Concentration cells

What is a Concentration Cell?

WHY does this work?

In a typical galvanic cell (like Zn|Cu), the driving force is the difference in $E°$ values. Zinc wants to give electrons more than copper does.
In a concentration cell, $E°_{\text{anode}} = E°_{\text{cathode}}$ because it's the same metal. So the standard cell potential $E°_{\text{cell}} = 0$ V.
But real cells operate under non-standard conditions. The Nernst equation tells us that potential depends on concentrations.
The Nernst equation creates a potential difference when concentrations differ, even if the electrodes are identical.

HOW does it work?

Set up two half-cells with the same metal electrode (say, copper).
One beaker has high $[\text{Cu}^{2+}]$ (concentrated), the other has low $[\text{Cu}^{2+}]$ (dilute).
Connect them with a salt bridge and wire.
At the dilute side (anode): $\text{Cu}(s) \to \text{Cu}^{2+}(aq) + 2e^-$ (oxidation increases dilute concentration)
At the concentrated side (cathode): $\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$ (reduction decreases concentrated solution)
Net result: Electrons flow from dilute → concentrated, and concentrations equalize.

Deriving the Cell Potential

Start with the Nernst equation for a half-cell: $E = E° - \frac{0.0592}{n} \log \frac{[\text{reduced form}]}{[\text{oxidized form}]}$

For a copper electrode in $\text{Cu}^{2+}$ solution ( $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ ): $E = E°_{\text{Cu}^{2+}/\text{Cu}} - \frac{0.0592}{2} \log \frac{1}{[\text{Cu}^{2+}]}$

(The $[\text{Cu}(s)] = 1$ by convention for pure solids.)

Now set up the concentration cell:

Cathode (reduction, concentrated): $[\text{Cu}^{2+}] = C_{\text{high}}$
Anode (oxidation, dilute): $[\text{Cu}^{2+}] = C_{\text{low}}$

WHY is the concentrated side the cathode? Because the Nernst equation gives a more positive potential when concentration is higher: $E_{\text{cathode}} = E° - \frac{0.0592}{2} \log \frac{1}{C_{\text{high}}}$ $E_{\text{anode}} = E° - \frac{0.0592}{2} \log \frac{1}{C_{\text{low}}}$

Cell potential: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$

Step-by-step derivation:

$E_{\text{cell}} = \left( E° - \frac{0.0592}{2} \log \frac{1}{C_{\text{high}}} \right) - \left( E° - \frac{0.0592}{2} \log \frac{1}{C_{\text{low}}} \right)$

The $E°$ terms cancel:

$E_{\text{cell}} = - \frac{0.0592}{2} \log \frac{1}{C_{\text{high}}} + \frac{0.0592}{2} \log \frac{1}{C_{\text{low}}}$

$E_{\text{cell}} = \frac{0.0592}{2} \left( \log \frac{1}{C_{\text{low}}} - \log \frac{1}{C_{\text{high}}} \right)$

Using $\log a - \log b = \log(a/b)$ :

$E_{\text{cell}} = \frac{0.0592}{2} \log \frac{C_{\text{high}}}{C_{\text{low}}}$

WHY this step? Because $\log(1/C_{\text{low}}) - \log(1/C_{\text{high}}) = \log(C_{\text{high}}/C_{\text{low}})$ by log properties.

Worked Examples

Worked example Example 1: Copper Concentration Cell

Problem: A concentration cell has copper electrodes. One half-cell has $[\text{Cu}^{2+}] = 1.0$ M, the other has $[\text{Cu}^{2+}] = 0.001$ M. Calculate $E_{\text{cell}}$ at 25°C. Solution:

$n = 2$ (Cu²⁺ + 2e⁻ → Cu)
$C_{\text{high}} = 1.0$ M, $C_{\text{low}} = 0.001$ M $E_{\text{cell}} = \frac{0.0592}{2} \log \frac{1.0}{0.001}$ WHY this step? We plug into the concentration cell formula with the ratio of concentrations.

