Visual walkthrough — Concentration cells

This page is the visual companion to Concentration cells. If a word here feels rushed, that note explains it in prose; here we draw it.

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Step 1 — Two beakers, one difference

WHAT. We label the crowding of ions with the symbol $C$, the concentration — literally "how many ions per litre of water." One beaker is crowded ($C_{\text{high}}$), one is nearly empty ($C_{\text{low}}$).

WHY. A normal battery (a galvanic cell) runs on different metals. We are deliberately removing that difference so we can isolate what a concentration difference alone can do. If the beakers were equally crowded, nothing at all would happen — so the crowding gap is our entire engine.

PICTURE. The two beakers below are the same colour of copper metal but different shades of blue solution — dark = crowded, pale = dilute.

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Step 2 — What a single electrode is "worth"

Before we compare two beakers, we need a number for one beaker: how strongly does that copper strip pull electrons? That number is the electrode potential $E$, measured in volts (V).

Let us read each piece, right where it sits.

• $E^{\circ }$ ("E-standard") — the voltage this electrode would have if $[{\text{Cu}}^{2+}]$ were exactly $1\text{M}$. It is a fixed property of the metal. See Electrochemical series.
• $n$ — the number of electrons the reaction moves. For ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, we have $n=2$.
• $0.0592$ — a constant (in volts) that bundles temperature and physical constants at 25 °C. It comes from $\frac{RT}{F}\ln 10$ — see Nernst equation for its birth.
• $\log \frac{1}{[{\text{Cu}}^{2+}]}$ — the crowding term. We use $\log$ (base 10) because each ten-fold change in concentration should shift the voltage by a fixed amount; that "equal steps for equal multiplications" behaviour is exactly what a logarithm does, and no simpler function does it.

WHY $\frac{1}{[{\text{Cu}}^{2+}]}$ and not just $[{\text{Cu}}^{2+}]$? The Nernst term is $\log \frac{[\text{reduced}]}{[\text{oxidised}]}$. The reduced form is solid copper, whose concentration is taken as $1$ by convention. The oxidised form is ${\text{Cu}}^{2+}$. So the fraction is $\frac{1}{[{\text{Cu}}^{2+}]}$.

PICTURE. The graph shows $E$ climbing as the solution gets more crowded — more ions, higher voltage.

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Step 3 — Two potentials, side by side

Now write the same formula twice, once per beaker.

Both share the same $E^{\circ }$ (same metal!) and the same $n=2$. Only the $C$ inside differs.

WHAT. We have turned "two beakers" into "two numbers," $E_{\text{high}}$ and $E_{\text{low}}$.

WHY. A voltage difference is what pushes current through a wire, just like a height difference pushes water. To find the push we must first have two heights to subtract.

PICTURE. Two vertical "voltage bars," the concentrated one taller. The gap between the tops is the voltage we're chasing.

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Step 4 — Which beaker is the cathode? (read it off the picture)

The cathode is where reduction happens (ions grab electrons and become metal); the anode is where oxidation happens (metal loses electrons and becomes ions).

WHAT. In Step 3 the concentrated beaker has the higher $E$. The higher-potential electrode always pulls electrons toward it — so it is the cathode.

WHY it must be this way. Follow the logic of the term $\log \frac{1}{C}$: a bigger $C$ makes $\frac{1}{C}$ smaller, makes the $\log$ more negative, so the whole subtracted term becomes less negative — meaning $E$ is larger. Crowded ⇒ higher $E$ ⇒ cathode. This also matches nature's goal: reducing ions at the crowded side removes ions there, and creating ions at the dilute side adds them there — both shrink the gap. (Related: Le Chatelier's principle.)

PICTURE. Electrons stream out of the pale (dilute) strip, through the wire, into the dark (concentrated) strip.

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Step 5 — Subtract, and watch $E^{\circ }$ vanish

Cell voltage is always cathode minus anode: $E_{\text{cell}}=E_{\text{cathode}}-E_{\text{anode}}=E_{\text{high}}-E_{\text{low}}$

Substitute Step 3: $E_{\text{cell}}=\left(E^{\circ }-\frac{0.0592}{2}\log \frac{1}{C_{\text{high}}}\right)-\left(E^{\circ }-\frac{0.0592}{2}\log \frac{1}{C_{\text{low}}}\right)$

WHAT just happened. The two $E^{\circ }$ terms are identical (same metal), so they cancel: $E_{\text{cell}}=-\frac{0.0592}{2}\log \frac{1}{C_{\text{high}}}+\frac{0.0592}{2}\log \frac{1}{C_{\text{low}}}$

WHY this matters. This cancellation is the whole personality of a concentration cell: the "baseline chemistry" $E^{\circ }$ contributes zero. What's left is pure concentration. This is why looking up $E^{\circ }$ tables is useless here — see Mistake 3 in Concentration cells.

