Spontaneity from E°_cell and ΔG = −nFE: Worked examples

This page is the "no surprises" drill for the parent topic. Everything runs on one bridge equation, but before we even write it down, let us make sure every symbol in it is already yours.

Definition The symbols, in plain words (read this first)

$E_{ce l l}^{\circ}$ = the "voltage hill" of the whole cell in volts (V). Positive = electrons roll downhill on their own. Note: this is always the cell potential (cathode minus anode), never a single half-cell potential — for a half-cell we write $E^{\circ} (Cu^{2 +} / Cu)$ with the couple named in brackets. See Standard Electrode Potentials.
$Δ G^{\circ}$ = Gibbs free energy change, in joules (J) per mole of reaction. Negative = the reaction is willing to go by itself. See Gibbs Free Energy Fundamentals.
$n$ = moles of electrons moved when the balanced equation runs once.
$F$ = Faraday's constant $= 96, 485$ C/mol — the charge carried by one mole of electrons.
The link between them: 1 volt × 1 coulomb = 1 joule. That single unit fact is why $n F E_{ce l l}^{\circ}$ (a charge times a voltage) comes out in joules, the same currency as $Δ G^{\circ}$ .

Now that all four symbols are defined and pictured, we may finally state the tool they build:

$Δ G^{\circ} = - n F E_{ce l l}^{\circ}$

We will push this through every kind of case an exam or a lab can hand you: positive voltage, negative voltage, exactly-zero voltage, big electron counts, backwards reactions, unit traps, and a real word problem. Each worked example is tagged with the matrix cell it kills. (To keep notation clean, we always write the full subscript $E_{ce l l}^{\circ}$ for the cell voltage; a bare $E^{\circ}$ never appears alone.)

The scenario matrix

Cell	Case class	What is special	Killed by
A	$E_{ce l l}^{\circ} > 0$ , small $n$	ordinary spontaneous cell	Ex 1
B	$E_{ce l l}^{\circ} < 0$ (reversed cell)	non-spontaneous, sign flips	Ex 2
C	Large / multi-electron $n$	must count all electrons	Ex 3
D	$E_{ce l l}^{\circ} = 0$ (degenerate)	limiting case, $Δ G^{\circ} = 0$ , $K = 1$	Ex 4
E	Reverse-direction check	reversing flips both signs	Ex 5
F	Unit trap (kJ vs J, rounding $F$ )	conversion discipline	Ex 6
G	Real-world word problem	build the cell yourself	Ex 7
H	Exam twist: solve for $n$ or $E_{ce l l}^{\circ}$ backwards	rearrange the equation	Ex 8

The figure below draws this whole matrix as a number line of $E_{ce l l}^{\circ}$ : green to the right (spontaneous, $Δ G^{\circ} < 0$ ), red to the left (non-spontaneous, $Δ G^{\circ} > 0$ ), and each worked example placed as a coloured dot at its own voltage. Keep it open as a map — every example below tells you which dot it is landing on, so you can watch the cases fill the line from left to right.

Figure — Spontaneity from E°_cell and ΔG = −nFE

Ex 1 — Cell A: ordinary spontaneous cell

Forecast: Guess the sign of $Δ G^{\circ}$ before computing. Zinc famously "wants" to give electrons to copper — so downhill — so what sign?

Identify cathode and anode. Cu²⁺ gains electrons → reduction → cathode. Zn loses electrons → oxidation → anode. Why this step? $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ only works if you label them correctly (Cathode Comes first).
Compute the voltage. $E_{ce l l}^{\circ} = (+ 0.34) - (- 0.76) = + 1.10 V$ Why this step? $E_{ce l l}^{\circ} > 0$ already announces spontaneity — this is the green dot Ex1 A at $+ 1.10$ V on the number-line figure.
Count electrons. Zn → Zn²⁺ + 2e⁻, so $n = 2$ . Why this step? $n$ scales the energy; miscounting halves or doubles $Δ G^{\circ}$ .
Apply the bridge. $Δ G^{\circ} = - n F E_{ce l l}^{\circ} = - (2) (96485) (1.10) = - 212, 267 J \approx - 212.3 kJ$ Why this step? Converts the voltage hill into a thermodynamic energy.

Verify: $E_{ce l l}^{\circ} > 0 \Rightarrow Δ G^{\circ} < 0$ ✓ (spontaneous, matches forecast). Units: $mol \times \frac{C}{mol} \times V = C\cdotV = J$ ✓.

