cathode, as a cap-like thing that has a direction pointing the way it does.

My issue is this: all cells want to dissolve. You may disbelieve this, but take a quick look at this link: http://en.wikipedia.org/wiki/Absolute_electrode_potential.

The article states that the "zero" hydrogen half-cell reaction does not have a voltage equal to zero, but in fact serves as a reference. You may have known this, but since the actual voltage of the oxidation of Hydrogen is 4.44 V, we must add this number to the oxidation reaction for the list of elements' oxidation reactions. Or, given the reduction reaction, we would subtract 4.44 V, because they are each measured in respect to a hydrogen half-reaction. Let us take a look at a table of reduction half-reactions:

enter image description here

After realizing that the actual reduction potentials need to have 4.44 V reduced from their total volts, we realize that not a single reaction wants to reduce at all! None of them!

So, my question is, given the equivalent circuit diagram, how can we say that any cap-like thing points the way it does? Its not like the positive ions in solution are attempting to form a metal (and get reduced) before the external circuit is closed. The electrons should be trying to escape the cathode just like the anode, you see what I mean?

Now let me be clear: I'm not talking about when the circuit is closed as it is in the diagram: I'm talking about before this moment, when the only thing connecting the cathode and anode is the electrolytic fluid. I get that when the circuit is closed the difference in potentials is all that matters (which would explain why two different metals are used for many galvanic cells). From my reasoning above, shouldn't even the metal in the right-hand area want to dissolve? Therefore, shouldn't both cap-like things point inwards? Basically, I'm saying that the direction of the cathodic cap-like thing should reverse.

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  • $\begingroup$ Also, credit to the open source MIT courseware for image 1: ocw.mit.edu/courses/chemical-engineering/… f $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 14:53
  • $\begingroup$ it's modeled as two capacitors using an equivalent circuit diagram: those capacitor like symbols are not, in fact, capacitor symbols. Batteries aren't modelled as capacitors since, for a capacitor, there is zero current through for constant voltage across. $\endgroup$
    – Alfred Centauri
    Commented Jul 22, 2014 at 14:59
  • $\begingroup$ Oh, those aren't capacitor symbols? Charge pump indicators perhaps? What are those called? $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 15:01
  • $\begingroup$ I'd better edit the caps out after I know what they're called. Thanks I've been wondering why they don't exactly look like caps. Thankfully I still have an issue with the direction. $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 15:03

2 Answers 2

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The two parallel lines of different length are called "batteries" and they're supposed to provide the circuit with a constant voltage for a long time. They're similar in effect to high-capacity capacitors which are charged to a very high charge that is constant, and so is the voltage. Except that normal capacitors are not that strong.

Moreover, capacitors don't really involve any chemistry, like batteries: they are passive elements.

enter image description here

There are different symbols for capacitors but there is no capacitor in your circuit. The wiggly teeth, like one next to $R_{\rm ext}$, are called resistors. They have voltage in between the ends only if a current is running through. Resistors are as different from batteries as well as capacitors as you can get.

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  • $\begingroup$ Ah, but what can explain the direction for the cathodic battery symbol? Have you read about the Hydrogen absolute cell voltage? A simple mistake though, I thought that the capacitor symbol was modified to mean capacitor-like for batteries. $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 15:19
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Sorry to say, but batteries aren't anything like capacitors. The two "capacitors" in your equivalent circuit diagram are just representations of a half-cell. In a formal definition, a battery is a combination of multiple chemical cells in series. The cells are represented as two parallel lines because they're often manufactured as layers of different materials. The battery is then represented as two cells stuck together, or 4 lines in parallel. One line is larger to represent the positive end, and also distinguish it from the capacitor symbol.

You're making a confusion somewhere when you conclude that not a single reaction wants to reduce, but it's hard to figure out what the misconception is from your question. Just to make it clear: for a reducing-oxidizing reaction to occur, the only thing that matters is whether the difference in potentials is positive or negative. Subtracting 4.44V (or any voltage) from all the numbers in the chart won't affect any difference in potential. As an example:

(A-4.44) - (B-4.44) = A - B - 4.44 + 4.44 = A - B

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  • $\begingroup$ (Good answer) Agreed, when the two cells are externally connected. But you have fallen into my trap. I made it clear that I was only analyzing the states before an external connection is made, looking at just the cathode. I like your A and B proof but I actually agreed with that. $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 16:01
  • $\begingroup$ If you're interested what happens before the cell is connected, I can give you this summary: a reaction will occur for a small amount of chemical material. Electrons will pile up on one end, while there will be a deficit of electrons on the other end. These will create an electric field with a voltage determined by the chemical reaction, with the potentials like in the chart. This electric field will counteract the diffusion occurring in the cell, such that the chemical reaction stops until electrons are added/removed. $\endgroup$
    – David
    Commented Jul 22, 2014 at 16:05
  • $\begingroup$ Right. A deficit of electrons, as opposed to say a surplus of protons. The reason a battery goes along is the difference, which is why the Hydrogen cell potential doesn't affect a full reaction. When it comes down to the half cells, is it not quite interesting that both actually attempt dissolving and forming an electron buildup at their respective electrode? They just want to with different strengths. Makes sense. $\endgroup$
    – Andres Salas
    Commented Jul 22, 2014 at 16:17
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    $\begingroup$ That's fairly well correct. I think I understand what your original concern was. It's true that most materials will in general attract spare electrons, due to effects like polarization, van der waals forces, etc. Even metals like lithium and sodium will attract spare electrons. It's just that certain chemicals attract them with far greater strength. $\endgroup$
    – David
    Commented Jul 22, 2014 at 16:39

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