Visual walkthrough — Spontaneity from E°_cell and ΔG = −nFE

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Step 1 — What is an electron actually doing?

WHAT. Before any formula, picture the raw event: a single electron leaving the zinc metal, sliding through a wire, and landing on a copper ion. That slide is the whole story.

WHY. Electricity is nothing mystical — it is charged things moving from where they "want to leave" to where they "want to arrive." Every symbol we later write ($E$, $n$, $F$) is just a way of counting or measuring this one motion. So we must see the motion first.

PICTURE. Look at the figure. The zinc side sits higher (electrons are unhappy there, eager to leave). The copper side sits lower (electrons are welcomed there). The drop from high to low is the "electron slide."

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Step 2 — Height of the slide = voltage $E$

WHAT. We name the steepness of that slide. That single number is the cell potential $E$, measured in volts (V).

WHY. We need a number that answers "how hard does each unit of charge get pushed down the slide?" A tall slide pushes hard; a flat one barely pushes. That is exactly what voltage means:

$$E=\underset{\text{joules per coulomb}}{\underbrace{\frac{\text{energy given to the charge}}{\text{amount of charge}}}}$$

Read the formula term by term:

• the top is energy handed out (joules),
• the bottom is how much charge received it (coulombs),
• so $1\text{V}=1\text{J/C}$ — one joule delivered per coulomb that slides through.

PICTURE. Same slide as Step 1, now with a height bar labelled $E$. A steeper bar = bigger $E$ = more energy per coulomb.

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Step 3 — How much charge slides? Count with $n$ and $F$

WHAT. One electron carries a tiny charge. We rarely move one — we move moles of them. So we count total charge as $n$ moles of electrons, each mole carrying $F$ coulombs.

WHY. Chemistry is written per mole of reaction, not per single atom. To turn "each electron" into "the whole reaction," we need two counters:

• $n$ = how many moles of electrons the balanced equation moves,
• $F$ = the charge of one mole of electrons = 96{,}485 C/mol (Faraday's constant, see Faraday's Laws of Electrolysis).

PICTURE. A stack of $n$ trays, each tray holding one mole of electrons worth $F$ coulombs. Total charge is the whole stack.

$$q=\underset{{\text{moles of e}}^{-}}{\underbrace{n}}\times \underset{\text{coulombs per mole}}{\underbrace{F}}=nF[\text{coulombs}]$$

• $n$ counts the trays (pure number, from the balanced equation),
• $F$ is the coulombs inside each tray,
• their product $nF$ is total charge that slides.

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Step 4 — Multiply height × amount = electrical work

WHAT. Combine Step 2 (energy $=E\times$ charge) with Step 3 (charge $=nF$). The energy the sliding electrons deliver to the outside world is:

$$w_{\text{by cell}}=\underset{\text{J per C}}{\underbrace{E}}\times \underset{\text{coulombs}}{\underbrace{nF}}=nFE[\text{joules}]$$

WHY. This is just the volt-definition rearranged: energy = voltage × charge. Nothing new is assumed — we simply plug in "how much charge" from Step 3. The units cancel perfectly: $\frac{\text{J}}{\text{C}}\times \text{C}=\text{J}$.

Term by term:

• $E$ = joules delivered per coulomb,
• $nF$ = number of coulombs that slide,
• $nFE$ = total joules of useful electrical work the cell hands to the surroundings.

PICTURE. The slide drives a tiny motor (the "surroundings"). Height $E$ times the pile of charge $nF$ fills the "work delivered" tank.

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Step 5 — The other name for that same energy: $-\Delta G$

WHAT. Thermodynamics measures the "useful energy budget" of a reaction with Gibbs free energy $\Delta G$ (see Gibbs Free Energy Fundamentals). Its rule at constant temperature and pressure:

$$w_{\text{by system, max}}=-\Delta G$$

WHY the minus sign here? $\Delta G$ is the change in the budget: (after) − (before). When a reaction runs on its own, the budget drops, so $\Delta G$ is negative. The energy that left the budget is what became usable work. "Energy that left" $=-(\Delta G)$ = a positive amount of work. Hence work done by the system $=-\Delta G$.

PICTURE. A fuel gauge. Before: gauge high. After: gauge lower. The drop (a downward arrow) is the work handed out. The drop is $-\Delta G$ because $\Delta G$ itself points downward (negative).

