Charge, chemical state and electrochemical potentials

Moving charge changes more than position

An ion crossing an interface experiences chemical surroundings and electric potential. Electrochemical potential combines these contributions for a charged species: it describes the change in free energy associated with adding that species under the stated conditions. A tendency inferred from concentration alone can be opposed or reinforced by an electric field.

Start with three different quantities. Charge q is counted in coulombs (C). Current is charge passage per time: at constant current, q = IΔt, with 1 A = 1 C/s. A potential difference is an energy-per-charge comparison, with 1 V = 1 J/C. A current is therefore not an amount of stored energy, and a voltage is not a flow rate. A cell may have a nonzero open-circuit voltage while its external current is zero.

For example, a constant 0.200 A for 50.0 s transfers 10.0 C through a chosen external cross-section. If the terminal voltage remains 1.04 V during that interval, the electrical work delivered to the external circuit is W_out = qV = 10.4 J. If current or voltage varies, split the record into intervals or use W_out = ∫VI dt. A voltage reading at just the start does not determine the energy of a long discharge.

The small models below supply exact fictional numerical inputs to isolate these meanings. Their values are not measured specifications for the copper/silver illustration, an operating recommendation or a prediction of a real battery. Conductors, insulators and semiconductors supplies related electrical vocabulary; each new electrochemical quantity is introduced here.

A reaction connects an electronic path to an ionic path

Consider the qualitative galvanic reaction Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), separated into two half-cells under conditions favouring that direction. At the copper anode, oxidation Cu → Cu²⁺ + 2e⁻ releases electrons into the metal. At the silver cathode, reduction 2Ag⁺ + 2e⁻ → 2Ag consumes electrons delivered through the external conductor. Anode means oxidation and cathode means reduction; the negative-anode/positive-cathode signs described here apply to this discharging galvanic case.

Electrons move externally from the copper side to the silver side. Conventional current in that conductor points in the opposite direction because electrons carry negative charge. Inside the electrolyte connection, ions move to compensate the local production of positive Cu²⁺ and removal of positive Ag⁺: anions can enter the anode solution and cations can enter the cathode solution. An electron is not required to travel through the bulk electrolyte to complete this charge-transfer cycle. This is a charge-path account, not a construction procedure.

One mole means a specified particle count. Use the rounded Faraday constant F = 96,500 C/mol, the magnitude of the charge of one mole of electrons. A written reaction transferring n electrons per reaction event transfers nF coulombs per mole of reaction progress. For the two-electron reaction, q/(2F) is the amount of reaction progress only if all the counted external charge belongs to that reaction. For 10.0 C this is 5.181 × 10⁻⁵ mol, while the electron amount is twice that: q/F = 1.036 × 10⁻⁴ mol.

Charge conservation does not establish that every coulomb advances the desired chemistry. Some current can charge an interface temporarily or support side reactions. A sustained external current also requires available internal ionic and interfacial paths. Detailed species transport is developed in Ionic, electronic and mixed transport; here the crucial distinction is the common charge account despite different moving species.

Add the signed energy contributions before judging a tendency

A chemical potential μ_chem describes how Gibbs free energy changes when a small amount of a species is added at fixed temperature, pressure and amounts of the other species. Its molar unit is J/mol. In a mixture it is an incremental quantity, not generally the total Gibbs energy divided by the total number of moles. Composition, interactions and phase affect this chemical contribution; concentration alone is not a universal measure of it.

For one species of charge number z, write μ̃ = μ_chem + zFφ. The tilde marks electrochemical potential, φ is electrical potential in volts and z is signed: +1 for a monovalent cation, −1 for a monovalent anion. The units agree because Fφ is (C/mol)(J/C) = J/mol. We use differences within a declared common convention: Δμ̃ = Δμ_chem + zF(φ_B − φ_A) for transfer from A to B. This φ is a local electrical potential in the model, not automatically a measured cell-terminal voltage.

