2.7.5 · D5Redox & Electrochemistry (Intro)
Question bank — Spontaneity from E°_cell and ΔG = −nFE
This is a rapid-fire trap bank for the parent topic. Every item targets a place where intuition quietly lies to you. Cover the answer, commit to a reason out loud, then reveal. If your reason was "yes/no" with no why, you have not passed the trap.
Before you start, remember the three anchor facts everything below leans on:
Recall The three anchors
Spontaneous means == which forces == (the minus sign flips the sign).
Related building blocks if a trap exposes a gap: Standard Electrode Potentials, Gibbs Free Energy Fundamentals, Galvanic vs Electrolytic Cells, Relationship between K_eq and ΔG°, Nernst Equation, Faraday's Laws of Electrolysis.
True or false — justify
A cell with V run twice as long gives a more negative ΔG°
False — ΔG° is per mole of reaction as written, an intensive-of-extent quantity fixed by the equation; running longer converts more moles but does not change ΔG° or E°.
Multiplying every coefficient in the balanced equation by 2 doubles ΔG°
True — doubling the equation doubles n (electrons), so doubles; but E° stays the same because voltage is intensive.
Multiplying the equation by 2 also doubles E°_cell
False — E° is an intensive property (energy per charge); both nF and ΔG° scale together, leaving unchanged.
A positive E°_cell guarantees the reaction is fast
False — E° and ΔG° describe thermodynamic spontaneity (whether it can go), not kinetics (how fast); a large positive E° can still sit stuck behind a huge activation barrier.
If ΔG° > 0 the reaction can never occur at all
False — ΔG° uses standard concentrations; under non-standard conditions ΔG (via the Nernst Equation) can still be negative, so the reaction may proceed until equilibrium.
For a spontaneous galvanic cell the anode is the negative terminal
True — oxidation dumps electrons at the anode, making it the electron-rich (negative) terminal that pushes electrons into the external wire.
A reaction with E°_cell = 0 is at equilibrium under standard conditions
True — , and gives , meaning neither direction is favoured at standard state.
Spot the error
"E°_cell = E°_anode − E°_cathode, so for Zn|Cu it's −0.76 − 0.34 = −1.10 V"
The formula is backwards; correct is V. Cathode Comes first.
"n electrons transferred here is 3 because each Al releases 3e⁻" (for 2Al + 3Cu²⁺)
n counts total electrons in the balanced equation: 2 Al × 3e⁻ = 6, matching 3 Cu²⁺ × 2e⁻ = 6, so n = 6 not 3.
"ΔG = +nFE, and since E = +1.10 V, ΔG is positive, so it's non-spontaneous"
The minus sign was dropped; for positive E, so the reaction is spontaneous.
"F = 96.5, so ΔG = −(2)(96.5)(1.10) ≈ −212 kJ"
F = 96 485 C/mol, giving ΔG in joules (−212 267 J); the −212 kJ answer is right only after dividing by 1000, so using 96.5 mixed up C and kC.
"I flipped the reaction so E° flips sign but ΔG° stays the same"
Both flip — reversing the reaction flips E° and ΔG° together, since ΔG° = −nFE° and n is unchanged.
"E°(Cu²⁺/Cu) = +0.34 V is bigger than E°(Zn²⁺/Zn), so Cu is a stronger reducing agent"
A higher (more positive) reduction potential means a stronger oxidizing agent / better at gaining electrons; Zn with the lower potential is the stronger reducing agent.
"To get ΔG in kJ I multiply the joule answer by 1000"
Divide by 1000 — there are 1000 J in a kJ, so −212 267 J = −212.3 kJ.
Why questions
Why does the sign of ΔG = −nFE come out negative rather than positive?
Because we define electrical work delivered by the cell as +nFE and maximum useful work as −ΔG; equating both "work-done-by-system" quantities gives nFE = −ΔG, hence the minus.
Why is n never zero for a real redox reaction?
A redox reaction by definition transfers electrons; n = 0 would mean no charge moves, so there is no cell voltage and no electrical work — it wouldn't be electrochemistry.
Why does F (Faraday's constant) appear instead of a single electron's charge?
ΔG is energy per mole of reaction, so we need the charge of a mole of electrons (), not one electron; F is the mole-scale conversion between charge and coulombs. See Faraday's Laws of Electrolysis.
Why can an electrolytic cell run a reaction with negative E°_cell?
Because an external power source supplies the missing energy; the applied voltage exceeds |E°_cell|, pushing the non-spontaneous reaction "uphill." See Galvanic vs Electrolytic Cells.
Why does E°_cell not depend on how the equation is scaled but ΔG° does?
E° is energy per unit charge (intensive); ΔG° is total energy for the written amount (extensive), so scaling the equation scales the total energy, not the energy-per-charge.
Why is standard potential measured at 1 M, 1 atm, 25 °C?
These are agreed reference conditions so every tabulated value is comparable; the Nernst Equation then corrects to any real concentration or pressure.
Why does a more negative ΔG° correspond to a larger equilibrium constant?
Because ; a more negative ΔG° needs a larger , so products dominate more heavily at equilibrium. See Relationship between K_eq and ΔG°.
Edge cases
E°_cell is exactly 0 V — what is K and which way does it go?
ΔG° = 0 gives K = 1; neither forward nor reverse is favoured at standard conditions, so the system sits balanced (any tiny push moves it via Nernst).
Concentrations are non-standard so E_cell = 0 even though E°_cell = +1.10 V — meaning?
The cell has reached equilibrium under those conditions (the Nernst term exactly cancels E°); ΔG = 0 and no more net electron flow occurs — the battery is "dead."
A reaction has huge negative ΔG° but nothing happens in the beaker — contradiction?
No — ΔG° governs spontaneity (feasibility), not rate; a high kinetic barrier can freeze a thermodynamically favourable reaction indefinitely.
Same metals but you swap which is cathode and which is anode — how do E° and ΔG° change?
Both flip sign (spontaneous becomes non-spontaneous); this is exactly the galvanic-vs-electrolytic distinction.
Two identical electrodes (a concentration cell) — is E°_cell zero, so ΔG° zero?
E°_cell is zero because cathode and anode have the same standard potential, but a real E_cell is nonzero due to the concentration difference (Nernst), driving a small spontaneous ΔG.
You double n but E° halves — does ΔG° stay the same?
Only if that scenario were physically real; in genuine chemistry doubling the written equation doubles n while E° is unchanged, so watch out — a "halving E°" claim usually signals a mis-scaled or misread equation.
Temperature rises well above 25 °C — is ΔG = −nFE still exactly valid?
The equation ΔG = −nFE stays valid at any T, but the value of E (and E°) shifts with temperature; the "°" tabulated values are strictly the 25 °C reference. See Gibbs Free Energy Fundamentals.