Electrochemistry

Why This Topic Belongs in Neuroscience

A chemistry topic appears in the nervous-system section because every action potential is driven by electrochemical gradients — ion concentration differences across a membrane that create an electrical potential. The math governing a zinc electrode in zinc sulfate is the same math governing a sodium channel in a neuron. The Nernst equation is the bridge between electrochemistry and physiology, and the MCAT tests both sides.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Electrochemical Cells: A Quick Foundation

Must know

For the full treatment of general electrochemistry — galvanic vs. electrolytic cells, electrode potentials, thermodynamics ($\Delta G = -nFE$), Faraday's laws, and batteries — see the 4C Electrochemistry guide. Here we need only the minimum to reach the biological payoff.

In a galvanic (voltaic) cell, two half-cells are connected by a wire (electron flow) and a salt bridge (ion flow to maintain neutrality). Oxidation happens at the anode (electrons leave); reduction at the cathode (electrons arrive) — mnemonic AN OX, RED CAT. The driving force is the cell potential $E_\text{cell}$: positive means spontaneous. The one fact we carry forward is that potential depends on concentration — which is exactly what the Nernst equation captures.

Quick check: If $\Delta G° < 0$, what can you say about $E°_\text{cell}$ and $K$?

Answer: $\Delta G° < 0$ means $E°_\text{cell} > 0$ (spontaneous) and $K > 1$ (products favored).

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The Nernst Equation

Must know

Standard potential $E°$ assumes 1 M, 1 atm, 25 °C — conditions real cells never hold. As a reaction proceeds, concentrations change and the actual driving force $E$ drifts from $E°$. The Nernst equation corrects for this via the reaction quotient $Q$:

$\boxed{E = E° - \frac{RT}{nF}\ln Q}$

At 25 °C this simplifies (converting $\ln$ to $\log$) to:

$E = E° - \frac{0.0592}{n}\log Q \quad \text{(at 25 °C)}$

The constant 0.0592 V (≈ 59.2 mV). Variables: $E$ = actual potential, $E°$ = standard potential, $n$ = electrons transferred, $Q = [\text{products}]/[\text{reactants}]$, $R$, $T$ (K), $F$ as above.

Know the logic

Sign intuition: If $Q < K$, the reaction is more spontaneous than standard, so $E > E°$ (the Nernst term subtracts a negative). As the reaction proceeds, $Q \to K$ and $E \to 0$ — the cell is dead (at equilibrium).

Quick check: A cell has $E° = +0.40$ V, $Q = 10^4$, $n = 2$. Is $E$ larger or smaller than $E°$? Estimate $E$.

Answer: $Q > 1 \Rightarrow \log Q > 0$, so $E < E°$.
$E = 0.40 - \frac{0.0592}{2}(4) = +0.282$ V. Still spontaneous, but weaker.

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Concentration Cells

The Core Concept

Must know

A concentration cell is a galvanic cell where both electrodes are the same material and both half-cells contain the same ion — only the concentration differs. There is no chemical driving force: $E° = 0$ because identical half-reactions cancel. The entire driving force is the concentration gradient.

The system spontaneously works to equalize concentrations. The lower-concentration side is the anode (metal dissolves, raising its ion concentration); the higher-concentration side is the cathode (ions plate out, lowering concentration). Electrons flow from low- to high-concentration half-cell — analogous to an ion flowing down its concentration gradient.

The Nernst Equation for a Concentration Cell

Must know

With $E° = 0$, for a metal $\ce{M}/\ce{M^{n+}}$ cell $\ce{M | M^{n+}(low) || M^{n+}(high) | M}$ the overall reaction is $\ce{M^{n+}(high) -> M^{n+}(low)}$, so:

$E = -\frac{0.0592}{n}\log\frac{[\text{M}^{n+}]_\text{low}}{[\text{M}^{n+}]_\text{high}} = \frac{0.0592}{n}\log\frac{[\text{M}^{n+}]_\text{high}}{[\text{M}^{n+}]_\text{low}}$

Since $[\text{low}] < [\text{high}]$, $E > 0$ (spontaneous). The bigger the concentration ratio, the larger $E$.

Worked Example

Must know

A copper concentration cell $\ce{Cu | Cu^2+(0.010\,M) || Cu^2+(1.0\,M) | Cu}$ at 25 °C. Here $E° = 0$, $n = 2$ (Cu²⁺), and $Q = [\text{anode}]/[\text{cathode}] = 0.010/1.0 = 10^{-2}$:

$E = -\frac{0.0592}{2}\log(10^{-2}) = -\frac{0.0592}{2}(-2) = +0.0592 \text{ V}$

Check: $E > 0$ ✓; electrons flow from the dilute (0.010 M) anode to the concentrated (1.0 M) cathode ✓.

