Reduction Potentials [closed]

Question

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

Answer 1

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 $\ce{B^{2+}}$.)

Answer 2

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 $\text{(4)}$.

Answer 3

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".

Answer 4

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.