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Standard electrode potential values can be used to judge the reducing ability of a metal. When I was introduced to this topic a simple picture was painted: you put a metal into a solution of its ions. An equilibrium is set up. You do so for another metal. If you now hook up the two electrodes and measure the voltage, you can tell which one is 'more negative' hence more electron donating. I was taught like it is covered here : https://www.chemguide.co.uk/physical/redoxeqia/introduction.html

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals?

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    $\begingroup$ You just changed the question completely. Please revert back to the original question and ask another. $\endgroup$
    – MaxW
    Commented Apr 7, 2019 at 16:01

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My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.

That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:

$$\ce{Ag+ + 1/2Mg -> Ag + 1/2 Mg^2+}$$

has the same cell potential as

$$\ce{2Ag+ + Mg -> 2Ag + Mg^2+}.$$

(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).

How has the voltage helped us determine the difference in the tendencies of these two metals?

What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.

You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.

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  • $\begingroup$ "That is not a problem because the tendency to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. " If one metal gives up electrons more easily than another metal, does that not have a bearing on their comparative reduction potential? This seems to be in contradiction to the link I've put in the question. $\endgroup$
    – Sal_99
    Commented Apr 7, 2019 at 14:29
  • $\begingroup$ @Sal_99 - "If metal A gives up one electron more easily than metal B" has bearing on it, "If metal A gives up two electrons more easily than metal B gives up one electron" has not bearing on it. $\endgroup$
    – Karsten ♦
    Commented Apr 7, 2019 at 14:33
  • $\begingroup$ I added a bit to that section because it was not clear. $\endgroup$
    – Karsten ♦
    Commented Apr 7, 2019 at 14:49
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    $\begingroup$ @KarstenTheis -- In your chemical formulas you forgot the 2+ exponent on Mg on the right hand side. $\endgroup$
    – MaxW
    Commented Apr 7, 2019 at 15:17
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    $\begingroup$ This is a great discussion. I would only add that reduction potentials are stated in V, which is J/C, an intensive property of the half cell. This may help in clearing up the “per electron” bit. It is per electron, but only because an electron’s charge is a fixed amount in coulombs. $\endgroup$
    – Withnail
    Commented Apr 7, 2019 at 18:28
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My question is that suppose we have to metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals?

The premise of this question is incorrect. As per your assumption, the electrode potential of the half cell as shown in tables in reduction form A(+) + e- -> A and B(2+) + 2e -> B is identical. So far this is fine. Let us see the concept of electroneutrality.

The gist is that overall there is no net charge in a galvanic or electrolytic cell. Therefore, if one B(2+) ion gains two electrons, two A atoms must lose electrons according to the equation 2A -> 2A+ + 2e in order to maintain electroneutrality.

This is true the other way round as well. So it is basically electron book-keeping to avoid net charge build up in a cell. Keep in mind that the interfaces (electrode surface and solution boundary) are electrostatically charged.

Coming to your voltage part of the question: The Ecell is defined as Ecathode-Eanode, if they are same numbers, Ecell = 0 V.

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  • $\begingroup$ yes in net charge is 0. But why does this mean plate B on its own cannot be more negatively charged than plate A? $\endgroup$
    – Sal_99
    Commented Apr 7, 2019 at 14:26
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    $\begingroup$ If B on it its own is more negatively charged than A, and they are connected, it means current must flow in the external circuit. From your very first condition that electrode potentials are equal, this is incompatible with the first statement under standard conditions. $\endgroup$
    – ACR
    Commented Apr 7, 2019 at 15:17
  • $\begingroup$ yes I see that now. $\endgroup$
    – Sal_99
    Commented Apr 7, 2019 at 15:34
  • $\begingroup$ please see this better-worded problem chemistry.stackexchange.com/questions/112316/… $\endgroup$
    – Sal_99
    Commented Apr 7, 2019 at 16:08
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Which plate would be negative depends on concentration of ions, as the plate with 1 electron transfer would have double rate of the potential change with the ion concentration at the electrode.

But if at particular concentrations both electrodes have the same potential, the cell voltage will be zero and neither of plates would be negative wrt the other.

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  • $\begingroup$ If both are put into a 1 M solution, how would it work? And what does "rate of potential change mean"? $\endgroup$
    – Sal_99
    Commented Apr 7, 2019 at 10:59
  • $\begingroup$ If the standard potential is the same ( we would then need activities, as activity of multicharge ions decrease faster), the the potentials would be the same. $\endgroup$
    – Poutnik
    Commented Apr 7, 2019 at 11:03
  • $\begingroup$ Rate of potential change with concentration change. $\endgroup$
    – Poutnik
    Commented Apr 7, 2019 at 11:03

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