How to Identify the Anode and Cathode in a Concentration Cell

A concentration cell produces electrical energy because the same electrochemical reaction occurs under different concentration conditions in its two compartments. To identify its electrodes, use the same rule as for any galvanic cell: oxidation occurs at the anode, and reduction occurs at the cathode.

For the common case of the same metal in contact with solutions containing different concentrations of its ions, the dilute-ion side is the anode and the concentrated-ion side is the cathode. The reason follows from the electrode potentials and from the cell’s tendency to reduce the concentration difference.

The rule for a metal–metal-ion concentration cell

Consider a cell containing the same metal electrode in both compartments, with different activities of metal ions. For example:

$\mathrm{Zn(s)\,|\,Zn^{2+}(dilute)\,||\,Zn^{2+}(concentrated)\,|\,Zn(s)}$

At the anode, metal atoms are oxidized and enter solution as ions:

$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$

At the cathode, ions are reduced and deposit as metal:

$\mathrm{Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)}$

The anode is therefore on the dilute side, where oxidation adds metal ions to the solution. The cathode is on the concentrated side, where reduction removes metal ions. Both changes tend to lessen the concentration difference.

Electrons travel through the external circuit from the dilute-side anode to the concentrated-side cathode. In a working galvanic cell, the salt bridge also completes the circuit: anions tend to migrate toward the anode compartment, where oxidation produces positive ions, and cations tend to migrate toward the cathode compartment, where positive ions are consumed.

Why concentration determines the polarity

For the reduction half-reaction

$\mathrm{M^{n+} + ne^- \rightarrow M(s)}$

the Nernst equation is

$E = E^\circ + \frac{RT}{nF}\ln a(\mathrm{M^{n+}})$

Here, $a(\mathrm{M^{n+}})$ is the ion activity, $n$ is the number of electrons transferred, and $R$, $T$, and $F$ have their usual meanings. For dilute solutions, concentration is often used as an approximation to activity.

For the same metal in both compartments, the standard reduction potential $E^\circ$ is the same. The compartment with the greater metal-ion activity therefore has the greater reduction potential. Reduction is favored there, so it is the cathode. The lower-activity compartment has the lower reduction potential and is the anode.

A useful general test is to compare the actual reduction potentials of the two electrodes. The higher one is the cathode; the lower one is the anode. This test also works when the cell is not the simple metal–metal-ion example.

Worked example: zinc ion concentrations

Suppose a zinc concentration cell has $0.010\ \mathrm{M}$ zinc ions on the left and $1.0\ \mathrm{M}$ zinc ions on the right, at $25^\circ\mathrm{C}$. Assume ideal behavior so that concentration approximates activity.

1. The right compartment has the higher zinc-ion concentration, so its reduction potential is higher. It is the cathode.
2. The left, more dilute compartment is the anode.
3. The half-reactions are:

\text{Anode: } &\mathrm{Zn(s) \rightarrow Zn^{2+}(0.010\ M) + 2e^-} \\ \text{Cathode: } &\mathrm{Zn^{2+}(1.0\ M) + 2e^- \rightarrow Zn(s)} \end{aligned}$$ The net reaction transfers zinc from the left electrode into the left solution while zinc ions from the right solution plate onto the right electrode. The cell potential is $$\begin{aligned} E_{\text{cell}} &= \frac{RT}{nF}\ln\left(\frac{a_{\text{high}}}{a_{\text{low}}}\right) \\ &\approx \frac{0.05916\ \mathrm{V}}{2}\log_{10}\left(\frac{1.0}{0.010}\right) \\ &= 0.05916\ \mathrm{V} \end{aligned}$$ The positive result is consistent with the spontaneous direction just assigned: electrons flow from the dilute side to the concentrated side. The factor of 2 is used because zinc transfers two electrons per zinc atom. ## When the simple concentration rule needs care “Dilute side is the anode” applies directly to a cell with the same metal electrodes and the same metal-ion redox couple, where the differing concentration is that of the metal ions. It is not a universal rule that the compartment with the lower concentration of any substance must be the anode. For another redox couple, write its reduction half-reaction and use the Nernst equation with the relevant activities. If both the oxidized and reduced forms have different concentrations, compare the resulting reduction potentials rather than looking at just one concentration. For concentration cells involving electrolyte solutions and specialized electrode or junction arrangements, details such as ion transport and liquid-junction potentials can also affect the measured voltage. ## Common mistakes - **Mixing up electrode names and reaction directions:** Anode means oxidation; cathode means reduction. Those definitions do not change between galvanic and electrolytic cells. - **Assuming the anode is always negative:** In a galvanic concentration cell, the anode is negative relative to the cathode, but electrode signs are not the definition of anode and cathode. - **Reversing the electron direction:** Electrons move through the external wire from anode to cathode, not from cathode to anode. - **Using the standard potential alone:** A concentration cell has the same standard electrode potential on both sides, so the difference comes from the actual concentrations or activities. - **Treating concentration as exact activity:** The simple calculation uses concentrations as an approximation. At higher ionic strengths, activities can differ appreciably from concentrations. When the two compartments reach equal ion activities under the idealized conditions, the concentration difference no longer drives the cell and its potential falls to zero.