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EDIT: The remainder was written after OP provided the link.

So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Applying this to the first law, some of the energy input for the reaction can be in the form of heat, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.

So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).

Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though).

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:

$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$

So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Applying this to the first law, some of the energy input for the reaction can be in the form of heat, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.