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The Biological Nernst Equation: Membrane Potentials

Must know

This is the payoff. The Nernst (equilibrium) potential for a single ion is the membrane voltage that exactly balances that ion's chemical gradient — the voltage at which there is no net driving force.

At physiological temperature (37 °C), the constant shifts so:

$E_\text{ion} = \frac{RT}{zF}\ln\frac{[\text{ion}]_\text{out}}{[\text{ion}]_\text{in}} \approx \frac{61.5\,\text{mV}}{z}\log\frac{[\text{ion}]_\text{out}}{[\text{ion}]_\text{in}}$

where $z$ is the ion charge with sign. Convention: membrane potential = inside minus outside; a resting neuron is ≈ $-70$ mV.

Passage-level

Typical equilibrium potentials: $E_\text{K} \approx -90$ mV, $E_\text{Na} \approx +60$ mV, $E_\text{Cl} \approx -70$ mV, $E_\text{Ca} \approx +125$ mV. Don't memorize the underlying concentration tables; recognize the values and know that $K^+$'s near $-90$ mV is why $\ce{K+}$ leak channels set the resting potential.

Know the logic

Direction: $\ce{K+}$ is more concentrated inside, so chemically it leaks outward, making the inside more negative; $E_\text{K}$ is the negative voltage that opposes this efflux. (Plugging $[\ce{K+}]_\text{out}=5$, $[\ce{K+}]_\text{in}=140$ into the equation gives $\approx -89$ mV.)

Quick check: $E_\text{Na} \approx +60$ mV. What does this mean for $\ce{Na+}$ at rest ($-70$ mV)?

Answer: At $-70$ mV both the electrical gradient (inside negative, attracting $\ce{Na+}$ in) and the chemical gradient (low $[\ce{Na+}]$ inside) favor entry. The cell is far from $+60$ mV, so there's a large inward driving force — why opening $\ce{Na+}$ channels causes rapid influx during depolarization.

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Direction of Electron Flow: Summary Rules

Must know

For a concentration cell:

1. The lower-concentration half-cell is the anode.
2. Electrons flow anode → cathode (low → high concentration) through the external wire.
3. In the salt bridge, cations migrate toward the cathode, anions toward the anode.

Quick check: In $\ce{Ag | Ag+(0.001 M) || Ag+(0.1 M) | Ag}$, which electrode is the anode and which way do electrons flow externally?

Answer: The 0.001 M side is the anode (oxidation: $\ce{Ag -> Ag+ + e-}$). Electrons flow from the 0.001 M electrode to the 0.1 M cathode — low to high concentration.

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Common Confusions & Tricks

1. Concentration cell anode = LOW concentration (not high). The dilute side wants to raise its ion concentration, so its metal dissolves (oxidation = anode).

2. Writing $Q$. Use $[\text{products}]/[\text{reactants}]$ from the overall reaction; for a concentration cell $Q = [\text{anode}]/[\text{cathode}] < 1$, giving $E > 0$.

3. Get $n$ right. $n = 1$ for $\ce{Ag+}$, $2$ for $\ce{Cu^2+}$, $3$ for $\ce{Al^3+}$. A wrong $n$ scales your answer.

4. 0.0592 V at 25 °C vs. 61.5 mV at 37 °C. Use the physiological constant when the passage specifies body temperature.

5. Nernst potential convention. $E_\text{ion} \propto \log([\text{out}]/[\text{in}])$ times $1/z$ (with the sign of $z$). Don't flip the ratio or drop the sign for anions.

6. $E = 0$ does NOT mean $E° = 0$. A cell is dead ($E = 0$) when $Q = K$. A concentration cell has $E° = 0$ but $E \neq 0$ until the concentrations equalize.

7. Fast shortcut: same metal, same ion, different concentration → $E° = 0$ → Nernst simplifies immediately.

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Key Equations

Equation	Variables & When to Use
$E = E° - \dfrac{RT}{nF}\ln Q$	Full Nernst; $T$ in K. Use when $T \neq 25$ °C.
$E = E° - \dfrac{0.0592}{n}\log Q$	Nernst at 25 °C; default for most MCAT problems.
$E_\text{cell} = E_\text{cathode} - E_\text{anode}$	From standard reduction potentials; positive = spontaneous.
$\Delta G = -nFE$, $\;\Delta G° = -nFE° = -RT\ln K$	Links cell potential, free energy, and $K$; $F \approx 96{,}500$ C/mol.
$E_\text{ion} \approx \dfrac{61.5\,\text{mV}}{z}\log\dfrac{[\text{ion}]_\text{out}}{[\text{ion}]_\text{in}}$	Single-ion Nernst (equilibrium) potential at 37 °C; $z$ = charge with sign.
$E_\text{conc. cell} = \dfrac{0.0592}{n}\log\dfrac{[\text{high}]}{[\text{low}]}$	Concentration cell at 25 °C ($E° = 0$); electrons flow low → high.