DAT General Chemistry: Concentration Cells

DAT High-Yield Concept

Identical Electrodes: Concentration cells use identical physical electrodes and identical aqueous ions in both half-cells; standard cell voltage is always zero ( $E^\circ_{\text{cell}} = \pu{0.00 V}$ ).
Oxidation at Dilute Side: Oxidation always occurs in the dilute half-cell (anode) to generate ions, while reduction occurs in the concentrated half-cell (cathode) to consume ions.
Equilibrium Voltage Decay: Non-standard cell voltage $E_{\text{cell}}$ is driven by the concentration ratio $Q = \frac{[\text{dilute}]}{[\text{concentrated}]}$ and decays to $\pu{0.00 V}$ as half-cell concentrations equalize at equilibrium ( $Q = 1$ ).
Biological Analogy: Resting membrane potentials across cell membranes act like microscopic concentration cells driven by ionic gradients (e.g., $\ce{Na^+}$ and $\ce{K^+}$ ).

1. Structure and Principles of Concentration Cells

A concentration cell is a specialized galvanic cell constructed from two half-cells that contain the same electrode material and the same active ionic species, but at differing concentrations. Because both compartments share identical standard reduction potentials, the standard cell potential is zero:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = \pu{0.00 V}

Despite having $E^\circ_{\text{cell}} = \pu{0.00 V}$ , the cell generates a non-zero cell voltage $E_{\text{cell}}$ under non-standard conditions due to the concentration gradient. Thermodynamic driving force arises from the system spontaneously tending toward equal concentration in both compartments, maximizing entropy and minimizing Gibbs free energy ( $\Delta G < 0$ ).

A concentration cell: same metal on both sides, voltage from the gradient alone — A concentration cell has the same metal and the same ion on both sides, so the two standard half-cell potentials are identical and the standard cell potential is exactly zero. Any voltage it produces comes from the concentration difference alone, through the Nernst equation. The cell runs in whichever direction moves the two solutions toward each other: the dilute half-cell is the anode, because oxidising the metal there adds ions to the side that is short of them, and the concentrated half-cell is the cathode, because plating metal out removes ions from the side that has too many. The voltage is small, on the order of tens of millivolts for a hundredfold difference in a two-electron system, and it falls steadily to zero as the two concentrations converge. A biological membrane potential is the same idea: one ion held at two different concentrations across a barrier.

Key Definition: Concentration Cell

An electrochemical cell with identical electrodes and solutions of the same electrolyte at different concentrations. Spontaneous current flows to reduce the concentration gradient between half-cells.

2. Half-Cell Reactions: Oxidation at the Dilute Compartment

To establish chemical equilibrium, the system must increase the concentration of ions in the dilute compartment and decrease the concentration of ions in the concentrated compartment:

Dilute Half-Cell (Anode): Oxidation occurs. Metal dissolves to yield ions, elevating the ion concentration toward equilibrium: $\ce{Cu(s) -> Cu^{2+}(aq, dilute) + 2e^-}$
Concentrated Half-Cell (Cathode): Reduction occurs. Ions gain electrons and deposit as solid metal, reducing ion concentration: $\ce{Cu^{2+}(aq, conc) + 2e^- -> Cu(s)}$

The net redox equation for the concentration cell conserves mass and charge while representing ion transfer down the gradient:

\ce{Cu^{2+}(aq, conc) -> Cu^{2+}(aq, dilute)}

DAT Memory Trick: "AN OX in Dilute, RED CAT in Concentrated"
ANode = OXidation happens in the Dilute compartment (to make more ions).
REDuction = CATHode happens in the Concentrated compartment (to remove ions).

Worked Example: Calculating Concentration Cell Potential

Problem: A concentration cell is prepared using two $\ce{Cu(s)}$ electrodes. Compartment A contains $\pu{0.010 M}$ $\ce{Cu^{2+}(aq)}$ and Compartment B contains $\pu{1.0 M}$ $\ce{Cu^{2+}(aq)}$ at $\pu{25 degC}$ . Identify the anode and cathode, write the overall reaction, and calculate $E_{\text{cell}}$ .

