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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.

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions.

A ---> A(+1) + e

B ---> B(2+) + 2e

Suppose these two processes have the same tendency ( it is just as energetically feasible for both to form their respective ions) So in a half-cell of B and its ions, plate B should be more negatively charged compared to plate A in its own half cell. Charge should flow from B to A in the external circuit. The electrode potentials will be different when measured practically. I think this is all true and fine. Now, we can use electrode potentials to tell us about the relative ability of a metal to lose or gain electrons. But in this case, the difference isn't due to a tendency difference: only the number of electrons is different.

How does this line up with the oft-taught idea that we can use reduction potentials to judge the ability to gain or lose?

I am aware I've asked a similar question, but apparently, I did not realize my wording was off and then it was too late to edit it all out

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closed as unclear what you're asking by Mithoron, Zhe, A.K., Todd Minehardt, Mathew Mahindaratne Apr 9 at 20:01

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  • $\begingroup$ Well please conform to standard terminology. (1) Write the equations as reductions since all electrode data tables are formatted that way. (2) What do you mean that when you say "Suppose these two processes have the same tendency." -- Do you mean the two half cells have the same half cell potential. $\endgroup$ – MaxW Apr 7 at 16:16
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    $\begingroup$ Please revise your question. "Suppose these two processes have the same tendency." Tendency to do what? "So in a cell, plate B should be more negatively charged. Charge should flow from B to A in the external circuit. The electrode potentials will be different when measured practically. I think this is all true and fine. " This is still not correct. $\endgroup$ – M. Farooq Apr 7 at 16:25
  • $\begingroup$ @Max, no that is exactly what I don't mean. . $\endgroup$ – Sal_99 Apr 7 at 16:49
  • $\begingroup$ @M.Farooq, as in it is just as energetically feasible for both to form their respective ions. $\endgroup$ – Sal_99 Apr 7 at 16:49
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    $\begingroup$ Also note that potential has nothing to do with charges on electrodes in a half reaction. Do not confuse the potential of a half-reaction as a direct measure of the spontaneity of the half-reaction. $\endgroup$ – Zhe Apr 8 at 0:51
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Let's write the half-cell equations in standard form as half-cell reductions.

In a table of standard electrode potentials it would silly to write both the reduction and oxidation reactions since that would needlessly double the size of the table.

$$ \begin{align} \ce{A+ + e- &-> A} &\quad V_\ce{A} \\ \ce{B^{2+} + 2e- &-> B} &\quad V_\ce{B} \end{align} $$

Now this assumes that each half cell has exactly a 1 molar solution of the cation. If the concentrations are not exactly 1 molar, then the voltage of the half-cell will vary according to the Nernst equation.

Now let's assume that we have used the Nernst equation and calculated the appropriate values for each half cell as $V^*_\ce{A}$ and $V^*_\ce{B}$. Now the values of $V^*_\ce{A}$ and $V^*_\ce{B}$ can either be positive or negative. But let's just convert the chemical equation for B to an oxidation and get the overall reaction;

$$\ce{2A+ + B -> 2A + B^{2+}}$$

The EMF for that reaction will be $V^*_\ce{A} - V^*_\ce{B}$.

  • If $V^*_\ce{A} - V^*_\ce{B} > 0 $ then the spontaneous reaction of the cell will be

$$\ce{2A+ + B -> 2A + B^{2+}}$$

  • If $V^*_\ce{A} - V^*_\ce{B} < 0$ then the spontaneous reaction of the cell will be

$$\ce{2A + B^{2+} -> 2A+ + B}$$

In either case the reaction will proceed until $V'_\ce{A} = V'_\ce{B}$ at which point no current will flow between the half cells. ($V'_\ce{A}$ and $V'_\ce{B}$ being calculated by Nernst equation again with final concentrations of $\ce{A+}$ and $\ce{B^{2+}}$.)

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  • $\begingroup$ I agree with everything you've said.That makes perfect sense. But my question is more about using these Voltage values to comment about the relative ease of forming ions for either. Let's say the second case in your example is what happens at 1M conc for both. My question is does this happening tell us anything about the relative ease with which A or B form their ions? $\endgroup$ – Sal_99 Apr 7 at 17:39
  • $\begingroup$ "The relative ease" is an ambiguous phrase. If the metal ion concentrations are exactly 1 molar then the EMF of each cell will start out at the reduction potential as listed in a table of standard reduction potentials. Then which reaction happens depends on which is greater $\ce{V_A}$ or $\ce{V_B}$. $\endgroup$ – MaxW Apr 7 at 17:45
  • $\begingroup$ I'll be more honest, but I'm risking confusing you totally. I wrote that the cells should have a concentration of exactly one molar. That isn't true. Rather the standard half cell potential is based on an activity of exactly 1. The activity is influenced by not only the concentration of the particular ion in the half-cell reaction, but also for example by the ionic strength of the solution. So a 1 molar solution of $\ce{BCl2}$ would have a different ionic strength than a solution of 1 molar $\ce{BSO4}$. Hence the activity of $\ce{B^{2+}}$ would be slightly different in the two solutions. $\endgroup$ – MaxW Apr 7 at 18:04
  • $\begingroup$ @Sal_99 I guess in part what I'm trying to say is that there is no function in chemistry to calculate relative ease. You have to be specific about what behavior you're trying to model mathematically. $\endgroup$ – MaxW Apr 7 at 18:14
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    $\begingroup$ @andselisk - Thanks for the edit. That did clean up the answer. $\endgroup$ – MaxW Apr 8 at 1:01
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The main thing to point out is that your question doesn't actually make sense the way you want it to.

