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Concentration Cells

General Chemistry • Electrochemistry

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Concentration Cells — Cell Potential at Nonstandard Conditions

Select a standard reduction half-reaction that defines the electrode (same chemistry in both half-cells). The anode uses this half-reaction in the oxidation direction, while the cathode uses it in the reduction direction. Enter the temperature and the activities (concentrations or gas pressures) for each half-cell. The calculator builds the reaction quotients for anode and cathode and applies the Nernst equation \(E_\text{cell} = \dfrac{RT}{nF}\ln \dfrac{Q_\text{anode}}{Q_\text{cathode}}\) to find the concentration cell potential and plot it as a function of the overall reaction quotient \(Q\).

Electrode (standard reduction half-reaction)

Concentration cell conditions

Default is 298.15 K (25 °C). \(R = 8.314~\text{J·mol}^{-1}\text{·K}^{-1}\), \(F = 96\,485~\text{C·mol}^{-1}\).

Activities in each half-cell (concentration cells)

Anode half-cell (oxidation)

Cathode half-cell (reduction)

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Frequently Asked Questions

What is a concentration cell in electrochemistry?

A concentration cell uses the same redox couple in both half-cells, but the ion activity or gas pressure differs between compartments. The voltage arises from the tendency to reduce the concentration gradient, not from different E° values.

Why is E°cell equal to 0 for a concentration cell?

Because the electrodes and redox couple are the same in both half-cells, their standard reduction potentials cancel. Under standard conditions with equal activities, there is no driving force and the standard cell potential is zero.

How does the calculator compute Ecell for a concentration cell?

It forms Q_anode and Q_cathode from the entered activities and applies Ecell = (R*T/(n*F)) ln(Q_anode/Q_cathode). The electron count n comes from the selected balanced half-reaction.

What should be included in the reaction quotient Q for concentration cells?

Use activities of variable species raised to stoichiometric powers, and omit pure solids and pure liquids because their activity is 1. Electrons are never included in Q.

What happens if the two half-cells have the same concentration or pressure?

If the activities are equal, the ratio in Q becomes 1 and ln(Q) = 0, so Ecell = 0. The plot and the computed value should both reflect zero driving force for that case.