Concentration cells: Question bank — Chemistry deep dive

The one formula this whole page leans on

Before you test yourself, let's earn every symbol in the formula the questions keep referring to. A cell is just two electrodes, each with its own electrode potential $E$ (a number in volts telling us how strongly that electrode pulls electrons). We always combine them by the sign convention

$E_{cell} = E_{cathode} - E_{anode}$

where the cathode is where reduction (electron gain) happens and the anode is where oxidation (electron loss) happens. This subtraction is a bookkeeping rule: it makes $E_{cell}$ come out positive exactly when the reaction is spontaneous.

Formula Where

$0.0592$ , $lo g$ , and $n$ come from Each electrode's potential obeys the Nernst equation. For the half-reaction $Cu^{2 +} + n e^{-} \to Cu$ , it reads $E = E^{\circ} - \frac{R T}{n F} ln \frac{1}{[ Cu ^{2 +} ]}$

$n$ = the number of electrons transferred in that half-reaction (for $Cu^{2 +} \to Cu$ , $n = 2$ ; for $Ag^{+} \to Ag$ , $n = 1$ ). More electrons per ion spreads the same energy over more charge, shrinking the voltage.
$R$ = gas constant, $T$ = temperature in kelvin, $F$ = Faraday's constant (charge per mole of electrons). The clump $\frac{R T}{F}$ has units of volts.
$ln$ is the natural logarithm (base $e$ ). Chemists prefer base-10 log, so we convert using $ln x = 2.303 lo g x$ , giving $\frac{2.303 R T}{F}$ .
Plugging in $R = 8.314$ , $T = 298 K$ (i.e. $25^{\circ} C$ ), $F = 96485$ gives the famous number: $\frac{2.303 R T}{F} \approx 0.0592 V$ . It is temperature-dependent — at other temperatures this constant changes, so $0.0592$ is a 25 °C only shortcut.

Subtracting the two electrodes' Nernst expressions, the $E^{\circ}$ terms cancel (same metal) and the two $lo g (1/ C)$ terms combine into one ratio, leaving $E_{cell} = \frac{0.0592}{n} lo g_{10} \frac{C _{high}}{C _{low}}$ The concentrated side sits in the numerator precisely because the sign convention $E_{cathode} - E_{anode}$ assigns the concentrated (higher- $E$ ) electrode as the cathode.

Now visualise the whole machine before quizzing yourself:

True or false — justify

Same metal on both sides means no current can ever flow.

False. Standard potentials cancel, but the actual potential depends on concentration via the Nernst equation, so unequal concentrations still produce a real voltage and a real current.

A concentration cell has

E_{cell}^{\circ} = 0 V

True. Both electrodes are the same material, so

E_{anode}^{\circ} = E_{cathode}^{\circ}

and their difference is exactly zero — the entire driving force is the non-standard concentration term.

E_{cell}

for a concentration cell can be negative if you set it up carefully.

False (for a spontaneous cell). Since the concentrated side is always the cathode, the ratio

C_{high} / C_{low}

exceeds 1, its log is positive, so

E_{cell} > 0

always.

As the cell runs, its voltage stays constant until the metal is used up.

False. Oxidation raises the dilute concentration and reduction lowers the concentrated one, so the ratio shrinks toward 1 and the voltage steadily falls toward zero.

At equilibrium the two half-cells have equal concentrations and

E_{cell} = 0

True. When

C_{high} = C_{low}

the ratio is 1,

lo g 1 = 0

, and the cell is dead — this is the mixed, maximum-entropy state.

The salt bridge is optional in a concentration cell.

False. Without it charge builds up in each beaker and stops the reaction; the salt bridge lets ions migrate to keep both solutions neutral, exactly as in any galvanic cell.

A concentration cell violates the rule "you need two different metals for a voltage."

True in appearance, but the rule was always incomplete — potential depends on both the metal identity and the ion concentration, and here only concentration differs.

Doubling both concentrations while keeping their ratio fixed changes

E_{cell}

False. Only the ratio

C_{high} / C_{low}

enters the formula, so scaling both by the same factor leaves the voltage unchanged.

The

0.0592 V

constant is a universal number valid at any temperature.

False. It equals

2.303 R T / F

evaluated at

298 K

; raise or lower the temperature and the constant changes, because

T

sits inside it.

Spot the error

"The concentrated side has more ions, so it must be the anode where oxidation happens."

Wrong. The cell equalizes concentrations, so oxidation must add ions at the dilute side (anode) and reduction must remove ions at the concentrated side (cathode).

"I'll look up

E_{Cu^{2 +} / Cu}^{\circ}

to get the cell potential."

Pointless — the standard potentials are identical and cancel. You must feed actual concentrations into the Nernst-derived formula

E_{cell} = \frac{0.0592}{n} lo g (C_{high} / C_{low})

"Electrons flow from the concentrated beaker to the dilute one in the wire."

Backwards. Electrons leave the anode (dilute side, where oxidation frees electrons) and travel through the wire to the cathode (concentrated side).

"Since

E_{cell}^{\circ} = 0

, the reaction is at equilibrium from the start."

No.