What does it mean if a reduction potential is zero? Or positive? Or negative? Is that favorable or not? It turns out we don't specify this at all!

As an example, consider the standard hydrogen electrode and it's reduction potential:

$$\ce{2H+ + 2e- -> H2},\quad \mathscr{E} = 0\ \mathrm{V}$$

Is this reaction on the borderline between spontaneous and nonspontaneous? I don't know the answer to this question, and importantly, this is the wrong question to ask. Why? Because we're doing electrochemistry by balancing an oxidation and reduction reactions. The individual potentials don't matter.

Take a reduction: $$\ce{2A+ + 2e- -> A2},\quad \mathscr{E}=V_{A}\tag{1}$$

Its potential is relative to that of the standard hydrogen electrode. Can you tell from this value if it will be reduced or not? No, you cannot! What you can tell is its tendency to be reduced relative to the standard hydrogen electrode.

Let's examine why everything works.

Take this reaction: $$\ce{2A+ + H2 -> 2H+ + A2},\quad \mathscr{E}=V_{A}\tag{2}$$

And if we have the reduction $$\ce{B+ + e- -> B},\quad \mathscr{E}=V_{B}\tag{3}$$

Then we can also get: $$\ce{2H+ + 2B -> H2 + 2B+},\quad \mathscr{E}=-V_{B}\tag{4}$$

Combining $\text{(3)}$ and $\text{(4)}$ and cancelling the hydrogens, we get: $$\ce{2A+ + 2B -> 2B+ + A2},\quad \mathscr{E} = V_{A}-V_{B}\tag{5}$$

This is exactly the same answer as taking the two reductions half reactions, converting one to an oxidation, and summing the half reactions.

Now, consider reaction $\text{(2)}$ again. There is a free energy change associated with this reaction that is:

$$\Delta G = -nF\mathscr{E}$$

Note that this says absolutely nothing about the free energy change of equation $\text{(1)}$ because we have no idea what the free energy change for the reduction hydrogen is. That is:

$$\Delta G = \Delta G_{\ce{A+|A}} - \Delta G_{\ce{H+|H2}}$$ $$\Delta G_{\ce{H+|H2}} =\ ?$$

But the point here is that we don't care because we were able to obtain equation $\text{(5)}$ without needing this information.

As an exercise, you can confirm that the total free energy change is indeed the expected value because $\Delta G_{\ce{H+|H2}}$ cancels from the sum of $\text{(2)}$ and $\text{(4)}$.

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    $\begingroup$ +1 - Very well stated. When doing electrochemisty focus on the cell EMF. $\endgroup$ – MaxW Apr 8 at 0:57
  • $\begingroup$ @Zhe, okay. So everything in electrochemistry is a comparison basically. Nothing absolute. My book says "the more negative the value of the standard electrode potential, the greater the tendency to give up electrons". I understand that they were all compared to SHE and then put into an order. My core concern while phrasing this question was that should we care about the fact that some metals from +2 ions, some +1, etc? $\endgroup$ – Sal_99 Apr 8 at 9:37
  • $\begingroup$ @Zhe, this became a point since I was introduced to this concept using an equilibrium between the metal and its ions. I was then told that those that have a higher tendency to give up electrons will have equilibria lying to the left, and for the others equilibria will be to the right. Then I was told that this results in a charges on the metal. Then it was that the metals with higher tendency to lose electrons will have comparatively more negative plates than their counterparts. Now. Two things with the same position of equilibrium but 1 losing 2 electrons, the other just losing 1. $\endgroup$ – Sal_99 Apr 8 at 9:56
  • $\begingroup$ How will they compare? $\endgroup$ – Sal_99 Apr 8 at 9:56
  • $\begingroup$ You've created more problems for yourself than you need to. On some level what you say is true because second ionization energies are higher than first ionization energies, but that's only the enthalpic component, and it's measured in the gas phase. We're in solution, so you need to account for solvation enthalpy and solvation entropy. That's extremely non-trivial. In order to not go crazy with those details, we abstract out what is important: spontaneity. And we use potentials and free energies to quantify it. @Sal_99 $\endgroup$ – Zhe Apr 8 at 13:34
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Quoting your comment

A better way to say it maybe that the two form ions just as readily. Then why should electrode B itself not be charged comparatively more negative?

Eo is not a direct measure of the tendency but rather Gibbs free energy. They are related as Delta Go= -nFEo; Write it the other way, you get

Eo = Delta Go/(-nF)

did you notice that, the number of electrons n, is already factored in? No matter how many electrons are gained or lost Eo is oblivious of the representation:

B(2+) + 2e = 2B 1/2 B (2+) + e = 1/2 B; all have the same electrode potential.

