Worked examples — Cell EMF E°_cell = E°_cathode − E°_anode

This page is the practice ground for the parent note Cell EMF: E°_cell = E°_cathode − E°_anode. There we built the formula. Here we throw every kind of case at it and grind each one out fully, so no exam surprise is left uncovered.

Before we start, let me re-anchor two symbols so you never have to scroll up:

The one habit that saves you everywhere: look up both half-cells as reductions, decide which is cathode (higher $E°$ in a natural cell), then subtract. Never hand-flip a sign.

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The scenario matrix

Every problem this topic can throw at you falls into one of these boxes. Each worked example below is tagged with the box it fills.

#	Case class	What's tricky about it	Example
A	Both $E°$ negative	subtracting two negatives	Ex 1
B	One positive, one negative (the classic)	sign of the difference	Ex 2
C	Both $E°$ positive	which one wins as cathode	Ex 3
D	Forced / "wrong" assignment → negative $E{°}_{\text{cell}}$	reading the minus sign correctly	Ex 4
E	Degenerate: identical half-cells → $E{°}_{\text{cell}}=0$	zero driving force, concentration rescue	Ex 5
F	Unequal electrons ($n$ differ)	you must NOT scale $E°$	Ex 6
G	Real-world word problem (corrosion / battery choice)	translate words → cells	Ex 7
H	Exam twist: work backwards (given $E{°}_{\text{cell}}$, find an unknown $E°$)	rearranging the formula	Ex 8
I	Limiting / spontaneity link ($\Delta G°$ sign)	connect voltage to feasibility	Ex 9

Standard values we reuse (all as reductions, from the table):

$E°({\text{Zn}}^{2+}/\text{Zn})=-0.76,E°({\text{Fe}}^{2+}/\text{Fe})=-0.44,E°({\text{Cu}}^{2+}/\text{Cu})=+0.34,E°({\text{Ag}}^{+}/\text{Ag})=+0.80,E°({\text{Al}}^{3+}/\text{Al})=-1.66$

Look at the number line above: it is the whole game in one picture. $E{°}_{\text{cell}}$ is always the gap between two dots, and the higher dot is always the cathode.

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[!example] Example 1 — Case A: both potentials negative (Zn | Fe)

Statement. Build a cell from Zn and Fe electrodes at standard conditions. $E°({\text{Zn}}^{2+}/\text{Zn})=-0.76\text{ V}$, $E°({\text{Fe}}^{2+}/\text{Fe})=-0.44\text{ V}$. Find $E{°}_{\text{cell}}$ and name the cathode.

Forecast: Two negatives — will the answer be negative? Guess a number before reading on.

1. Rank the two $E°$ values. $-0.44>-0.76$ (on the number line, $-0.44$ sits to the right). Why this step? In a natural (galvanic) cell the higher-$E°$ half is the cathode — this decides who reduces.
2. Assign roles. Cathode = Fe ($-0.44$), anode = Zn ($-0.76$). Why this step? We need to know which number to plug where before subtracting.
3. Subtract. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=(-0.44)-(-0.76)=+0.32\text{ V}$ Why this step? Subtracting a negative adds it back, so two negatives still give a positive gap. The gap is the driving force.

Verify: The gap on the number line between $-0.76$ and $-0.44$ is $0.32$ — a distance is always positive, matching our $+0.32\text{ V}$. Units: V $-$ V $=$ V. ✓ Positive ⇒ spontaneous, Zn dissolves (as it does — Zn protects steel).

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[!example] Example 2 — Case B: one positive, one negative (the Daniel cell revisited)

Statement. Zn and Cu electrodes. $E°({\text{Zn}}^{2+}/\text{Zn})=-0.76$, $E°({\text{Cu}}^{2+}/\text{Cu})=+0.34$. Find $E{°}_{\text{cell}}$. See Daniel Cell (Detailed Mechanism).

Forecast: Positive minus negative — bigger or smaller than either dot? Guess.

1. Rank. $+0.34>-0.76$, so Cu is cathode, Zn is anode. Why this step? Cu holds electrons more tightly → reduction happens there.
2. Subtract. $E{°}_{\text{cell}}=0.34-(-0.76)=+1.10\text{ V}$ Why this step? The minus of a negative adds, so the gap spans across zero — that's why it's larger than either single number.

Verify: Distance from $-0.76$ to $+0.34$ on the line $=0.76+0.34=1.10\text{ V}$. ✓ This is the textbook Daniel-cell voltage. Galvanic, positive, spontaneous.

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[!example] Example 3 — Case C: both potentials positive (Cu | Ag)

Statement. Cu and Ag. $E°({\text{Cu}}^{2+}/\text{Cu})=+0.34$, $E°({\text{Ag}}^{+}/\text{Ag})=+0.80$. Find $E{°}_{\text{cell}}$.

Forecast: Both positive — is the answer positive?