Ex 2 — Cell B: the reversed, non-spontaneous cell

Forecast: This is Ex 1 run backwards. If you believe the "flip the sign" rule, you already know the answer's sign.

Relabel. Now Zn²⁺ is reduced (cathode), Cu is oxidized (anode). Why this step? The chemistry decided the direction; we just read who gains electrons.
Voltage. $E_{ce l l}^{\circ} = (- 0.76) - (+ 0.34) = - 1.10 V$ Why this step? $E_{ce l l}^{\circ} < 0$ = uphill = the red dot Ex2 B at $- 1.10$ V, the mirror image of Ex 1 across zero on the number line.
$n = 2$ (same two electrons move).
Bridge. $Δ G^{\circ} = - (2) (96485) (- 1.10) = + 212, 267 J \approx + 212.3 kJ$ Why this step? Two minus signs multiply to a plus — positive $Δ G^{\circ}$ .

Verify: $E_{ce l l}^{\circ} < 0 \Rightarrow Δ G^{\circ} > 0$ ✓ non-spontaneous. Numerically it is exactly $- 1 \times$ Ex 1, confirming the reversal rule. To force it you'd need an external supply — see Galvanic vs Electrolytic Cells.

Ex 3 — Cell C: multi-electron transfer

Forecast: Same $F$ , same voltage size as before-ish, but $n$ is bigger. Will $∣Δ G^{\circ} ∣$ be much larger than Ex 1?

Voltage. $E_{ce l l}^{\circ} = (+ 0.34) - (- 1.66) = + 2.00 V$ Why this step? Note $E_{ce l l}^{\circ}$ does not get multiplied by stoichiometry — it is an intensive property, unchanged by scaling the equation. This lands as the green dot Ex3 C at the far right ( $+ 2.00$ V) of the number line.
Count total electrons. Each Al loses 3e⁻ → 2 Al lose 6e⁻; each Cu²⁺ gains 2e⁻ → 3 Cu²⁺ gain 6e⁻. They match: $n = 6$ . Why this step? This is exactly the trap in the parent note's Mistake 1 — count across coefficients.
Bridge. $Δ G^{\circ} = - (6) (96485) (2.00) = - 1, 157, 820 J \approx - 1158 kJ$ Why this step? Big $n$ × big $E_{ce l l}^{\circ}$ → very negative → strongly spontaneous.

Verify: $Δ G^{\circ} < 0$ ✓. Sanity: it's about $5.4 \times$ the magnitude of Ex 1 — makes sense since $6 \times 2.00$ vs $2 \times 1.10$ gives $12.0/2.2 \approx 5.45$ ✓.

Ex 4 — Cell D: the degenerate case $E_{ce l l}^{\circ} = 0$

Forecast: If the voltage hill is perfectly flat, does the reaction want to go either way?

Voltage is zero. $E_{ce l l}^{\circ} = 0$ V. Why this step? Identical standard electrodes have identical $E^{\circ}$ , so their difference is exactly zero — this is the blue dot Ex4 D sitting right on the boundary between the green and red regions of the number line.
Bridge. $Δ G^{\circ} = - (2) (96485) (0) = 0 J$ Why this step? Zero times anything is zero — no standard free-energy drive.
Equilibrium constant. Using $Δ G^{\circ} = - R T ln K$ (see Relationship between K_eq and ΔG°): $0 = - R T ln K \Rightarrow ln K = 0 \Rightarrow K = 1$ Why this step? $K = 1$ means products and reactants are equally favoured — perfectly balanced.

Verify: $E_{ce l l}^{\circ} = 0 \Leftrightarrow Δ G^{\circ} = 0 \Leftrightarrow K = 1$ ✓. This is the limiting "neither spontaneous nor forced" state. (Any real drive would then have to come from unequal concentrations via the Nernst Equation, which is a separate mechanism from the standard potential being zero.)

Ex 5 — Cell E: proving the reversal rule numerically

Forecast: One line, no calculator, if you trust the rule.

Apply reversal. Reversing a reaction negates $E_{ce l l}^{\circ}$ , hence negates $Δ G^{\circ}$ : $Δ G_{r e v er se}^{\circ} = - Δ G_{f or w a r d}^{\circ} = - (- 212.3) = + 212.3 kJ$ Why this step? $Δ G^{\circ}$ is a state-function difference; swapping products and reactants flips its sign. On the number line, Ex 1 (green, right) and its reverse (red, left) are the same distance from zero — exactly the Ex 1 / Ex 2 mirror pair.