You are imposing a condition (1) now that there is a divalent metal B and a monovalent metal A. You have two beakers, one contains A dipping in A(+), and the other beaker containing B dipping in B(2+). We have not connected them as yet via an external circuit. Assume we have a electrostatic charge meter. You can test A and B separately. As per your logic, B should be more "electrostatically negative" than A because B is divalent. Let us assume this is true for a moment. The moment you complete the circuit, electricity should flow from B to A, because B had more electrostatic charge.

Now you impose another condition (2), which is quite ambiguous, i.e. both have the same tendency to "form ions just as readily". Translating this mathematically,

Delta Go values of the half cell with respect to SHE, A(+) + e -> A and B(2+) + 2e -> B are numerically the same. Hence Ecell = zero, which implies that there must be no charge build-up on B with respect to A.

Do you notice that both (1) and (2) are mutually incompatible?

Also think of the pair Zn(2+)/Zn and Cu(2+)/Cu, by condition (1), this cell should not work because both metals gain or lose equal number of electrons. In reality this pair works perfectly because the condition no 2 is not imposed on them.

It is good to have a thought process, however, when it defies the experiment, chances are that the line of reasoning is not correct. As I.M. Kolthoff used to say "Theory guides, experiment decides".

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  • $\begingroup$ Unfortunately, this is not correct. You cannot translate a reduction potential to a simple $\Delta G$ because everything is referenced against the standard hydrogen electrode, and the $\Delta G$ you compute lost the value of the free energy change from oxidation in the standard hydrogen electrode. As a concrete example, what is the free energy change associated with the reduction of a hydrogen ion to hydrogen gas? This process does not have $\Delta G = 0$. $\endgroup$ – Zhe Apr 7 at 23:55
  • $\begingroup$ You raised a good point. We should always specify "with respect to" for Eo and delta Go. For the OP see chemistry.stackexchange.com/questions/4355/… $\endgroup$ – M. Farooq Apr 8 at 3:41
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I was introduced to this concept using an equilibrium between the metal and its ions. I was then told that those that have a higher tendency to give up electrons will have equilibria lying to the left, and for the others equilibria will be to the right. Then I was told that this results in a charges on the metal. Then it was that the metals with higher tendency to lose electrons will have comparatively more negative plates than their counterparts. Now. Two things with the same position of equilibrium but 1 losing 2 electrons, the other just losing 1.

@Sal_99, You are trying to quantify the word tendency. If someone says that this vegetable soup tastes better than that chicken soup, how can we quantify this tendency to prefer vegetable soup? It is a qualitative descriptor at this end. Don't use electrode potential to correlate the amount of charge on accumulated an electrode at this moment and also don't impose the condition that both A(+) and B(2+) have the same tendency to lose electrons. Potential is work done per unit charge.

Now imagine another thought experiment, two metallic rods, A and B dipping separately in two beakers of A(+) and B(2+). They are not connected and the metal is in equilibrium with its own ion. Assume we have a charge meter (they exist in reality and they have two leads) and we can measure the amount of charge by connecting one lead to the test electrode the other end to the ground. In a pure electrostatic sense, if one A(+) ion is reduced or one A is oxidized at the A electrode, similarly, one B(2+) is reduced or one B is oxidized at B electrode, then the amount of charge developed on B will be higher. We cannot guess the sign of A or B with this thought nor we can say anything about this electrode potential, however, you are right in this pure electrostatic sense.

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  • $\begingroup$ I went over what @Zhe and you said. I understand this now: The discussion with redox potentials is always in comparative terms. We use the SHE as a common halfcell to compare everything else to. The more negative E values simply suggest that that half cell has relative tendency to be oxidised when hooked up to the SHE. This is simply from energetics. Now about electrostatics: e.g. a half cell with negative E value hooked up to a SHE. Electrons flow from this half cell to the SHE. This must mean that the electrode is more negative (or less +ive) than the electrode in SHE. $\endgroup$ – Sal_99 Apr 11 at 3:19
  • $\begingroup$ One can rougly deal with this with position of equilibria and "tendency" to lose electrons ( @M.Farooq, you got exactly my mind's picture in your last para), but this is not a rigorous approach and should not be expected to stand rigorous critique since it is not really defined well. I use 2 texts and they both take this appraoch. Maybe it's simple and builts some intuition. But I now realise that I was attempting to define tendency on my own and then question theory. But why is it that the half cell that is relatively going to be oxidised always has an electrode more negative -or less +ive? $\endgroup$ – Sal_99 Apr 11 at 3:24
  • $\begingroup$ I can't seem to find a rigorous answer to this. Maybe dG is a better way of thinking what E values tell us ( not absolute dG, since we don't know the value for SHE) And now there is one thing I would like clarfication on: this equation linking E to G was new to me.Suppose we have the metals A and B as in my question. Suppose each half reaction has the same dG. So the dG when combined with SHE should come to be the same. How do I use this equation to show how the E s wil compare? $\endgroup$ – Sal_99 Apr 11 at 3:29

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