1. Rank. $+0.80>+0.34$, so Ag is cathode, Cu is anode. Why this step? Ag wants electrons more; it reduces.
2. Subtract. $E{°}_{\text{cell}}=0.80-0.34=+0.46\text{ V}$ Why this step? Two positives: the gap is just the ordinary difference.

Verify: $0.80-0.34=0.46>0$ ⇒ spontaneous with Ag${}^{+}$ being reduced and Cu dissolving. ✓ (Matches Example 2 of the parent note.)

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[!example] Example 4 — Case D: the "forced/wrong" assignment gives a negative sign

Statement. Same Cu | Ag pair, but a student insists Cu is the cathode. What does the formula report, and what does the sign mean?

Forecast: Will $E{°}_{\text{cell}}$ come out $+0.46$, $-0.46$, or $0$?

1. Plug the student's (wrong) roles. Cathode = Cu ($0.34$), anode = Ag ($0.80$). $E{°}_{\text{cell}}=0.34-0.80=-0.46\text{ V}$ Why this step? The formula never argues — it just subtracts what you give it.
2. Read the sign. Negative $E{°}_{\text{cell}}$ = non-spontaneous: electrons will not flow that way on their own. Why this step? Sign = feasibility signal, not a computational error.
3. Fix the assignment and you recover Example 3's $+0.46\text{ V}$. Why this step? Nature picks the assignment that makes $E{°}_{\text{cell}}>0$; to run the reverse you'd need an external battery (electrolytic — Galvanic vs Electrolytic Cells).

Verify: $-0.46=-(+0.46)$: the wrong assignment is exactly the negative of the right one. ✓ The magnitude is unchanged; only the sign flips — that is the whole information content of the minus sign.

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[!example] Example 5 — Case E: degenerate, identical half-cells (concentration cell)

Statement. Two Cu electrodes, both in Cu${}^{2+}$ solution. (a) Standard $E{°}_{\text{cell}}$? (b) If the cathode side has $[{\text{Cu}}^{2+}]=1.0\text{ M}$ and the anode side $0.010\text{ M}$, what is the actual $E_{\text{cell}}$? ($n=2$.)

Forecast: Part (a) has a special value — guess it. Part (b): positive, negative, or still zero?

1. Part (a): subtract identical numbers. $E{°}_{\text{cell}}=0.34-0.34=0\text{ V}$ Why this step? Identical half-cells ⇒ zero gap ⇒ zero driving force. This is the degenerate case.
2. Part (b): standard formula can't help (it gave 0), so switch tools to the Nernst Equation: $E_{\text{cell}}=E{°}_{\text{cell}}-\frac{0.0592}{n}\log Q,Q=\frac{[{\text{Cu}}^{2+}{]}_{\text{anode}}}{[{\text{Cu}}^{2+}{]}_{\text{cathode}}}$ Why this step? When $E{°}_{\text{cell}}=0$, concentration is the only thing left that can create voltage — the Nernst term is exactly that concentration correction.
3. Insert numbers. $Q=0.010/1.0=0.010$, $\log (0.010)=-2$. $E_{\text{cell}}=0-\frac{0.0592}{2}(-2)=+0.0592\text{ V}$ Why this step? Dilute side wants to concentrate → it oxidizes (anode), driving a small positive voltage.

Verify: $-\frac{0.0592}{2}\times (-2)=+0.0592$. Positive, as predicted from "$Q<1\Rightarrow \log Q<0\Rightarrow E_{\text{cell}}>0$." ✓ Units: dimensionless log, so V. This is how biological ion sensors read concentration.

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[!example] Example 6 — Case F: unequal electron counts — do NOT scale $E°$

Statement. Al | Cu cell. ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$, $E°=-1.66$; ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, $E°=+0.34$. Find $E{°}_{\text{cell}}$.

Forecast: Since Al uses 3 electrons and Cu uses 2, do we multiply any $E°$ by 3 or 2? (Trap!)

1. Rank as reductions. $+0.34>-1.66$ ⇒ Cu is cathode, Al is anode. Why this step? Ranking uses the tabulated $E°$ regardless of $n$.
2. Subtract — no scaling. $E{°}_{\text{cell}}=0.34-(-1.66)=+2.00\text{ V}$ Why this step? $E°$ is a per-charge quantity (volts = joules per coulomb); it does not change when you multiply a reaction. Only $\Delta G°$ (an energy) scales with $n$.
3. Balance electrons for the reaction, not the voltage. LCM$(3,2)=6$: $2\text{Al}+3{\text{Cu}}^{2+}\to 2{\text{Al}}^{3+}+3\text{Cu}$, $n=6$. Why this step? You need matched electrons to write the equation and later compute energy — but $E{°}_{\text{cell}}$ stays $2.00\text{ V}$.

Verify: $0.34-(-1.66)=2.00$. Independent check via energy: $\Delta G°=-nFE°=-(6)(96485)(2.00)=-1.158\times 10^6\text{ J}=-1158\text{ kJ}$ — large and negative ⇒ very spontaneous. ✓ Note we used $n=6$ only for the energy, never for the volts.