Suppose a fictional monovalent cation has Δμ_chem = −2,000 J/mol for A → B, but B is electrically 0.0300 V above A. The chemical contribution favours B, whereas the electrical term is +(96,500)(0.0300) = +2,895 J/mol. The sum is +895 J/mol, so a permitted uncoupled small transfer in that direction raises free energy under the stated conditions. The opposing electrical term outweighs the chemical preference.

In this same fixed chemical-state model the contributions balance when φ_B − φ_A = 2,000/96,500 = 0.02073 V. This is not a universal threshold for ions. Reversing the electrical difference or changing the sign of z changes the electrical term; changing chemical state changes the other term. An uncoupled permitted transfer tends toward lower electrochemical potential, but a favourable sign supplies no rate and guarantees no permeability. A species can also move uphill when coupled to another favourable process or driven externally.

The separate chemical/electrical terms for individual ions require a consistent convention; this example supplies them rather than claiming an independent absolute measurement of each. Measurements of complete cells compare terminal potentials and include interfaces. We therefore keep the local transfer model separate from the cell/reference account that follows.

A cell potential is a comparison

A measured electrode potential uses a reference and a stated chemical state. Equilibrium cell voltage relates to reaction free energy; a working cell can differ because of kinetic and transport losses. State reference, composition, temperature and whether current flows.

A quoted electrode value must include its comparison. In an original reference example, the left electrode is −0.400 V and the right electrode +0.800 V relative to the same reference R, at specified fixed chemical states and temperature. With the positive meter lead on the right, E_cell = E_right − E_left = +1.200 V. These are consistent reduction-potential comparisons, not one oxidation value mixed with one reduction value.

Now choose a reference R′ whose potential is 0.250 V above R on the same convention. Subtract 0.250 V from both quotes: the left becomes −0.650 V and the right +0.550 V. Their difference remains +1.200 V. A reference shift changes the two coordinates, not the cell’s physical driving force. Values reported against different references cannot simply be subtracted before converting to a common one.

A change in composition, phase or temperature can instead change the equilibrium comparison itself. Standard-state electrode tables refer to specified activity conventions and reactions; standard does not mean that every real solution behaves ideally or that temperature need not be reported. Our fictional values are supplied for stated states, not imported standard potentials.

Open circuit means zero net external current under the measurement conditions. A sufficiently relaxed, reversible electrode comparison can approach an equilibrium voltage for the present states. An immediate reading can retain concentration gradients, phase-history effects or slow interfacial relaxation, and internal reactions may persist. Even a well-relaxed nonzero cell voltage does not mean the overall cell reaction has reached chemical equilibrium: its advance can be blocked by the open external path.

Relate reaction progress to reversible electrical work

At fixed temperature and pressure, the Gibbs energy decrease of a reversible cell reaction supplies its maximum non-expansion work. In an ideal cell where that work is electrical, E_rev = −Δ_rG/(nF). Here Δ_rG is the Gibbs-energy change per mole of the reaction as written, and n is its stoichiometric electron count. This is a reversible voltage for the present chemical state. It is not the same quantity as the total energy that an entire finite cell can deliver.

The sign follows an energy account. For a small amount of forward reaction progress dξ, the reversible positive work out is −Δ_rG dξ. The charge magnitude is dq = nF dξ. Dividing gives E_rev = (−Δ_rG dξ)/(nF dξ). A negative reaction free energy therefore corresponds to a positive reversible discharge voltage in that chosen direction. If the reaction equation is doubled, Δ_rG and n both double, so the voltage does not.

In our separate fictional two-electron dossier, Δ_rG = −231.6 kJ/mol at the specified state. Thus E_rev = 231,600/(2 × 96,500) = 1.200 V. For a short interval carrying 10.0 C while that state is approximated as unchanged, reversible electrical work is 12.0 J. This approximation is appropriate only within the stipulated small state window. When state and reversible voltage change appreciably, sum E_rev dq along the path.

