Nernst equation E = E° − (RT - nF) ln Q: Question bank

This page is a rapid-fire trap detector. Each line is a claim or question; read it, decide your answer before revealing, then compare with the reasoning. These target the sneaky misconceptions the Nernst equation $E = E° - \tfrac{RT}{nF}\ln Q$ invites — no heavy arithmetic, just clean thinking.

Definition Symbols used on this page (quick reminder)

Before the traps, pin down every symbol so no line ambushes you:

$E$ — the actual cell voltage measured under the present conditions (volts).
$E °$ — the standard cell potential: the voltage when every species is at its standard state (all activities $= 1$ , i.e. $1$ M solutes / $1$ atm gases, at $298$ K). It's the fixed "starting height."
$Q$ — the reaction quotient: the ratio $\frac{products}{reactants}$ right now (see Reaction Quotient Q vs Equilibrium Constant K). At equilibrium $Q = K$ .
$R$ — the gas constant, $8.314 J mol^{- 1} K^{- 1}$ .
$T$ — absolute temperature in kelvin ( $298$ K at 25°C).
$n$ — the count of moles of electrons transferred in the balanced reaction (always a positive whole number).
$F$ — Faraday's constant, $96, 485$ C per mole of electrons.

The equation reads $E = E ° - \frac{R T}{n F} ln Q$ , or at 25°C the base-10 form $E = E ° - \frac{0.0592}{n} lo g_{10} Q$ .

True or false — justify

A concentration cell always has

E ° = 0

True — both half-cells use the identical electrode/couple, so the standard potentials cancel exactly; all the voltage comes from the concentration difference through the

ln Q

term.

Q = 1

, then

E = E °

regardless of temperature.

True —

ln (1) = 0

kills the correction term entirely, so

E = E °

no matter what

R

T

, or

n

are. This is the definition-of-standard sanity check.

At equilibrium the cell potential

E

equals the standard potential

E °

False — at equilibrium

E = 0

(dead battery), while

E °

is a fixed nonzero number. The relation is

E ° = \frac{R T}{n F} ln K

, not

E ° = E

Increasing product concentration always lowers the cell voltage.

True — more products raise

Q

, so

ln Q

grows, the term

- \frac{R T}{n F} ln Q

becomes more negative, and

E

drops. This is Le Chatelier's Principle speaking in volts.

A negative

E °

guarantees the reaction is nonspontaneous under all conditions.

False —

E °

only fixes the standard-state behaviour. A large enough reactant excess can make

Q

tiny, pushing

E = E ° - \frac{R T}{n F} ln Q

positive. Standard nonspontaneous ≠ always nonspontaneous.

Doubling

n

(electrons transferred) halves the size of the concentration correction.

True —

n

sits in the denominator of

\frac{R T}{n F}

, so a larger

n

makes the cell's voltage less sensitive to concentration changes for the same

Q

The Nernst equation only applies to galvanic cells, not electrolytic ones.

False — it applies to any electrochemical cell at any moment; it's just thermodynamics. The sign of

E

tells you which type you have, it doesn't restrict where the equation is valid.

If you write the cell reaction backwards, the value of

E

you compute flips sign.

True — reversing the reaction flips

E °

's sign and inverts

Q

(so

ln Q \to - ln Q

), and both changes flip

E

. The physics is unchanged; you just chose the opposite "downhill" direction.

Q

can be built from raw molarities with no loss of accuracy in any solution.

False —

Q

is defined from activities. Molarities are fine only in dilute, near-ideal solutions; in concentrated ones the activity coefficient departs from 1 and raw concentrations give a wrong

E

Spot the error

"For

Zn (s) + Cu^{2 +} \to Zn^{2 +} + Cu (s)

, I wrote

Q = \frac{[ Zn ^{2 +} ] [ Cu ]}{[ Zn ] [ Cu ^{2 +} ]}

Error: pure solids Zn and Cu have activity

= 1

and must be dropped. Correct

Q = \frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]}

— only the aqueous ions appear (see Reaction Quotient Q vs Equilibrium Constant K).

"I used

E = E ° - \frac{0.0592}{n} ln Q

at 25°C."

Error:

0.0592

is the base-10 prefactor and pairs with

lo g_{10} Q

; the natural-log form uses

0.0257

. Mixing them inflates the correction by the factor

2.303

"The reaction transfers electrons from right to left, so I set

n = - 2

Error:

n

is a count and is always positive. Direction is already encoded in the sign of

E °

and in

Q

; a negative

n

double-counts the direction.

"Concentrations are

0.01

M, not

1

M, but I'll just read

E

straight off the reduction-potential table."

Error: the table gives

E °

, valid only at standard state. Off-standard concentrations demand the Nernst correction — that's the whole point of the equation.

"At equilibrium

ln Q = 0

, so

E = E °

Error: at equilibrium

Q = K

(usually far from 1), not

Q = 1

. Instead

E = 0

, which forces

E ° = \frac{R T}{n F} ln K

"I included the solvent water as

[H_{2} O]

Q

for an aqueous redox reaction."

Error: pure liquid water is the solvent with activity

\approx 1

and is omitted, just like a pure solid — unless it's a gas-phase reactant or a solute.

"I plugged in

2

M concentrations straight into

Q

and trusted the answer to three decimals."