PICTURE. The two voltage bars from Step 3, now stacked so their equal $E^{\circ }$ bases line up and cancel — only the tips (the $\log$ parts) survive as the shaded difference.

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Step 6 — Fold two logs into one

We now clean up the leftover. Factor out $\frac{0.0592}{2}$: $E_{\text{cell}}=\frac{0.0592}{2}\left(\log \frac{1}{C_{\text{low}}}-\log \frac{1}{C_{\text{high}}}\right)$

WHY a log rule now? Two logs subtracted is exactly what the identity $\log a-\log b=\log \frac{a}{b}$ collapses into one. We want one clean ratio, because a ratio of concentrations is the physical quantity we care about.

$\log \frac{1}{C_{\text{low}}}-\log \frac{1}{C_{\text{high}}}=\log \left(\frac{1/C_{\text{low}}}{1/C_{\text{high}}}\right)=\log \frac{C_{\text{high}}}{C_{\text{low}}}$

giving

PICTURE. A curve of $E_{\text{cell}}$ against the ratio $C_{\text{high}}/C_{\text{low}}$ on a log axis — a straight climbing line, one bump of $\frac{0.0592}{n}$ volts per ten-fold ratio.

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Step 7 — The degenerate case: equal beakers

WHAT. Set $C_{\text{high}}=C_{\text{low}}$. Then the ratio is $1$, and $\log 1=0$, so $E_{\text{cell}}=\frac{0.0592}{n}\log 1=0\text{V}.$

WHY show this. It is the sanity check. If nothing distinguishes the beakers, there is nothing to drive electrons — the voltage is exactly zero, and the cell is dead. As the running cell equalises its two concentrations, it slides down this curve toward zero and stops. That endpoint is the cell's death, and it is why concentration cells slowly run flat on their own.

Edge behaviour at the other extreme: as $C_{\text{low}}\to 0$ (one beaker nearly pure water), the ratio $\to \infty$ and $\log \to \infty$, so $E_{\text{cell}}$ grows without bound in theory — though in practice you run out of measurable ions long before. The curve rises but never flattens.

PICTURE. The same curve as Step 6, with the point at ratio $=1$ circled in amber at $E=0$, and an arrow showing a live cell sliding leftward toward it as it discharges.

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Worked check — plug in real numbers

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The one-picture summary

Every step lives in this single diagram: two beakers → two potentials → subtract → cancel $E^{\circ }$ → one log ratio → one climbing line that hits zero when the beakers match.

Recall Feynman retelling — say it back in plain words

I have two cups of blue copper water, one strong, one weak, each with a copper spoon in it. Each spoon has a "wanting-electrons" number, and the stronger cup's spoon wants electrons more. So electrons crawl through the wire from the weak cup's spoon to the strong cup's spoon. To find how hard they're pushed, I write down each spoon's number using the Nernst rule, subtract them, and the "same metal" part cancels perfectly — leaving only a number that depends on how many times more crowded one cup is than the other, squished through a logarithm and scaled by $\frac{0.0592}{n}$. When both cups are equally crowded, the number is zero and nothing flows. As the cell runs, it evens out the two cups until they match — then it dies.

Recall Quick self-test

Why does $E^{\circ }$ cancel in a concentration cell? ::: Both electrodes are the same metal, so $E_{\text{cathode}}^{\circ }=E_{\text{anode}}^{\circ }$; subtracting identical terms gives zero. Which beaker is the cathode, and why? ::: The concentrated one, because higher $C$ gives a higher electrode potential $E$, and the higher-potential electrode attracts electrons (reduction). What is $E_{\text{cell}}$ when the two concentrations are equal? ::: Exactly $0$ V, because $\log (C_{\text{high}}/C_{\text{low}})=\log 1=0$. For a silver cell ($n=1$), a ten-fold ratio gives what voltage? ::: $\frac{0.0592}{1}\log 10=0.0592$ V.

See also: Membrane potentials and pH meters are concentration cells in disguise — the same one-log-ratio formula runs a nerve cell and a pH probe.