A reversible voltage is an energy limit, not a promise of finite power. A current needs reaction and transport rates; finite current can create irreversible losses. Nor is all reaction heat equal to the electrical loss: a full thermal account can also include reversible entropy-related heat and changing stored energy. The next account concerns the electrical work shortfall for a declared state and charge.

Keep terminal work separate from the reversible comparison

For a bounded discharge account write V_term = E_rev − IR_ohmic − ℓ_other. Current I is positive during discharge. R_ohmic represents a supplied ohmic series contribution, and ℓ_other is a positive aggregate voltage loss from the other included electrochemical processes. The symbol ℓ_other is a voltage, not a length or a measured universal material constant. Its value may depend on current, state, temperature and history. We have defined a positive-loss convention so no half-cell overpotential sign has to be guessed.

The table gives evaluations of an invented same-state model with E_rev = 1.200 V and R_ohmic = 0.500 Ω. Each ℓ_other is separately supplied; the three entries happen to scale with I in this limited table. That does not prove that reaction and transport losses behave as one constant resistor at other currents.

At 0.200 A for 50.0 s, q = 10.0 C and V_term = 1.200 − 0.100 − 0.060 = 1.040 V. Work out is 10.4 J rather than the reversible 12.0 J. The electrical shortfall is 1.60 J: qIR_ohmic = 1.00 J plus qℓ_other = 0.600 J. Equivalently the powers are 0.240 W reversible, 0.208 W terminal, 0.0200 W ohmic and 0.0120 W other; multiplying each by 50.0 s gives the same account.

Calling the entire 0.160 V drop an ohmic drop would give R_apparent = 0.160/0.200 = 0.800 Ω. Terminal voltage alone has combined the supplied 0.500 Ω ohmic part and a 0.300 Ω ratio from other losses at this point. It has not identified their mechanisms. Conversely, a larger voltage drop at higher current does not require a changed equilibrium reaction; it is already possible with the fixed E_rev used here.

These equations and data are discharge-specific. Charging reverses reaction direction and requires a suitable sign and energy account, not blind substitution of a negative number into an unsigned loss table. We make no charge/discharge efficiency, full capacity, lifetime or material operating-window claim.

Use a changed observation to choose the next comparison

Suppose two specimens give different loaded voltages. First ask whether they share composition, temperature, phase history, current direction, elapsed time, reference and measurement connections. A difference could reflect a changed reversible chemical state, a changed ohmic path, a changed interfacial/transport loss, or more than one. Equal external current alone is not a matched material comparison.

An interruption-and-relaxation record can distinguish an immediate response from slower relaxation, but a particular time trace is not a unique mechanism proof. A separate justified ohmic measurement, controlled chemical-state comparison and electrode-specific referenced observations provide different information. Conductivity, voltammetry and impedance evidence explains the additional limits of voltammetry and impedance; this lesson establishes why more than one observable is needed.

A useful report therefore states what crossed the boundary, the two points defining voltage, the charge and duration, the chemical and thermal state, and whether the number is an equilibrium estimate, a loaded reading or a model output. Keep the energy account and its assumptions alongside the result. A repeatable voltage is evidence about that comparison, not an absolute potential of one isolated electrode.

Worked examples

Try the reasoning

1. Is an electrode potential meaningful without a reference convention?

   Show the answer and reasoning

   It is incomplete.

   Potentials are measured as differences.

2. Does open circuit guarantee every internal region is equilibrated?

   Show the answer and reasoning

   No.

   Slow relaxation or internal reactions may remain.

3. Two constant-current records are A: 0.100 A for 100 s at 1.12 V, and B: 0.400 A for 25.0 s at 0.880 V. Compare charge, terminal work and what the difference does not establish.

   Show the answer and reasoning

   Both carry 10.0 C. A delivers 11.2 J and B 8.80 J, a 2.40 J difference.

   q = IΔt is equal, while W = qV differs. These supplied short-interval readings do not establish total cell capacity, a changed reaction or a general high-current law.

