Why doesn't electrode potential depend on amount of metal?

[D O]

When a metal (electrode) is in contact with water, an equilibrium forms:
Mg ⇔Mg²⁺+e^-
Adding more of the metal Mg should convert some of it into more electrons so that the ratio of Mg to Mg2+ remains the same (as the equilibrium requires).
Therefore twice the amount of metal in contact with water will lead to twice the number of electrons.
Electric Potential Energy is the amount of work done by a test charge coming from infinity distance to a certain point - it is given by kqQ/d^2 where k is 1/4πε0. This is proportional to Q, which would be the charge on the metal (which is constant once dynamic equilibrium has been reached), so the electric potential energy depends on the amount of metal initally added.

I don't understand why the electrode potential doesn't depend on the amount of metal added - where am I going wrong? Surely the potential difference between two points is the work done by 1 coulomb of charge moving between these points, which would depend only on the electric potential energy, which as shown above does depend on the amount of metal.

Thanks,
Dan

 

[gleem]

What you are also asking is similar to why doesn't 2 kg of Mg burn twice as hot as 1 kg.. Clearly you get twice the heat but not twice the temperature.

Work and therefore energy depend on the quantity of charge being move as well as the potential difference. Since twice the amount of a material supplies twice the charge, the work done or energy used is therefore doubled as we would expect keeping the potential difference constant.

 

[Andy Resnick]

Similarly, the OP is asking why a AAA battery has the same voltage as a D, even though the D cell is much larger. The voltage is the same (given by electrochemical properties), but the number of available charges is not.

 

[D O]

But surely you can have a potential difference without any charge being moved (very high resistance circuit)? I don't understand how voltage depends on the amount of charge being moved (in what time?)[PLAIN]http://en.wikipedia.org/wiki/Test_charge[/PLAIN]

 

[NascentOxygen]

Adding more of the metal Mg should convert some of it into more electrons so that the ratio of Mg to Mg2+ remains the same (as the equilibrium requires).
Therefore twice the amount of metal in contact with water will lead to twice the number of electrons.

An analagous scenario would be:

If you have stirred more sugar into a glass of water than will dissolve at equilibrium, will the addition of still more sugar cause more to dissolve?

 

[gleem]

But surely you can have a potential difference without any charge being moved (very high resistance circuit)? I don't understand how voltage depends on the amount of charge being moved (in what time?)

Yes you can. The potential difference does not depend on the amount of charge moved but on the material that is used. Some elements as zinc have electrons that are not as tightly bound as say for example Copper. If an electrode of each metal are placed in an electrolyte containing ions of the metals and connected by a wire current will flow as electrons from the zinc will travel to the copper. Zinc from the electrode will become positively charged and dissolve in the electrolyte while the extra electron flowing to the copper electrode will reduce the copper ions into new metal on the copper electrode. It is as if the copper were oxidizing the zinc. This is chemical energy which can be equated to electrical energy in the form of the current generated and an inherent potential difference between the two metals.

Older physics text often describe the chemical oxidation processes ( heat) in terms of electrical energy by reducing an oxidized material using electrolysis. For example. Water is formed by the oxidation (burning) of hydrogen which releases 69 kcal/mole. You can reverse this i.e. dissociate the water or reduce the hydrogen with electrolysis. If we believe in conservation of energy then the energy used in electrolysis to split water into Hydrogen and Oxygen must equal to the energy released in the burning of hydrogen. This comes out to be 3.0 eV/molecule. Where an eV (electron volt) is the energy gained by an electron traveling between potential difference of 1 volt. and thus the origin of the voltage associated with various materials. In the real world the only thing that is important is potential differences so using hydrogen as a reference (Set= 0 volts) The potential difference of metals can be determined as this list is called the Electrochemical Series.with Lithium being the lowest a -2.96 V and Gold the highest at +1.36 V relative to hydrogen.

 

[D O]

Thanks for the responses.
The way I'm visualising the electrons produced on the electrodes is as spherical concentrations of charge; the potential is given by kQ/r where Q is the charge on the electrode and r is the distance to the electrode; as I understand it the amount of charge on the electrode Q is what matters to the potential, not the charge density, so wouldn't twice as much charge on the electrode produce a potential of double the magnitude?

Am I using the wrong expression for electrical potential (V= kQ/r)? If not I can't understand why a higher Q (charge on the electrode) would not provide a higher V.

 

[willem2]

Am I using the wrong expression for electrical potential (V= kQ/r)? If not I can't understand why a higher Q (charge on the electrode) would not provide a higher V.

You're indeed using using the wrong expression. A battery would be more like a parallel plate capacitor, and the capacitance would rise if you made the plates bigger, so it could still have the same potential, even with a bigger charge.

No matter what the geometry of the plates is however, the potential will be the same anyway, because the reaction will run until the potential produced by the charge on the plates is equal to the chemical potential of the battery.

The only thing that can affect the chemical reaction somewhat is the concentration of the reagents, but the concentration of the metal and the electrons is really constant here because:
- if you insert more metal in the electrolyte, it will touch more electrolyte as well
- the extra amount of electrons because of the charge on the plate is tiny compared to the total amount of conduction electrons.
The only concentration that can affect the reaction rate is that of Mg²⁺. If there's a lot of it around the cell voltage wil be lower. Look up the Nernst equation.

 

[gleem]

Am I using the wrong expression for electrical potential (V= kQ/r)? If not I can't understand why a higher Q (charge on the electrode) would not provide a higher V.

AS @willem2 has stated you are using the wrong formula to try and analyze a battery. The potential difference between the electrodes exists like the potential difference between the top of a hill and the bottom in a gravitational field. But the actual description of the forces as a body rolls down the hill is not given simply by the fundamental potential expression GM/r as it rolls, bounces etc down the hill. although the gain in energy (neglecting friction of course) is easily understood.

The other point is that for a battery the net charge of the system is zero right? for every electron there is a proton. So a battery is not like a capacitor. What is happening ( and this explanation is a bit clumsy ) is that electron must be freed from one electrode to be "pulled" to the other and absorbed but the pull is not just an easily described electric field between the two electrodes although we know there is some field at every point in the system to move the electrons around the circuit. It turns out that the actual emf of a cell is different than one would expect from the electrochemical series differences. On discharge a certain amount of energy is used or liberated in purely chemical processes. The microscopic description is complex and needs thermodynamics to work through the process. As often happen a seemingly simple question has a very involved explanation.