Concentration cells: Worked examples — Chemistry deep dive

This page is the drill-ground for Concentration cells. The parent note built the one master formula. Here we throw every kind of situation at it — normal ratios, the sneaky "both beakers equal" degenerate case, tiny and huge ratios, an unknown hiding on the wrong side, a real-world corrosion story, and an exam twist where the numbers try to fool you.

Before you touch a single example, let us re-earn the one tool we lean on the whole page.

The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example is tagged with the cell it hits.

#	Case class	What is special about it	Example
A	Standard ratio, find $E$	ordinary numbers, $n = 2$	Ex 1
B	Degenerate: equal concentrations	$C_{high} = C_{low}$ , so $E = 0$	Ex 2
C	Limiting: huge ratio	ratio $\to$ very large, one side almost empty	Ex 3
D	$n = 1$ metal (silver)	electron count changes the prefactor	Ex 4
E	Unknown concentration, on the anode side	must solve backwards, watch which is high	Ex 5
F	Direction only (which is anode?)	no arithmetic, pure reasoning + sign	Ex 6
G	Real-world word problem	corrosion / differential aeration	Ex 7
H	Exam twist: numbers given "backwards"	dilute listed first, must not blindly subtract	Ex 8

The worked examples

Worked example Example 1 — Case A: ordinary copper cell

Statement. Copper concentration cell. Left beaker $[Cu^{2 +}] = 0.10$ M, right beaker $[Cu^{2 +}] = 0.010$ M. Find $E_{cell}$ at 25 °C.

Forecast: A ten-fold ratio with $n = 2$ — guess: will it be bigger or smaller than $0.03$ V?

Identify high and low: $C_{high} = 0.10$ , $C_{low} = 0.010$ . Why this step? The formula demands the larger number on top; mislabelling flips the sign.
$n = 2$ because $Cu^{2 +} + 2 e^{-} \to Cu$ . Why this step? $n$ is set by the ion's charge, not by the concentrations.
Plug in: $E_{cell} = \frac{0.0592}{2} lo g \frac{0.10}{0.010} = 0.0296 \times lo g (10)$ . Why this step? Direct substitution into the one formula.
$lo g 10 = 1$ , so $E_{cell} = 0.0296 \times 1 = 0.0296 V \approx 29.6 mV$ . Why this step? $lo g$ of a power of ten is just the exponent.

Verify: A per-decade (10×) voltage step for $n = 2$ is always $0.0592/2 = 0.0296$ V. Small, positive — matches the forecast that concentration cells give little voltages.

Worked example Example 3 — Case C (limiting): a huge ratio

Statement. Copper cell, $C_{high} = 1.0$ M, $C_{low} = 1.0 \times 1 0^{- 6}$ M. Find $E_{cell}$ .

Forecast: A million-fold gap. Will the voltage be huge like a battery (≈1.5 V)?

Ratio $= \frac{1.0}{1 0 ^{- 6}} = 1 0^{6}$ . Why this step? Divide high by low; the tiny denominator makes a big ratio.
$lo g (1 0^{6}) = 6$ . Why this step? Exponent of a power of ten.
$E_{cell} = \frac{0.0592}{2} \times 6 = 0.0296 \times 6 = 0.1776 V \approx 0.178 V$ . Why this step? Substitute the log value.

Verify: Even a million-fold gap gives only ~0.18 V — that is the whole point of the logarithm (Ex-A intuition). To hit a real battery's ~1.5 V you would need $lo g (\dots) \approx 50$ , i.e. a $1 0^{50}$ ratio — impossible in a real beaker. So concentration cells are inherently low-voltage. Sanity: 6 decades × 0.0296 V/decade = 0.178 V. ✓

Worked example Example 5 — Case E: unknown on the

anode side Statement. Silver concentration cell. The anode (dilute) half-cell is $[Ag^{+}] = 0.010$ M and the measured $E_{cell} = 0.118$ V. Find the cathode concentration.

Forecast: Will the answer be a bit above 0.010 M, or dramatically larger?

Anode is dilute, so $C_{low} = 0.010$ ; the unknown cathode is $C_{high}$ . $n = 1$ . Why this step? "Anode = oxidation = dilute side" — the unknown is therefore the larger number.
$0.118 = \frac{0.0592}{1} lo g \frac{C _{high}}{0.010}$ . Why this step? Substitute knowns into the one formula, leaving the unknown inside the log.
Divide: $\frac{0.118}{0.0592} = 1.993 = lo g \frac{C _{high}}{0.010}$ . Why this step? Isolate the log term before undoing it.
Undo the log (antilog): $\frac{C _{high}}{0.010} = 1 0^{1.993} = 98.4$ . Why this step? $lo g x = y$ means $x = 1 0^{y}$ — the antilog reverses base-10 log.
$C_{high} = 0.010 \times 98.4 = 0.984 M \approx 1.0 M$ . Why this step? Multiply through to free the unknown.