Error: at

2

M the solution is far from ideal, so activity

\neq =

concentration; the "concentration"

Q

is only an approximation and the computed

E

carries real error unless activity coefficients are applied.

"Raising temperature always increases

E

because

T

appears in the numerator of

\frac{R T}{n F}

Error:

T

scales the whole correction term, which is subtracted. If

ln Q > 0

, higher

T

makes

E

smaller; if

ln Q < 0

, higher

T

makes

E

larger. Sign of

ln Q

decides — and

E °

itself also drifts with

T

"Solids don't appear in

Q

, so a cell built from solids alone has no defined voltage."

Error: the ions in solution are what appear in

Q

; the electrodes may be solids but the couple always involves a dissolved (or gaseous) species whose concentration sets

Q

Why questions

Why does the driving-force term carry a minus sign in

E = E ° - \frac{R T}{n F} ln Q

Because

Δ G = - n F E

flips the sign when converting free energy to voltage: a rising

Q

(more products) raises

Δ G

toward zero, which lowers

E

. The minus keeps "less spontaneous" = "less voltage."

Why does a concentration cell produce any voltage at all if

E ° = 0

Because dilution is itself spontaneous: driving ions from high to low concentration lowers free energy, and the

- \frac{R T}{n F} ln Q

term (with

Q < 1

) supplies a positive

E

from pure entropy of mixing.

Why can we replace

ln

with

2.303 lo g_{10}

and change the prefactor from

0.0257

0.0592

Because

ln x = 2.303 lo g_{10} x

exactly, and

0.0257 \times 2.303 \approx 0.0592

. It's a change of logarithm base, not new physics — pick the one that matches your

Q

expression.

Why does the voltage of a real battery sag as it discharges?

As reactants convert to products,

Q

climbs, so

- \frac{R T}{n F} ln Q

grows more negative and

E

falls; when

Q = K

the cell is dead. This is exactly the shape of a discharge curve.

Why does a large positive

E °

imply a large equilibrium constant

K

Setting

E = 0

at equilibrium gives

E ° = \frac{R T}{n F} ln K

, so

lo g K = \frac{n E °}{0.0592}

; big

E °

(or big

n

) pushes

K

up exponentially — the reaction runs nearly to completion. See Gibbs Free Energy and Spontaneity.

Why does pH show up inside

Q

for many half-reactions?

Because

H^{+}

is an aqueous species in the balanced half-reaction, its concentration enters

Q

directly, so

E

depends on pH — the foundation of pH and Half-Cell Potentials.

Why is

F

(Faraday's constant) the right conversion factor between charge and moles of electrons?

F = 96, 485

C is the charge carried by exactly one mole of electrons, so

n F

converts "moles of electrons per reaction" into total coulombs, letting

Δ G = - n F E

balance joules against volt-coulombs.

Why do textbooks write

Q

with concentrations if it's really about activities?

Because in the dilute solutions of most examples the activity coefficient is

\approx 1

, so activity

\approx

molarity; concentrations are a convenient stand-in that only fails in concentrated, non-ideal media.

Edge cases

What is

E

the instant a cell is assembled with pure products and no reactants (

Q \to \infty

ln Q \to + \infty

, so

E \to - \infty

formally — meaning the reaction runs strongly in reverse. Physically the labelled "cell reaction" simply proceeds backwards until

Q

drops to

K

What is

E

when there is essentially no product yet (

Q \to 0

ln Q \to - \infty

, so

E \to + \infty

formally — maximum forward driving force. Real cells never reach this because trace product always exists, but it explains why fresh cells start above

E °

when reactant-rich.

For a concentration cell, what happens to

E

as the two concentrations become equal?

Q \to 1

ln Q \to 0

, and since

E ° = 0

already,

E \to 0

. The cell dies exactly when the concentration gradient vanishes — no gradient, no drive.

n

could hypothetically be zero, what would the Nernst equation say?

It would divide by zero — nonsense, which correctly signals that a redox process with no electron transfer isn't an electrochemical cell at all. A real cell always has

n \geq 1

At the exact standard state (all activities 1, at 298 K) is the Nernst correction ever nonzero?

No — every activity is 1, so

Q = 1

ln Q = 0

, and

E = E °

precisely. Standard state is the single point where the correction term is guaranteed to vanish.

What voltage does an ideal reversible cell show at open circuit (no current drawn)?

The full Nernst

E

for the present

Q

— with no current there's no resistive loss, so the measured voltage is the true thermodynamic potential. Drawing current pulls the terminal voltage below this.

Recall One-line self-test

Cover everything: state in your own words why $E$ falls as a battery runs, and why it hits exactly zero (not $E °$ ) at the end. Answer ::: Products accumulate → $Q$ rises → $- \frac{R T}{n F} ln Q$ grows more negative → $E$ drops; at equilibrium $Q = K$ makes $E = 0$ , because $Δ G = 0$ there and $E = - Δ G / n F$ .

Connections

Parent: Nernst equation
Reaction Quotient Q vs Equilibrium Constant K — what belongs in $Q$
Le Chatelier's Principle — the concentration-shift intuition behind the sign
Gibbs Free Energy and Spontaneity — where $Δ G = - n F E$ comes from
Battery Discharge Curves — the sagging-voltage edge case in practice
pH and Half-Cell Potentials — why $H^{+}$ enters $Q$