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[!example] Example 7 — Case G: real-world word problem (galvanized nail)

Statement. A steel (≈iron) nail is coated with zinc. A scratch exposes both metals to salty water, forming a tiny cell. Which metal corrodes (oxidizes), and with what standard driving voltage? Use $E°({\text{Fe}}^{2+}/\text{Fe})=-0.44$, $E°({\text{Zn}}^{2+}/\text{Zn})=-0.76$.

Forecast: Does the zinc coating or the iron rust first?

1. Translate words to a cell. Two exposed metals in an electrolyte = a galvanic cell; the one that oxidizes is the anode. Why this step? Corrosion is oxidation at the anode.
2. Lower $E°$ becomes the anode. $-0.76$ (Zn) $<-0.44$ (Fe), so Zn oxidizes, Fe is protected (cathode). Why this step? The metal more eager to give up electrons sacrifices itself — "sacrificial protection."
3. Compute the driving voltage. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=(-0.44)-(-0.76)=+0.32\text{ V}$ Why this step? Positive ⇒ this sacrificial process is spontaneous; the zinc keeps corroding instead of the iron.

Verify: Same maths as Example 1 (Zn|Fe): $+0.32\text{ V}$, positive. ✓ Real chemistry: this is exactly why galvanized steel resists rust — the zinc is the anode and dies first.

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[!example] Example 8 — Case H: exam twist, work backwards for an unknown $E°$

Statement. A cell made of an unknown metal M (as ${\text{M}}^{2+}/\text{M}$) versus a copper cathode gives $E{°}_{\text{cell}}=+1.10\text{ V}$, with M as the anode. $E°({\text{Cu}}^{2+}/\text{Cu})=+0.34$. Find $E°({\text{M}}^{2+}/\text{M})$ and identify M.

Forecast: Will the unknown $E°$ be positive or negative? Guess before solving.

1. Write the formula with the unknown in place. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}\Rightarrow 1.10=0.34-E°({\text{M}}^{2+}/\text{M})$ Why this step? M is the anode, so its $E°$ sits in the "anode" slot — no sign flip, just place it.
2. Rearrange. $E°({\text{M}}^{2+}/\text{M})=0.34-1.10=-0.76\text{ V}$ Why this step? Algebra: move the unknown across, subtract. A big positive cell voltage forces the anode metal to sit far below Cu.
3. Identify from the table. $-0.76\text{ V}$ is zinc. Why this step? Matching the recovered $E°$ to Standard Reduction Potentials (Table) names the metal.

Verify: Plug back: $0.34-(-0.76)=1.10\text{ V}$ ✓. This is just the Daniel cell (Ex 2) read in reverse — consistency confirms the method.

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[!example] Example 9 — Case I: limiting case, does the reaction go? (voltage ↔ ΔG°)

Statement. For the Al | Cu cell of Example 6 ($E{°}_{\text{cell}}=+2.00\text{ V}$, $n=6$), decide whether the reaction is feasible, and how the sign of $\Delta G°$ tracks the sign of $E{°}_{\text{cell}}$ in general. Link: Gibbs Free Energy and Spontaneity.

Forecast: Positive voltage — will $\Delta G°$ be positive or negative?

1. State the bridge equation. $\Delta G°=-nFE{°}_{\text{cell}}$ Why this step? This is the only line that converts "volts" (electrical push) into "kJ" (energy), and its minus sign is what pairs $E{°}_{\text{cell}}>0$ with $\Delta G°<0$.
2. Substitute. $\Delta G°=-(6)(96485)(2.00)=-1.158\times 10^6\text{ J}=-1158\text{ kJ}$. Why this step? A negative $\Delta G°$ is the textbook flag for spontaneous.
3. Read the general rule by inspecting the sign of $-nF$ (always negative, since $n,F>0$):
   • $E{°}_{\text{cell}}>0\Rightarrow \Delta G°<0\Rightarrow$ spontaneous (galvanic).
   • $E{°}_{\text{cell}}=0\Rightarrow \Delta G°=0\Rightarrow$ at equilibrium (the Ex 5 degenerate case).
   • $E{°}_{\text{cell}}<0\Rightarrow \Delta G°>0\Rightarrow$ non-spontaneous (needs external power — Ex 4). Why this step? This single sign map covers all limiting behaviours in one glance.

Verify: $-(6)(96485)(2.00)=-1157820\text{ J}$; negative as predicted for $E{°}_{\text{cell}}>0$. ✓ Uses Faraday's constant $F=96485\text{ C/mol}$.

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[!recall]- Rapid self-check (reveal after guessing)

Connections

• Parent: 2.7.03 Cell EMF E°_cell = E°_cathode − E°_anode (Hinglish)
• Standard Reduction Potentials (Table)
• Nernst Equation · Gibbs Free Energy and Spontaneity
• Galvanic vs Electrolytic Cells · Daniel Cell (Detailed Mechanism) · Faraday's Laws of Electrolysis