cell voltage question

[theonlytycrane]

The answer given is (B). I'm looking for clarification on the reasoning behind choices I and III.

I) Even if the quantity of Zn(s) is doubled, equilibrium will eventually be restored such that the cell voltage stays at 1.5V?

III) Na(s) will ionize easier due to a lower Zeff which will result in a larger voltage than 1.5 at equilibrium?

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[Astra]

Lets look at the equation for voltage.

V = kQ/r

k: constant
Q: charge
r: distance


Increasing the quantity of Zn doesn't cause the charge of each particle to increase does it?

It simply means that the battery can operate for longer by supplying a larger number of charged molecules.



Lets look at Na.

The trend for ionization energy is that it increases as you go up and to the right of the periodic table.

Na is only 1 above but Zn is several columns to the right so Zn will have a larger ionization energy.


If we compare Na and Zn ionization in terms of spontaneity, Na is more spontaneous due to having a smaller amount of energy input required.

If we look at the half reaction for Na oxidation then, it will be more spontaneous, and there fore have a higher positive voltage ( more positive voltage, more spontaneous)

The cathode doesn't change but the anode gains a more positive voltage. Therefore, it increases the net voltage difference.

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The answer given is (B). I'm looking for clarification on the reasoning behind choices I and III.

I) Even if the quantity of Zn(s) is doubled, equilibrium will eventually be restored such that the cell voltage stays at 1.5V?

III) Na(s) will ionize easier due to a lower Zeff which will result in a larger voltage than 1.5 at equilibrium?

 

[aldol16]

I) Even if the quantity of Zn(s) is doubled, equilibrium will eventually be restored such that the cell voltage stays at 1.5V?

One key thing to remember is that a cell potential is an intrinsic property that does not change with the quantity of material you have present. Think of it like a waterfall. No matter how much water you have on one side of the waterfall, it's still gonna fall down the same height.

III) Na(s) will ionize easier due to a lower Zeff which will result in a larger voltage than 1.5 at equilibrium?

Zn is placed at the anode, which tells you that it will be oxidized. So Zn --> Zn2+ most likely. Now, how would you make the reaction more spontaneous and therefore have a greater cell potential? Well, you replace Zn with something that wants to be oxidized even more. Na(s) has only one valence electron and thus really wants to be oxidized. Thus, its redox potential will be lower, making the cell potential higher. This is why Na(s) is almost never found in nature.