Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I know, that voltage is analogous to pressure for charge, but analogies lie. I don't see charge pressing anything and I don't understand definition of $U=A/q$ (voltage = work/charge), cause I can't see what is that work.

enter image description here

Voltage in a circuit is determined by an accumulator. I suppose, that accumulator is characterized with voltage output and current output. I don't get how they are related, but can imaging some tweaks to the accumulator:

1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output? Intuitively it will cause twice the amount of Ions bump into plates and give their charges to them, so it would double the current and won't change any "charge pressure", I can think of.

2) If I double the concentration of each solution, would it double the current or charge. My intuition that it is the same as if I doubled the area of plates - it would just double the amount of ions transferring their charge per moment of time. Nernst equation suggests a change of voltage in this case, but I can't see what it is.

What I want to understand, is what is voltage in terms of ball-like ions, forces and such intuitive entities. Thanks.

share|improve this question
    
If you short-circuit the electrodes like that, there can't be a potential difference (voltage). –  RedGrittyBrick Nov 12 '12 at 11:47
    
All right, there is something in between. I don't want to say, that there is a resistance in between, cause my next question is going to be what is resistance. –  Bob Nov 12 '12 at 11:57

3 Answers 3

What I want to understand, is what is voltage in terms of ball-like ions, forces and such intuitive entities.

First you'd need to clarify the situation in your drawing: where exactly does the voltage appear?

Answer: it's all in the microscopic gap between each metal surface and the electrolyte.

In other words, if you dip a chunk of copper into a bowl of salt solution, a constant voltage appears, with the metal charged negative and the liquid charged positive.

In other words, all batteries actually contain two batteries: one on the surface of one electrode, and the other on the other electrode surface. Google "half-cell" to find info on this.

Simplified intuitive explanation: when metal touches water, the polar water atoms start rapidly dissolving the metal. The water atoms surround metal atoms and yank them out of the surface. The metal starts dissolving rapidly, as rapidly as salt or sugar. But something else happens too, otherwise metals in water would vanish within minutes.

When water pulls a metal atom out of the metal surface, it ends up with a positive metal ion. Metal atoms have a 'loose' outer electron which becomes part of the bulk electron-sea of the metal and does not adhere to each atom. So, as metal atoms (positive metal ions) leave the surface and move into the water, the water acquires an enormous positive charge. The metal surface is full of left-behind electrons and develops an enormous negative charge.

In other words, when a blob of metal touches a blob of water, we end up with a self-charging capacitor!

Now obviously this capacitor cannot charge up to infinite voltage. The "dielectric" is a thin layer of water molecules separating the positively-charged water from the negatively charged metal. A few volts across a nanometers-wide dielectric will give us an e-field of megavolts/cm. This strong field pushes backwards on the positive metal ions and slows the corrosion process. Fewer ions leave the metal, the voltage rises more slowly, and finally we have equilibrium where a few ions are randomly falling back to the metal, and an equal number are still being dissolved.

So, a hunk of metal in a bowl of water doesn't dissolve in minutes, but instead develops a significant constant voltage where the metal is negative and the water is positive. (And, if you could somehow reduce this voltage, the corrosion process would run wild, and the metal would rot away as you watched. Or if you INCREASED the voltage instead, you could force any ions already in the water to plate back onto the surface, and the hunk of metal would start growing larger.)

What then is a battery? If you put two hunks of copper in a bowl of water, both hunks charge up to the same negative volts (a few volts wrt the water,) so if you touched them together, nothing would happen. Ah, but different types of metal will spontaneously charge up to different voltage. Touching them together then decreases the water-metal voltage at one surface, and increases it for the other. Try copper and zinc. WOW! They get hot, and one hunk starts dissolving furiously, while the other starts growing larger.

This would be very magical if noticed in centuries past, since the metal-eating reaction only happens when the two metal objects bump against each other ...or when both are touched against a 3rd piece of metal.

share|improve this answer

1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output? Intuitively it will cause twice the amount of Ions bump into plates and give their charges to them, so it would double the current and won't change any "charge pressure", I can think of.

Remember that the potential is energy per charge. So it doesn't what the absolute number of ions bumb into the surface, the thing that matters is only the energy given per electron. Therefore if you make the area twice as large, you don't get twice as much energy given to each electron, you just get twice as many electrons with the same energy. So you get twice the current, but the same potential.

2) If I double the concentration of each solution, would it double the current or charge. My intuition that it is the same as if I doubled the area of plates - it would just double the amount of ions transferring their charge per moment of time. Nernst equation suggests a change of voltage in this case, but I can't see what it is.

You need to be careful with your intuition when considering this. The potential is dependent on what goes on at the surface which isn't just linearly linked to the bulk concentration. If you for instance have a surface with a fixed low number of catalytic sites, once you move beyond a concentration that allows all of them to be occupied, you might not see that much change in current as you increase concentration. Also you need to remember that a solution with double the concentration isn't just something with double the amount of ions, you have a change in the solvent/ion ratio, which may cause other things to take place, perhaps changing how ions are organised on the surface.

share|improve this answer
    
I agree about the first point. I disagree about the second. Nernst equation holds even for low concentrations of solutions (e.g. micromole/litre is a typical concentration of a substance in living cell and Nernst equation is applied to it), so that interactions between ions can be neglected and doubling/halving its concentration won't have any spatial effect. –  Bob Nov 12 '12 at 20:06
    
@Bob You seem to be confused. Are you trying to because there does exist a special case where the concentration differences will obey Nernst equation that my assessment that you cannot assume a general linearity between potential and concentration is wrong? If that is the case, please remember that Nernst equation is not a linear relationship firstly, secondly, even if it was, you could not argue against a trend not being general true with one special case where it is. eg. arguing against; All humans are not male, with: yes they are, this one is! –  Vincent Nov 13 '12 at 11:30

1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output?

No, the voltage is dependent on the chemistry. "D" and "AAA" sized zinc-carbon cells both produce about 1.5 volts.

Bigger batteries can produce larger currents. This is presumably because a larger electrode surface area allows for a greater number of ions to be reacted with.

If I double the concentration of each solution, would it double the current or charge.

It would increase the charge. Empirically you measure the charge in a lead-acid battery by measuring the specific gravity of the electrolyte. There are limits to how concentrated a solution can be. A fully-charged lead-acid battery contains sulphuric acid, a fully discharged lead-acid battery contains water.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.