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In a galvanic cell, current is driven because the species at the cathode is reduced by grabbing some electrons from the cathode. The cathode now lacks electrons so it pulls them from the anode. Now the anode needs electrons, so it pulls them from the species that will be oxidized in the other half cell. So, now we have a current!

Now here is a concentration cell:

enter image description here

As you can see. the anode and cathode are made of the same metal. Only the concentrations of the solutions are different. Using my logic from above, since the metals are the same, they should be pulling on the electrons with equal magnitude. So how come current still flows? In other words, why are the nickel ions at the cathode more "hungry" for electrons than the nickel ions at the anode?

My attempt: It has to do with the rates of the metal dissolving. If I were to cut the wire connecting the two half-cells, I would notice that the beaker with the lower concentration solution would be increasing in concentration (from solvating the metal) faster than the beaker with the higher concentration. This is because the lower concentration one is "further" from the solubility equilibrium.

When I connect them back together the metal at the anode dissolves faster, thus demanding electron recompense from the cathode metal.

Attempt 2: The solution with higher Ni ion concentration has overall more ions dissolved in it. Maybe all these ions create an electric field greater than the one on the other side, thus prompting electron movement.

Attempt 3: It has to do with kinetics. Lets assume that everytime a nickel cation collides with an electrode, it steals some electrons. The half-cell with the higher concentration will have a higher amount of collisions per second. Thus, more electrons are being "consumed" at the cathode than anode per second, creating a charge differential, which in turn generates a current.

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  • $\begingroup$ @EdV Thanks for the answer. Do you know if my physical explanation (difference in solvation rates) was correct? Don't want to be building on faulty premises. $\endgroup$ – Nova Jul 27 '20 at 0:11
  • $\begingroup$ Actually, when the Nernst equation is applied to more concentrated solutions, the reaction quotient Q should be expressed as 'effective concentrations' or in terms of the 'activities of the electroactive ionic species'. So, adding MgSO4 to one of the two Nickel sulfate ion concentration cells may alter the solution's activity and be reflected in a voltage change (an interesting experiment). $\endgroup$ – AJKOER Jul 28 '20 at 19:48
  • $\begingroup$ Note, per my comment, I would further comment that adding an excess of MgSO4 could salt out some of the Nickel sulfate, resulting in a significant change in dissolved Nickel. $\endgroup$ – AJKOER Jul 28 '20 at 20:05
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Start with a simple thought experiment: pour 100 mL of 1 M nickel (II) sulfate solution into a beaker and very carefully layer 100 mL of 0.01 M nickel (II) sulfate solution on top of the more concentrated layer. Then, even without convection or deliberate mixing, diffusion will, sooner or later, result in the solution having concentration of 0.55 M. In what follows, it is assumed that evaporation is negligible, even on long time scales.

Now consider Fig. 1 below:

Ni concentration cell

In this concentration cell, it is assumed, for simplicity, that the solution volumes are equal in the two electrode solution reservoirs. The salt bridge is 0.55 M nickel (II) sulfate solution, so its concentration is exactly half-way between that of the anode solution and cathode solution.

With the concentrations shown in the figure, the cell potential is +0.02958 V, and the anode is the nickel electrode in the more dilute solution at the left. On the cathode side, the log term is zero, in the Nernst ewuation at upper right in Fig. 1, so the potential is -0.236 V. But on the anode side, the potential is lower by 29.58 mV, due to the log term. So the anode is more negative, the cathode is more positive and electron flow is always from more negative to more positive. Therefore, Ni is spontaneously oxidized at the anode, yielding the necessary electrons, and nickel ions are spontaneously reduced at the cathode.

As a direct consequence of having an external connection, i.e., load or voltmeter, between the electrodes, net oxidation will spontaneously take place at the anode, resulting in an increase in the nickel ion concentration in that solution reservoir. Likewise, nickel ions in cathode solution reservoir will be spontaneously reduced at the nickel cathode and the nickel ion concentration will decrease. The external connection affords the opportunity for something useful to happen, i.e., using the cell as a power source. Otherwise, only diffusion happens (see below).

The cell will be "dead", i.e., have no more free energy to tap, when all three solutions are 0.55 M. There will be no concentration gradient left.

Now consider Fig. 2:

Open circuit Ni conc cell

This is the same as Fig. 1 except that the cell is open circuited, i.e., there is no way for electrons to transfer from one electrode to the other. So all that is going to happen is spontaneous diffusion through the salt bridge. Eventually, there will be just one final concentration at 0.55 M.

During this process, no net oxidation occurs at the left electrode and no net reduction occurs at the right electrode. If a voltmeter is subsequently attached, the reading will be less than +0.02958 V because the concentrations are no longer the original ones: both concentrations are closer, thanks to diffusion, to the 0.55 M center value.

The measured voltage, even using a high impedance voltmeter, is not actually the open circuit voltage. It may be very close, but there is a conceptual issue. In the open circuit situation, no electrons can flow between the electrodes and no net redox processes occur at the electrodes. So the open circuit potential is an ideal potential and the measurement goal is to estimate accurately the open circuit potential while allowing negligible current to flow.

With a high input impedance voltmeter attached to the electrodes, a very small current, i.e., nA or pA, can flow. This only slightly loads down ("perturbs") the cell potential, so it provides an accurate estimate of the true open circuit potential. And it means that oxidation occurs at the anode, to a very slight extent, and reduction at the cathode, likewise to a very slight extent. Solution concentrations are negligibly perturbed.