4. The fictional reaction has Δ_rG = −231.6 kJ/mol and n = 2. Rewrite it with every coefficient doubled. Is its reversible voltage now 2.400 V?

   Show the answer and reasoning

   No. The doubled equation has −463.2 kJ/mol of the rewritten reaction and n = 4; E_rev remains 1.200 V.

   Both energy per mole of written reaction and charge per mole of that reaction double. Their ratio is unchanged. The amount assigned to one mole of reaction progress has changed.

5. A record carries 10.0 C, but independent evidence assigns only 8.00 C to the intended two-electron reaction. What reaction progress is supported, and can the remaining 2.00 C be ignored?

   Show the answer and reasoning

   ξ = 8.00/(2 × 96,500) = 4.145 × 10⁻⁵ mol. The remaining 2.00 C needs a separate account.

   Using total 10.0 C would overestimate intended progress by 25% relative to the supported 8.00 C result. The charge may belong to interfacial storage or other reactions; the total alone cannot identify which.

6. For a monovalent cation keep Δμ_chem = −2,000 J/mol but make φ_B − φ_A = −0.0100 V. Find the total and decide whether it establishes rapid transfer.

   Show the answer and reasoning

   The electrical term is −965 J/mol and Δμ̃ = −2,965 J/mol. It does not establish rapid transfer.

   The two terms now favour the same direction. A thermodynamic tendency still needs a permitted path and kinetics; a large barrier can make transfer slow.

7. At the specified E_rev = 1.200 V, a new same-state reading is 0.300 A and 0.960 V. A separate justified measurement gives R_ohmic = 0.400 Ω. What aggregate other loss follows?

   Show the answer and reasoning

   Total drop is 0.240 V, ohmic drop is 0.120 V and ℓ_other = 0.120 V.

   The electrical other-loss power is 0.300 × 0.120 = 0.0360 W. Without the independent ohmic information, terminal data would only give an apparent total ratio 0.240/0.300 = 0.800 Ω and would not separate the mechanisms.

8. A constant 0.200 A runs for 50.0 s. Voltage is 1.040 V for the first 25.0 s and 0.960 V for the next 25.0 s. Is 10.4 J the terminal work?

   Show the answer and reasoning

   No. Each interval carries 5.00 C, giving 5.20 J + 4.80 J = 10.0 J.

   The first reading is not valid over the whole interval. Sum q_kV_k, or integrate VI over time. This record alone also does not reveal whether the change arose from chemical state or operating losses.

9. Both electrodes are quoted versus R as +0.200 V and +0.700 V. A new reference is 0.100 V below R. What are the new quotes and right-minus-left difference?

   Show the answer and reasoning

   The new quotes are +0.300 V and +0.800 V; their difference is still +0.500 V.

   Subtracting a reference shift of −0.100 V adds 0.100 V to both coordinates. Changing the reference is different from reversing the meter leads, which would reverse the measured difference’s sign.

10. A cell shows 1.18 V immediately after interruption and 1.21 V later, with zero external current throughout. Does the later number prove every internal region is equilibrated, or that the cell reaction has no remaining driving force?

    Show the answer and reasoning

    Neither conclusion follows.

    The relaxation demonstrates that the initial reading was time-dependent. Slow internal processes may remain, and an open external path can block overall reaction progress even when interfaces approach their present-state equilibrium values. State the elapsed time and the evidence for the claimed equilibrium estimate.

What to carry forward

• Include chemical and electric driving forces.
• Distinguish equilibrium voltage from operating voltage.
• Current counts charge per time; voltage compares energy per charge; terminal energy requires both over the actual interval.
• Use signed zFΔφ with the chemical contribution and its declared convention, not concentration alone.
• Keep electrode reference, reaction direction, chemical state and temperature with every voltage.
• A reversible reaction-energy account and a loaded terminal-work account describe different limits and boundaries.
• An electrical shortfall is not a complete heat balance, and one voltage trace does not uniquely identify its mechanism.