$E_{\text{cell}} = 0.0296 \log(1000) = 0.0296 \times 3 = 0.0888 \text{ V}$

WHY this step? Because $\log(1000) = \log(10^3) = 3$ .

Answer: 0.089 V (or 89 mV)

Worked example Example 2: Silver Concentration Cell

Problem: Ag|Ag⁺ concentration cell has $[\text{Ag}^+]_{\text{anode}} = 0.01$ M and generates $E_{\text{cell}} = 0.118$ V. Find $[\text{Ag}^+]_{\text{cathode}}$ .

Solution:

$n = 1$ (Ag⁺ + e⁻ → Ag)
$C_{\text{low}} = 0.01$ M

$0.118 = \frac{0.0592}{1} \log \frac{C_{\text{high}}}{0.01}$

WHY this step? We use the concentration cell formula and solve for the unknown concentration.

$\frac{0.118}{0.0592} = \log \frac{C_{\text{high}}}{0.01}$

$1.993 = \log \frac{C_{\text{high}}}{0.01}$ WHY this step? Divide both sides by 0.0592 to isolate the log term.

$\frac{C_{\text{high}}}{0.01} = 10^{1.993} \approx 98.4$

WHY this step? Take antilog (10 to the power) to undo the logarithm.

$C_{\text{high}} = 0.01 \times 98.4 = 0.984 \text{ M}$

Answer: Approximately 1.0 M

Common Mistakes

Common mistake Mistake 2: Putting concentrated solution as anode

Wrong thinking: "More ions means more oxidation, so the concentrated side should be the anode." Why it feels right: More reactant usually means more reaction.

The fix: The cell operates to equalize concentrations. To reduce the concentration gap, oxidation must happen at the dilute side (to increase its concentration) and reduction at the concentrated side (to decrease its concentration).

Steel-man: The logic "more reactant → more reaction" applies to kinetics, but electrochemical direction is determined by thermodynamics (which way lowers free energy). Equalizing concentrations increases entropy (mixing is disorder), and this entropy gain is what lowers the free energy $\Delta G < 0$ . The Nernst equation tells us exactly how large the resulting potential is.

Memory Aids

Recall Explain to a 12-year-old

Imagine you have two rooms connected by a door. One room is super crowded with people (concentrated), the other is nearly empty (dilute). If you let things balance out, people will move from the crowded room to the empty one until both rooms have about the same number.

A concentration cell is like that, but with metal ions. One beaker has tons of copper ions (Cu²⁺), the other has very few. The cell "wants" to balance them out. Here's the clever part: to move ions from one beaker to the other, the cell uses electrons flowing through a wire!

In the crowded beaker, copper ions grab electrons and become solid copper (they leave the solution). In the empty beaker, solid copper gives up electrons and becomes ions (they join the solution). The electrons travel through the wire from the empty side to the crowded side, and we can use that electron flow as electricity!

So even though both sides have the same type of metal, the concentration difference creates a battery. It's nature's way of mixing things up.

Connections

Nernst equation — the mathematical foundation; concentration cells are a special case where $E°_{\text{cell}} = 0$
Galvanic cells — concentration cells are a type of galvanic cell with unusual symmetry
Le Chatelier's principle — concentration changes shift equilibrium; the cell operates to restore balance
Entropy and free energy — concentration cells are driven by entropy increase (mixing), not enthalpy
Membrane potentials — biological concentration cells (ion gradients across cell membranes generate nerve signals)
pH meters — use a concentration cell principle with H⁺ gradients
Electrochemical series — helps understand why some metals work better than others, though concentration cells use the same metal

Practice Problems

#flashcards/chemistry

What is a concentration cell? :: An electrochemical cell where both electrodes are made of the same material, but immersed in solutions of different concentrations of the same ion. The potential arises purely from the concentration difference.

In a concentration cell, which solution acts as the anode?

