In a galvanic cell, why does an electrode with a higher potential "pull" the electrons from the electrode with a lower potential?

Question

I'm doing Grade 12 Chemistry and I'm unsure of this, taking the hydrogen electorde as having a potential of 0 volts.

I'll use the copper-zinc cell as an example. Just let me know if I've got the full logic right:

1. Zinc has higher electronegativity, so due to the equilibrium $\ce{Zn <=> Zn^{2+} + 2e-}$, we have more charge building up in the Zinc electrode than in the Copper electrode. (Given both half-cells are of same concentration and in same temp)

2. We have a potential difference between the two electrodes. This means some charge from the Zinc electrode will move to the Copper electrode in such a way that both electrodes will momentarily have equal charges.

3. Since the charge in the Copper electrode has gone up, we have added electrons to the equilibrium reaction $\ce{Cu^{2+} + 2e- <=> Cu}$. We therefore create more copper.

Is this why an electrode with a higher potential will "pull" the electrons from the electrode with a lower potential?

Answer 1

There are not many problems where electronegativity helps you find the answer. It is popular for some reason amongst newer chem students but it shouldn't be.

Here's what happens in the Daniell cell:

Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons. Both of these processes are endothermic.

Zinc ions are hydrated. This is exothermic.

Copper ions are dehydrated. This is endothermic.

Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.

Each of the above processes also involves an entropy change.

Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.

By the way, silver's electronegativity is 1.93 but the reduction potential for $\ce{Ag+}$ is $\pu{0.8 V}$ while $\ce{Cu^2+}$ is $\pu{0.34 V}$.

In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.