Step-by-Step Solution:

Identify Anode and Cathode: Oxidation occurs at the lower concentration. Compartment A ( $\pu{0.010 M}$ ) is the anode (dilute); Compartment B ( $\pu{1.0 M}$ ) is the cathode (concentrated).
Determine Electron Transfer ( $n$ ): $\ce{Cu^{2+}}$ involves $n = 2$ electrons per ion.
Set Reaction Quotient ( $Q$ ): $Q = \frac{[\ce{Cu^{2+}}]_{\text{dilute}}}{[\ce{Cu^{2+}}]_{\text{concentrated}}} = \frac{\pu{0.010 M}}{\pu{1.0 M}} = 1.0 \times 10^{-2}$
Calculate Voltage (using the Nernst relationship covered in depth in its own lesson): $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{\pu{0.0592 V}}{n} \log Q$ $E_{\text{cell}} = \pu{0.00 V} - \frac{\pu{0.0592 V}}{2} \log(1.0 \times 10^{-2})$ $E_{\text{cell}} = -(\pu{0.0296 V})(-2.00) = +\pu{0.0592 V}$

Common Exam Trap: Inverting the reaction quotient

Q

by placing the concentrated value in the numerator. If

Q > 1

\log Q

is positive, yielding a negative cell potential (non-spontaneous). Spontaneous galvanic cells always require

E_{\text{cell}} > 0

, meaning

Q = \frac{[\text{dilute}]}{[\text{concentrated}]}

must be less than 1.

3. Voltage Decay to Zero as Concentrations Equalize

As electrons flow through the external circuit, $[\text{dilute}]$ steadily rises while $[\text{concentrated}]$ falls. The ratio $Q$ approaches 1. When the concentrations in both half-cells become equal:

[\text{dilute}] = [\text{concentrated}] \implies Q = 1

\log(1) = 0 \implies E_{\text{cell}} = \pu{0.00 V}

At this stage, chemical equilibrium is achieved ( $\Delta G = 0$ ), and net electron transfer ceases. This process models battery discharge to complete exhaustion.

Property	Dilute Half-Cell (Anode)	Concentrated Half-Cell (Cathode)
Process	Oxidation ( $\ce{M(s) -> M^{n+}(aq) + n e^-}$ )	Reduction ( $\ce{M^{n+}(aq) + n e^- -> M(s)}$ )
Ion Concentration Trend	Increases over time	Decreases over time
Electrode Mass Trend	Decreases (electrode dissolves)	Increases (metal deposits)
Role in $Q$ Expression	Numerator ( $[\text{dilute}]$ )	Denominator ( $[\text{concentrated}]$ )

4. Biological Membrane Potential as a Concentration Cell Analogy

Living cells maintain distinct intracellular and extracellular ion concentrations using selective membrane transport pumps. For example, a resting neuron maintains a high intracellular concentration of potassium ions ( $\ce{K^+}$ ) relative to the extracellular fluid, while sodium ions ( $\ce{Na^+}$ ) are kept predominantly outside.

Esquema sobre el potencial de Nernst; cubetas con KCl separadas por una membrana semipermeable. (Source: Retama · CC BY-SA 3.0 · Wikimedia Commons)

Selective ion channels in the lipid bilayer allow specific ions to diffuse down their concentration gradients. As positive ions diffuse outward across the semipermeable membrane, an electrical potential difference (voltage) builds across the membrane. Diffusion continues until the electrical force opposing ion movement balances the chemical concentration gradient, producing a steady-state resting membrane potential (typically $\pu{-70 mV}$ inside relative to outside). This biological voltage generation directly mirrors the physics of an inorganic concentration cell.

DAT Critical Thinking Questions

System Dynamics: If a zinc concentration cell with $\pu{0.050 M}$ $\ce{Zn^{2+}(aq)}$ and $\pu{0.50 M}$ $\ce{Zn^{2+}(aq)}$ is diluted by adding pure water to the dilute side, how will the cell voltage $E_{\text{cell}}$ change initially? Explain in terms of $Q$ .
Equilibrium State: Why does a concentration cell have $E^\circ_{\text{cell}} = \pu{0.00 V}$ , yet generate a positive measured voltage $E_{\text{cell}} > 0$ at start-up?
Biological Application: If a neurotoxin selectively blocks potassium leak channels, preventing $\ce{K^+}$ movement across the neuronal membrane, what immediate effect does this have on the membrane potential analogy?

References

Sources returned by Google Search grounding during lesson verification.

19.6: Cell Potential and Concentration - Chemistry LibreTexts — chem.libretexts.org
Concentration cell (video) | Nernst equation - Khan Academy — en.khanacademy.org

Concentration Cells

Summary

Practice Questions

AI Tutor