Accelerated corrosion?

Hello,
when building a simple voltaic cell, made of two common electrodes in a solution of their ions (let's take Cu/Cu2+ half cell coupled with a Zn/Zn2+ half cell), and we close the circuit, electrons will flow from the half-cell with lower potential (Zn/Zn2+) to the one with higher potential (Cu/Cu2+), effectively leading to the corrosion of Zn and reduction of Cu2+.

This reaction will have its own rate of reaction and once the Zn electrode is consumed, the voltaic cell is fully spent.

What I was wondering was: what would happen if we connected a battery in series with this voltaic cell ? Would the extra potential accelerate the consumption of the Zn electrode, effectively accelerating the rate of corrosion of Zn? Or that extra potential would not affect this rate of reaction? Would this be equivalent to shorting the battery, or the voltaic cell still somewhat behaves like a resistance?

In my head I pictured the extra potential "helping" push the electrons across the two half-cells, therefore accelerating the rate of reaction. On the other hand, I'm inclined to think that the increased potential difference would only affect the thermodynamics of the process but not the kinetics.

I drew a sketch of the two different systems

Quote: Originally posted by Rainwater

Depends on how you connect it. With the positive of the cell connected to the positive of the battery. This is the same as connecting 2 batterys in parrallel, their. Charges will flow current and attempt to equalize. With the positive of the battery connected to the negitave of the cell, your battery/cell becomes the load. Maxium current will be drawn and your battery and cell will discharge [Edited on 19-1-2024 by Rainwater]

Thank you!

Yes, the second case is what happens in electrolysis, where you spend energy to make non-spontaneous reactions happen (like when you want to reduce otherwise irreducible species like Li+, or electroplate something or to recharge batteries), but in this case I was wondering what would have happened in the other case, where you double down on the potential to further accelerate an already spontaneous reaction.

About the positive/negative connections in my drawing, the Zn anode is where electrons are produced and sent to the Cu cathode which happily accepts them to reduce the Cu2+ to Cu. So the positive end of the battery should pull those electrons "better" and accelerate them towards the Cu electrode which should be positively charged, that's the intention. I also get confused with the conventional currents directions, that's why I'm speaking in terms of electrons moving (the conventional current would have the opposite sign otherwise).

In such case, the schematic would be equivalent to two batteries connected in series with no apparent resistance, so this "device" should just discharge at a rate dictated by.... by what? In my head, it's like I'm shorting these 2 batteries, but maybe the redox rate is so slow that rapid discharge and heating up is avoided.

Or maybe, since the ions concentrations at a given time depend on the amount of electrons (and therefore of charge) that has flown in the circuit, the rate of reaction is more dependent on the current intensity rather than the total voltage.

In this case, would connecting the battery in parallel increase the current intensity in this circuit? Or would this just result in the battery bypassing the voltaic cell and just shorting itself?

[Edited on 22-1-2024 by Metallus]