So there are two parallel processes taking place. First, diffusion spontaneously acts to equalize the solution concentrations. Second, the concentration cell provides a way of hurrying the equalization process along while extracting useful electrical energy. So the former wastes the available free energy while the latter extracts much of it.

Finally, consider Fig. 3 below:

Cannot work Ni conc cell

Now the salt bridge is gone and nothing happens: there is no DC cell potential and the solution concentrations are constant. In figures 2 and 3, there is no net redox happening: for every Ni atom that might get oxidized, there is a nickel ion that gets reduced. Everything is balanced at both electrodes. (So if the electrodes start with highly polished "mirror" finishes, they may get surface roughened by this zero net redox process.)

But once a load or voltmeter is attached to the two electrodes in Fig. 2, this opportunity causes the equilibria to become unbalanced: the solution concentrations must change by increasing (in the anode reservoir) and decreasing (in the cathode reservoir), and that immediately implies that electrons flow from the anode (at the left), through the load or voltmeter, to the cathode (at the right).

Perhaps the following will also help clarify matters. Consider Fig. 4 below:

Ni conc cell 4

This shows two beakers, each made of solid nickel, and containing the indicated nickel (II) sulfate solutions. Obviously, this does nothing: it is just two fancy beakers of solutions. Next consider Fig. 5:

Ni conc cell 5

Now the two nickel beakers are touching, i.e., in electrical contact. Does this change anything? No. Nothing happens. Next, consider Fig. 6 below:

Ni conc cell 6

Now there is simply a wide glass beaker with a solid nickel partition that entirely prevents the two solutions from making contact. Does this change anything? No. Nothing happens. Finally, consider Fig. 7 below:

Ni conc cell 7

This differs from the previous figure in having a salt bridge. It is exactly equivalent to using a nickel wire to short the two nickel electrodes in Fig. 2. So the cell is shorted out, there is no cell potential, and the cell will try to supply its short circuit (i.e., maximum) current, limited by the kinetics at the electrodes. So the nickel ion concentration will increase in the anode reservoir and decrease in the cathode reservoir, as expected.

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  • $\begingroup$ Hi Ed, this is a great answer, thank you. One thing I still don't understand is why the Nickel at the cathode is more likely to be reduced than the Nickel at the anode. The nickel cation floating around the cathode is "dumb" and can't sense the concentration of nickel in the other cell, yet it still is more likely to be reduced. I updated my answer with another guess that has to do with kinetics. $\endgroup$ – Nova Jul 27 '20 at 13:38
  • $\begingroup$ Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$ – Ed V Jul 28 '20 at 2:18
  • $\begingroup$ Up vote(+1), understandable. But, per a source, chem.libretexts.org/Bookshelves/General_Chemistry/… , "Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10–3 M.". Suggestion: reduce conc or introduce activity coefficent. $\endgroup$ – AJKOER Jul 28 '20 at 20:55
  • $\begingroup$ Thanks for the upvote and information! Funny thing, though: when I used to do the Daniell cell demo in lecture, I used 1 M copper sulfate and zinc sulfate solutions (they were already available in the lecture prep room) and the Nernst equation worked like a charm. I would routinely get +1.100 V or within 5 mV of that. This was one of the lecture demos I never had to worry would screw up at showtime! Still, in my overly long answer, I regret not discussing the (obvious) ion flows in the open circuit situation (Fig. 2) and in the externally loaded situation (Fig. 1). $\endgroup$ – Ed V Jul 28 '20 at 21:29
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The potential of each electrode is given by Nernst law:

$$E = E^\circ + \pu{0.0296 V}\cdot\log[\ce{Ni^2+}]$$

After my tables, $E^\circ(\ce{Ni^2+}/\ce{Ni}) = \pu{-0.23 V}.$

So, in the $\pu{1 M}$ solution, the potential of the nickel electrode $E = \pu{-0.23 V}.$

In the $\pu{0.001 M}$ solution

$$E = \pu{-0.23 V} + \pu{0.0296 V}·(-3) = \pu{-0.23 V} - \pu{0.09 V} = \pu{-0.32 V}.$$

This value is more negative than in the $\pu{1 M}$ solution. As a consequence, this electrode is the anode, and $\ce{Ni}$ gets oxidized, producing electrons going to the $\pu{1 M}$ electrode.

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    $\begingroup$ Please don't use \ce{…} for anything other than chemical expressions and definitely not for an entire math formula. Also, keep an eye on units: they are not supposed to vanish and then appear out of nowhere. $\endgroup$ – andselisk Jul 27 '20 at 13:15
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    $\begingroup$ @ andselisk. OK. And thank you for correcting the units. But why not use \ce{...} for math expression ? $\endgroup$ – Maurice Jul 27 '20 at 21:09
  • $\begingroup$ No prob. \ce{…} is a macro designed and tested for typesetting chemistry only. Used improperly, things might break over time and hinder readability of your posts, not to mention it breaks code semantics (a big deal for parsers and converters like pandoc). Also, \ce{} converts every Roman symbol in math mode to the upright form (analogous to \mathrm{…}), but the variables such as $E$ must be italicized. On top of that, it uses different spacing rules, which are not standardized for math equations and make them look off. $\endgroup$ – andselisk Jul 28 '20 at 3:08

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