The dilute solution. Oxidation occurs here to increase the ion concentration and equalize the two sides.

What is the standard cell potential (E°) for a concentration cell?

Zero (0 V), because both electrodes are identical, so E°_cathode = E°_anode.

Formula for concentration cell potential

E_{\text{cell}} = \frac{0.0592}{n} \log \frac{C_{\text{high}}}{C_{\text{low}}}

(at 25°C, using log base 10)

A Cu concentration cell has [Cu²⁺] = 1.0 M at cathode and 0.01 M at anode. Calculate E_cell.

E_{\text{cell}} = \frac{0.0592}{2} \log \frac{1.0}{0.01} = 0.0296 \times 2 = 0.0592

Why does a concentration cell generate voltage even though both electrodes are identical?

Because the Nernst equation shows that electrode potential depends on ion concentration, not just the standard potential. Different concentrations create different actual potentials, producing a cell voltage.

In a Ag|Ag⁺ concentration cell, if [Ag⁺] at cathode is 10 times that at anode, what is E_cell?

E_{\text{cell}} = \frac{0.0592}{1} \log(10) = 0.0592 \times 1 = 0.0592

What drives electron flow in a concentration cell?

The thermodynamic drive to equalize concentrations. Mixing increases entropy, which lowers free energy (ΔG < 0). This is achieved by oxidation at the dilute side and reduction at the concentrated side.

Direction of electron flow in concentration cell

From the dilute solution (anode) to the concentrated solution (cathode) through the external circuit.

Why is the concentrated solution the cathode in a concentration cell?

Because reduction here decreases the ion concentration, helping to equalize the two solutions. The Nernst equation also gives a more positive potential to the higher concentration side.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, concentration cell ka core idea bahut simple aur elegant hai. Normal galvanic cell (jaise Zn-Cu) mein do alag metals hote hain aur electron flow hota hai kyunki ek metal doosre se zyada electron dena chahta hai—yaani unke standard potentials ( $E°$ ) alag hote hain. Lekin concentration cell mein dono electrodes SAME metal ke hote hain (jaise dono copper), bas unke solutions ki concentration alag hoti hai. Toh yahan $E°_{cell} = 0$ ho jaata hai, phir bhi cell kaam karta hai! Kaise? Kyunki nature hamesha concentrations ko equal karna chahti hai—yeh entropy ka kamaal hai, exactly jaise paani upar se neeche behta hai, waise hi ions barabar hone ki koshish karte hain.

Yeh equalization kaise hota hai, yeh samajhna important hai. Dilute wale side pe metal oxidize hota hai (anode), matlab wahan naye ions bante hain jisse concentration badh jaati hai. Concentrated wale side pe ions reduce hoke metal ban jaate hain (cathode), jisse wahan concentration kam ho jaati hai. Result? Electrons dilute se concentrated ki taraf flow karte hain, aur dono taraf ki concentration slowly barabar ho jaati hai. Cell potential nikalne ke liye hum Nernst equation dono half-cells pe lagate hain, aur jab subtract karte hain toh $E°$ terms cancel ho jaate hain, bacha rehta hai clean formula: $E_{cell} = \frac{0.0592}{n} \log \frac{C_{high}}{C_{low}}$ .

Yeh topic isliye matter karta hai kyunki yeh tumhe dikhata hai ki concentration akela hi voltage generate kar sakta hai—koi different material ki zaroorat nahi! Exams mein yeh numerical bahut aate hain aur formula seedha hai: jitna zyada concentration ka ratio, utna zyada potential. Practically bhi yeh concept important hai—jaise pH meters aur ion-selective electrodes isi principle pe kaam karte hain. Toh Nernst equation aur log properties acche se yaad rakho, phir yeh questions tumhare liye easy marks ban jaayenge.

What is a Concentration Cell?

Deriving the Cell Potential

Worked Examples

Common Mistakes

Memory Aids

Connections

Practice Problems

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Redox & Electrochemistry (Intro)

Connections