A concentration difference tells only part of the story for a charged particle. An ion moving toward a higher concentration may still move down its electrochemical gradient if the electrical contribution is sufficiently favorable. Likewise, the larger tabulated electrode potential does not, by itself, describe every possible operating condition of a cell.
The useful habit is to write the proposed direction before calculating. The AAMC content outline includes electrochemical gradients, cell potentials and the Nernst equation. The examples below are original teaching problems, not official exam questions.
Fix the potential convention before choosing signs
Electrode tables usually list reduction potentials. For a cell reaction written in its proposed forward direction:
E°cell = E°reduction,cathode − E°reduction,anode.
Reduction occurs at the cathode; oxidation occurs at the anode. The subtraction already reverses the anode half-reaction energetically. Reversing its sign and then subtracting that reversed value would count the reversal twice. OpenStax’s electrode-potential treatment establishes this convention.
Consider two fictional metal couples, with both entries written as reductions:
| Half-reaction | Standard reduction potential |
|---|---|
| X²⁺ + 2e⁻ → X(s) | −0.40 V |
| Y²⁺ + 2e⁻ → Y(s) | +0.20 V |
For X(s) + Y²⁺ → X²⁺ + Y(s), Y²⁺ is reduced and X is oxidized. Therefore E°cell = 0.20 − (−0.40) = +0.60 V. Two electrons are transferred per reaction as written.
Do not multiply a potential by a stoichiometric coefficient when balancing electrons. Potential is energy per charge. Doubling the entire reaction doubles both its free-energy change and its electron transfer; their ratio stays unchanged.
Apply the reaction quotient to the written reaction
At 25°C, use:
Ecell = E°cell − (0.0592 V/n) log₁₀Q.
The general form uses RT/(nF) and a natural logarithm; the 0.0592 V version incorporates temperature and the logarithm conversion. Q is dimensionless and formally uses activities. Here we assume ideal solution behavior, approximate each dissolved-ion activity by concentration divided by 1 M, and assign pure solid activities a value of one. These are model assumptions, not universal properties of concentrated solutions. OpenStax relates Nernst to free energy and reaction composition.
Suppose [X²⁺] = 0.010 M and [Y²⁺] = 1.0 M. For the reaction above:
- Q = a(X²⁺)/a(Y²⁺) = 0.010.
- log₁₀Q = −2.
- Ecell = 0.60 − (0.0592/2)(−2) = 0.6592 V, approximately 0.66 V.
The sign makes chemical sense. Relative to standard activities, there is less dissolved product compared with reactant, increasing the forward driving force.
Using ΔG = −nFE and F = 96,485 C/mol gives approximately −127 kJ per mole of reaction as written. The negative value describes thermodynamic favorability under these conditions. It does not establish a reaction rate, remove an activation barrier, or calculate the voltage lost when substantial current flows through a real cell.
A common wrong answer, 0.5408 V, comes from using Q = 100 instead. That quotient belongs to the reversed concentration ratio. Before repairing arithmetic, compare each species in Q with its position in the balanced reaction.
An ion gradient is a different accounting problem
For movement of an ion from outside to inside, define Vm = ψinside − ψoutside. Its molar free-energy change is:
ΔGoutside→inside = RT ln(cin/cout) + zFVm.
The first term represents the concentration contribution; the second represents electrical work. The two terms can reinforce or oppose each other. This is the electrochemical-gradient distinction described in OpenStax’s transport discussion.
Imagine a monovalent positive ion with cin/cout = 10 at 25°C. Take Vm = −0.090 V, R = 8.314 J/(mol·K), T = 298 K, and ideal activities.
The concentration contribution is approximately +5.70 kJ/mol: inward movement goes toward the higher concentration. The electrical contribution is approximately −8.68 kJ/mol: the positive ion moves toward the more negative interior. Their sum is about −2.98 kJ/mol. Inward movement is favorable despite opposing the concentration gradient.
For that ion, the equilibrium voltage is near −59 mV, where the two contributions cancel. A membrane at −90 mV is more negative than this equilibrium value, favoring inward movement. A membrane at −30 mV would leave the concentration contribution dominant, favoring outward movement.
The free-energy calculation identifies the thermodynamically favorable direction regardless of membrane permeability. Actual ion flux also requires a pathway; an impermeable membrane can maintain a gradient without appreciable ion movement. Also, this ion’s equilibrium potential is not automatically the membrane potential of a cell containing several permeant ion species.
Transfer exercise: reverse a condition, not your convention
Return to the fictional metal cell. Now let [X²⁺] = 1.0 M and [Y²⁺] = 0.010 M. Keep the forward reaction, temperature and activity assumptions unchanged.
Which conclusion follows?
A. The forward potential becomes −0.6592 V.
B. The forward potential becomes +0.5408 V.
C. The potential remains +0.60 V because the metals have not changed.
D. The forward reaction must reverse whenever product concentration exceeds reactant concentration.
Answer: B. Q = 100, so Ecell = 0.60 − (0.0592/2)(2) = +0.5408 V. The concentration change weakens the forward driving force but does not overcome the positive standard potential.
A negates the earlier potential without updating the concentrations. Reversing the reaction at the new concentrations would instead give −0.5408 V. C discards the composition term. D treats Q > 1 as sufficient for reversal, although the relevant comparison is with the equilibrium constant. At equilibrium Ecell = 0 and Q = K; there is no net thermodynamic driving force in either direction.
As a second check, reverse the entire balanced reaction at these same concentrations. Its E° becomes −0.60 V and its quotient becomes 1/100. Applying both changes produces −0.5408 V, exactly the negative of the forward value. Changing only one would be inconsistent.
Make the next problem auditable
On your next electrochemistry passage, write three short lines before calculating: the proposed reaction or transport direction, the voltage convention, and the relevant quotient. Then predict whether changing a concentration should strengthen or weaken that direction.
If the numerical result contradicts that prediction, inspect the reaction quotient and charge sign first. This is a learning suggestion for making reasoning visible, not a claim that one procedure guarantees exam performance.