Verify: Matches Ex 2's $+ 212.3$ kJ computed the long way ✓. So you never need to redo the arithmetic for a reversed cell.

Ex 6 — Cell F: the unit trap

Forecast: Guess whether the student's number is off by roughly a factor of 1000.

Spot the unit slip. The student used $F \approx 96.5$ (kilo-coulombs, sloppily) instead of $96, 485$ C/mol. Why this step? $F$ must be in coulombs so that C·V = J.
Correct computation. $Δ G^{\circ} = - (1) (96485) (0.50) = - 48, 242.5 J$ Why this step? Now everything is in base SI units.
Convert to kJ properly. $- 48, 242.5 J \div 1000 = - 48.24 kJ$ Why this step? The student's " $- 48.25$ " was numerically near the kJ answer but for the wrong reason — they'd have called it joules. Land the units on purpose, not by accident.

Verify: $- 48242.5$ J $= - 48.24$ kJ ✓; sign negative ⇒ spontaneous ⇒ green side of the number line (a small $E_{ce l l}^{\circ} = + 0.50$ V dot, close to the zero boundary) ✓. Lesson: keep $F$ in C, then divide by 1000 only at the end for kJ.

Ex 7 — Cell G: real-world word problem

Forecast: Silver visibly deposits on the iron in reality — so, before any math, what sign do you expect for $Δ G^{\circ}$ ?

Build the cell from the story. Ag⁺ is being reduced to Ag metal (cathode); Fe is dissolving, i.e. oxidized (anode). Why this step? A word problem gives you chemistry, not labels — you assign cathode/anode from who gains vs loses electrons.
Voltage. $E_{ce l l}^{\circ} = (+ 0.80) - (- 0.44) = + 1.24 V$ Why this step? Positive ⇒ the observed spontaneous plating is confirmed. This is the orange dot Ex7 G at $+ 1.24$ V on the green side of the number line.
Count electrons. Fe → Fe²⁺ + 2e⁻ and 2Ag⁺ + 2e⁻ → 2Ag, so $n = 2$ . Why this step? Two silver ions each take one electron: total 2.
Bridge. $Δ G^{\circ} = - (2) (96485) (1.24) = - 239, 282.8 J \approx - 239.3 kJ$

Verify: $Δ G^{\circ} < 0$ ✓ — agrees with the fact that silver really does plate onto iron. Units C·V = J ✓.

Ex 8 — Cell H: exam twist, solve backwards for $n$

Forecast: We know $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ . Which symbol are we isolating, and roughly how big will $n$ be?

Rearrange the bridge for $n$ . $n = \frac{- Δ G ^{\circ}}{F E _{ce l l}^{\circ}}$ Why this step? $n$ , $F$ , $E_{ce l l}^{\circ}$ multiply together — divide the energy by the other two to free $n$ .
Put $Δ G^{\circ}$ in joules. $- 386$ kJ $= - 386, 000$ J. Why this step? $F$ is in coulombs, so energy must be in joules or the number is off by 1000.
Plug in. $n = \frac{- ( - 386000 )}{( 96485 ) ( 2.00 )} = \frac{386000}{192970} \approx 2.00$ Why this step? The negatives cancel, leaving a clean positive integer — as $n$ must be. It shares the $+ 2.00$ V position with Ex 3 (the grey dot Ex8 H stacked below the green Ex3 dot on the number line), showing two different reactions can sit at the same voltage.

Verify: Forward-check: $- (2) (96485) (2.00) = - 385, 940$ J $\approx - 386$ kJ ✓. $n = 2$ is a whole number, as physics demands.

Recall Self-test on the matrix

Which case gives $Δ G^{\circ} = 0$ ? ::: The degenerate case $E_{ce l l}^{\circ} = 0$ (Cell D), where also $K = 1$ . Reversing a reaction does what to $Δ G^{\circ}$ ? ::: Flips its sign (Cell E). Why must $F$ be in coulombs? ::: So that C·V = J, matching $Δ G^{\circ}$ 's units (Cell F). In $2 A l + 3 C u^{2 +} \to \dots$ , what is $n$ ? ::: $6$ , the total electrons across all coefficients (Cell C).