Verify: Plug back: $\frac{0.0592}{1} lo g \frac{0.984}{0.010} = 0.0592 \times lo g (98.4) = 0.0592 \times 1.993 = 0.118$ V. ✓ Answer sits above the dilute value (as it must, since $C_{high} > C_{low}$ ).

Worked example Example 6 — Case F: direction only (see figure)

Statement. Two zinc electrodes: Beaker A has $0.50$ M Zn²⁺, Beaker B has $0.050$ M Zn²⁺. Which is the anode, and which way do electrons flow in the wire?

Forecast: Point your finger: which beaker loses metal?

Dilute beaker (B, $0.050$ M) = anode: $Zn (s) \to Zn^{2 +} + 2 e^{-}$ . Why this step? Oxidation makes more ions where they are scarce, shrinking the gap.
Concentrated beaker (A, $0.50$ M) = cathode: $Zn^{2 +} + 2 e^{-} \to Zn (s)$ . Why this step? Reduction removes ions where they are crowded, shrinking the gap from the other side.
Electrons travel through the wire from anode to cathode: B → A. Why this step? Electrons always leave the oxidation site (they are the product of oxidation) and go to where reduction consumes them.

Verify: Both processes push toward $[Zn^{2 +}]$ equal — entropy increases, $Δ G < 0$ , so the arrow is correct. Sanity number: $E_{cell} = \frac{0.0592}{2} lo g \frac{0.50}{0.050} = 0.0296 V > 0$ , confirming spontaneous flow. Compare the layout with an ordinary galvanic cell — same wiring rule, only the reason differs.

Worked example Example 7 — Case G: real-world (differential aeration corrosion)

Statement. An iron pipe runs through soil. One spot sits in well-aerated soil ( $[Fe^{2 +}]$ kept low, $1.0 \times 1 0^{- 4}$ M, because oxygen sweeps ions away) and a buried spot is oxygen-poor with $[Fe^{2 +}] = 1.0 \times 1 0^{- 2}$ M. Model this as an $n = 2$ Fe concentration cell: which spot corrodes, and estimate the driving voltage?

Forecast: Corrosion = the metal dissolving = oxidation. Which spot is the anode?

Anode = dilute side = the low- $[Fe^{2 +}]$ well-aerated spot. Why this step? Oxidation happens where ions are scarce; that spot loses metal — it corrodes.
Ratio $= \frac{1.0 \times 1 0 ^{- 2}}{1.0 \times 1 0 ^{- 4}} = 1 0^{2}$ , $lo g = 2$ . Why this step? High over low; two decades apart.
$E_{cell} = \frac{0.0592}{2} \times 2 = 0.0592 V \approx 59 mV$ . Why this step? Substitute; the driving voltage of the corrosion cell.

Verify: Real "differential aeration" corrosion indeed eats the aerated (accessible) metal while the buried part is protected — matching our anode = dilute conclusion. Voltage ~0.06 V, tiny but enough over years to pit a pipe. Same physics underlies Membrane potentials across a cell wall. ✓

Worked example Example 8 — Case H: the exam twist (numbers listed backwards)

Statement. "A copper concentration cell has $[Cu^{2 +}]_{left} = 0.001$ M and $[Cu^{2 +}]_{right} = 1.0$ M. A student writes $E_{cell} = \frac{0.0592}{2} lo g \frac{0.001}{1.0}$ and gets a negative number. Fix it and give the correct $E_{cell}$ ."

Forecast: Can a real concentration cell's $E_{cell}$ ever be negative?

The formula needs $C_{high}$ on top, regardless of left/right labels. Here $C_{high} = 1.0$ , $C_{low} = 0.001$ . Why this step? The student subtracted in reading-order, not size-order — the classic trap.
$\frac{C _{high}}{C _{low}} = \frac{1.0}{0.001} = 1000$ , $lo g 1000 = 3$ . Why this step? Correct orientation gives a ratio $> 1$ , hence a positive log.
$E_{cell} = \frac{0.0592}{2} \times 3 = 0.0296 \times 3 = 0.0888 V \approx 0.089 V$ . Why this step? Substitute the corrected log.

Verify: The student's $- 0.089$ V is exactly the negative of the right answer — a sign-flip from putting the small number on top. Measured $∣ E_{cell} ∣$ from a voltmeter is $0.089$ V; the sign just tells you which electrode you called "+". A real spontaneous cell always reports positive. ✓ (This is the same setup as the parent note's Example 1, deliberately re-scrambled.)

Recall Quick self-test

Ten-fold ratio, copper ( $n = 2$ ) — what is $E_{cell}$ ? ::: $0.0296$ V (one decade × 0.0296 V/decade). Equal concentrations — what is $E_{cell}$ ? ::: Exactly 0 V (log 1 = 0); no anode or cathode. Same 10× ratio but silver ( $n = 1$ ) — bigger or smaller than copper's? ::: Bigger, $0.0592$ V (double), because you divide by 1 not 2. Which side is the anode? ::: Always the dilute side (oxidation makes ions where they are scarce). Voltmeter reads a negative number — likely mistake? ::: Put $C_{low}$ on top; swap so $C_{high}$ is on top.