E^{\circ} = 0

only means standard conditions give no drive; the actual non-standard concentrations give

E_{cell} > 0

, so it is far from equilibrium until the concentrations meet.

E_{cell} = \frac{0.0592}{n} lo g (C_{low} / C_{high})

The ratio is upside-down. Written this way the log is negative, implying a negative voltage; the correct form is

lo g (C_{high} / C_{low})

, which is positive because the sign convention

E_{cathode} - E_{anode}

puts the concentrated side on top.

"More reactant means faster reaction, so the concentrated side reacts first."

This confuses kinetics with thermodynamic direction. Direction is set by which way lowers free energy (mixing), not by which side has more reactant.

"I used

ln

with the

0.0592

constant."

Mismatched. The

0.0592

prefactor already absorbed the

2.303

conversion, so it goes with base-10

lo g

; if you want

ln

, use the constant

R T / F \approx 0.0257 V

instead.

Why questions

Why does mixing the two concentrations lower the free energy?

Mixing increases entropy (disorder), and at constant temperature a positive

Δ S

contributes a negative

- T Δ S

Δ G

, making the process spontaneous — see Entropy and free energy.

Why is the concentrated side the cathode, in Nernst terms?

The Nernst equation gives a more positive electrode potential for higher

[Cu^{2 +}]

, and the more positive electrode is by definition the cathode under

E_{cell} = E_{cathode} - E_{anode}

Why do only ratios of concentration matter, not absolute amounts?

Because the two

E^{\circ}

and the constant Nernst terms subtract; what survives is

lo g C_{high} - lo g C_{low} = lo g (C_{high} / C_{low})

, a pure ratio.

Why does the number

0.0592

appear so often in these cells?

It is

2.303 R T / F

at 25 °C — the natural-to-base-10 conversion

2.303

times

R T / F

— the volts-per-decade "exchange rate" between a factor-of-ten concentration change and potential.

Why does the voltage decay as the cell operates?

Running the cell moves the two concentrations toward each other, shrinking the ratio; since

E_{cell} \propto lo g (ratio)

, the voltage drops and hits zero when they meet.

Why is Le Chatelier's principle a good intuition here?

The system is "stressed" by a concentration difference and responds by consuming the excess (concentrated) and generating more where it is scarce (dilute) — Le Chatelier's relief of the imbalance.

Why can a tiny concentration ratio still give a usable voltage?

The log function grows with each factor of ten in the ratio; a ratio of 1000 gives

lo g = 3

, and with

n = 1

that is already

\approx 0.18 V

— small ratios, but not negligible potentials.

Why should precise work use activity instead of concentration?

At high ionic strength ions shield one another so their effective (activity) concentration falls below the measured value, and plugging raw

C

into the Nernst equation then overestimates the potential.

Why are living cells' Membrane potentials essentially concentration cells?

Ion concentrations differ across a membrane, and that difference alone sets up a Nernst-type potential — the same physics, no different metals needed.

Why does a pH meter rely on this idea?

It measures the potential of a concentration cell built on a

H^{+}

difference across a glass membrane, converting the log of the

H^{+}

ratio directly into pH.

Edge cases

What is

E_{cell}

when the two concentrations are exactly equal?

Zero. The ratio is 1,

lo g 1 = 0

, so no potential exists — this is the equilibrium / dead-cell state.

What happens to

E_{cell}

C_{low} \to 0

(one side almost pure water)?

The ratio

C_{high} / C_{low} \to \infty

, so

lo g \to \infty

and

E_{cell}

grows without bound in the formula — physically the voltage climbs steeply but real limits (finite ions, kinetics) cap it.

If you accidentally used identical concentrations, would any electrode still be "anode"?

No meaningful anode or cathode — with no concentration difference there is no drive, no defined direction, and no current.

Does swapping which beaker you call "cell 1" change the physics?

No. Nature always oxidizes the dilute side and reduces the concentrated side regardless of labels; only your bookkeeping sign flips if you compute

C_{low} / C_{high}

by mistake.

For a cell with

n = 3

electrons transferred, is the voltage larger or smaller than for

n = 1

at the same ratio?

Smaller. The prefactor

0.0592/ n

shrinks as

n

grows, so more electrons per ion means a lower voltage for the same concentration ratio.

Would the same cell give a different voltage on a hot day versus 25 °C?

Yes. The prefactor is

2.303 R T / F

, so raising

T

raises the volts-per-decade constant above

0.0592

, nudging the voltage up for the same ratio.

Could you build a concentration cell that runs forever?

No. It is a spontaneous, entropy-driven process that ends when the concentrations equalize; running it drains the difference, so it stops at equilibrium.

If both half-cells are concentrated but by different metals' ions, is it a concentration cell?

No — different metals mean different

E^{\circ}

, so it is an ordinary galvanic cell driven by the Electrochemical series, not a pure concentration cell.

Recall Quick self-test

The concentrated side is always the ::: cathode (reduction removes its excess ions). $E_{cell}$ depends only on the ::: ratio of the two concentrations, not their absolute values. The number $0.0592 V$ is ::: $2.303 R T / F$ evaluated at 298 K, so it is temperature-dependent. In the Nernst equation, $n$ stands for ::: the number of electrons transferred in the half-reaction.