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Step 6 — Two names, one energy: equate them

WHAT. Step 4 and Step 5 both measured work done BY the system on the surroundings — Step 4 through wires (electrical), Step 5 through thermodynamics (free-energy drop). One event, two descriptions ⇒ set them equal:

$$\underset{\text{electrical work out (Step 4)}}{\underbrace{nFE}}=\underset{\text{max useful work out (Step 5)}}{\underbrace{-\Delta G}}$$

Multiply both sides by $-1$:

$$\boxed{\Delta G=-nFE}$$

WHY it's legal to equate them. Maximum useful work equals the electrical work only when the cell runs reversibly (infinitely slowly, no energy wasted as heat). We assume that ideal, so the two "work out" numbers are the same energy.

PICTURE. Two pipes — "electrical work" and "free-energy drop" — merge into one tank. Because both fill the same tank, whatever is in one equals what is in the other; the minus flips as we cross the equals sign.

Under standard conditions (1 M, 1 atm, 25 °C) we tag everything with a °:

$$\boxed{\Delta G^{\circ }=-nF{E}_{\text{cell}}^{\circ }}$$

• $\Delta G^{\circ }$ = standard free-energy change, per mole of reaction as written,
• $E_{\text{cell}}^{\circ }$ = standard cell potential (from Standard Electrode Potentials).

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Step 7 — Reading all three signs at once

WHAT. Because $n$ and $F$ are always positive counters, the sign of $\Delta G$ is fixed entirely by the sign of $E$. Three cases — cover them all:

$E_{\text{cell}}^{\circ }$	$\Delta G^{\circ }=-nF{E}^{\circ }$	Slide direction	Cell type
$E^{\circ }>0$	$\Delta G^{\circ }<0$	downhill on its own	spontaneous (galvanic)
$E^{\circ }<0$	$\Delta G^{\circ }>0$	must be pushed uphill	non-spontaneous (electrolytic)
$E^{\circ }=0$	$\Delta G^{\circ }=0$	flat — no net slide	at equilibrium

WHY the $E^{\circ }=0$ (degenerate) case matters. When the slide is flat, no direction is preferred: forward and reverse rates match. This is exactly $\Delta G^{\circ }=0$, the doorway to the link with $K_{eq}$ and the Nernst Equation (which tells you what happens once concentrations drift from standard).

PICTURE. Three slides side by side: tilted down (green, spontaneous), tilted up (red, forced), and perfectly flat (grey, equilibrium).

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Worked check — Zn/Cu, all the way through

Reaction: $\text{Zn}(s)+{\text{Cu}}^{2+}(aq)\to {\text{Zn}}^{2+}(aq)+\text{Cu}(s)$

$$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=(+0.34)-(-0.76)=+1.10\text{V}$$

$n=2$ (Zn loses 2 e⁻, Cu²⁺ gains 2 e⁻). Then:

$$\Delta G^{\circ }=-nF{E}^{\circ }=-(2)(96,485)(1.10)=-212,267\text{J}\approx -212.3\text{kJ}$$

$E^{\circ }>0\Rightarrow \Delta G^{\circ }<0\Rightarrow$ spontaneous. Every arrow in every figure agrees.

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The one-picture summary

Here is the whole chain compressed: one electron slide → count charge ($nF$) → times height ($E$) → that is the work → same as the free-energy drop ($-\Delta G$) → equate → $\Delta G=-nFE$.

Recall Feynman retelling — the whole walkthrough in plain words

Picture one electron bored on the zinc, itching to jump to a copper ion sitting lower down. That jump is the reaction and the electricity, all at once.

Two questions: how far down does it fall? and how many fall? The fall-per-charge is the voltage $E$ (volts = joules given out per coulomb). The count is $nF$ — $n$ moles of electrons, each mole worth $F=96,485$ coulombs. Multiply how far by how many and you get the total useful energy the cell hands out: $nFE$ joules.

Now the chemists' side. Every reaction has an "energy budget" called $\Delta G$. When it runs by itself, the budget shrinks, so $\Delta G$ is negative; the amount that got spent as useful work is $-\Delta G$. But that "useful work spent" is the same energy the electrons delivered through the wire. Two names, one energy — so $nFE=-\Delta G$, i.e. $\Delta G=-nFE$.

The punchline lives in the minus sign: $n$ and $F$ can never be negative, so a downhill slide (positive $E$) forces a negative $\Delta G$ — the reaction runs on its own. Flat slide ($E=0$) means a stalled budget ($\Delta G=0$): equilibrium. Uphill slide (negative $E$) means positive $\Delta G$: you must plug in a battery to push it.