Worked examples — Standard electrode potentials — SHE reference, electrochemical series

This page is the worked-examples floor for the parent topic. The parent gave you the tools: the half-reaction idea, the SHE zero, and the master formula. Here we drill every kind of question the topic can throw at you — every sign, the tricky zero cases, the degenerate "same electrode" case, the limiting agents at the ends of the series, a real-world word problem, and an exam trap. You will never meet a scenario on this topic that isn't a copy of one of these.

Before we start, one promise from the parent kept in plain words:

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The scenario matrix

Think of every exam question as landing in one of these boxes. The figure below is the single picture that organises the whole topic: a vertical potential axis (volts) with each half-reaction drawn as a coloured tick at its own $E°$.

Figure s01 — reading it: the white vertical line is the $E°$ axis in volts, arrow pointing up toward more positive values. The dashed yellow line at $E°=0$ is the SHE reference. Each coloured dot-and-dash is one half-reaction, labelled with its value: blue dots (${\text{Li}}^{+}/\text{Li}$ at $-3.04$, ${\text{Zn}}^{2+}/\text{Zn}$ at $-0.76$) sit below SHE — the strong reducers. Green dots (${\text{Cu}}^{2+}/\text{Cu}$ at $+0.34$, ${\text{Ag}}^{+}/\text{Ag}$ at $+0.80$) sit just above; the red dot (${\text{F}}_2/{\text{F}}^{-}$ at $+2.87$) sits at the very top — the strongest oxidiser. The red caption at the top and blue caption at the bottom tell you the rule: whoever sits higher on this line is the cathode (gets reduced); whoever sits lower is the anode (gets oxidised). Every example below is just "pick two dots and read the vertical gap between them."

In the table, the last column "Hit by" means "the worked example on this page that fully demonstrates this case" — so Case A is worked out in Example 1, Case B in Example 2, and so on. Together, Examples 1–9 cover every row.

#	Case class	What's tricky	Hit by (worked example)
A	Anode below SHE (negative $E°$)	measured cell gives a minus sign	Example 1
B	Cathode above SHE (positive $E°$)	reference is the anode	Example 2
C	Both electrodes on the same side of SHE	SHE not even in the cell	Example 3
D	Ordering three+ species	which pair reacts, which just sits	Example 4
E	Non-spontaneous direction ($E{°}_{\text{cell}}<0$)	the reaction runs backwards	Example 5
F	Limiting / extreme ends (F₂ strongest oxidiser, Li strongest reducer)	end-of-series behaviour	Example 6
G	Degenerate case: same half-reaction both sides ($E{°}_{\text{cell}}=0$)	zero output, equilibrium	Example 7
H	Real-world word problem (a battery / corrosion)	translate words → half-cells	Example 8
I	Exam twist: $n$ does not scale $E°$ (and where $\Delta G°$ does use $n$)	doubling a half-reaction	Example 9

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[!example] Example 1 — Case A: an anode below SHE (the sign trap)

Statement. A cell is built: $\text{Pt}|{\text{H}}_2(1\text{bar})|{\text{H}}^{+}(1\text{M})||{\text{Ni}}^{2+}(1\text{M})|\text{Ni}$. Experiment shows Ni dissolves and H₂ gas forms. The voltmeter reads $E{°}_{\text{cell}}=+0.25\text{V}$. Find $E{°}_{{\text{Ni}}^{2+}/\text{Ni}}$.

Forecast: guess the sign first — will $E{°}_{{\text{Ni}}^{2+}/\text{Ni}}$ come out positive or negative? (Ni is being eaten, so...)

1. Assign electrodes from the observation. Ni dissolves → Ni is oxidised → Ni is the anode. H₂ forms → H⁺ is reduced → SHE is the cathode. Why this step? Signs are decided by what physically happens, not by which side you drew.
2. Write the master formula. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=E{°}_{{\text{H}}^{+}/{\text{H}}_2}-E{°}_{{\text{Ni}}^{2+}/\text{Ni}}$ Why this step? It's the only equation linking the reading to the unknown.
3. Insert the SHE zero and solve. $+0.25=0.00-E{°}_{{\text{Ni}}^{2+}/\text{Ni}}\Rightarrow E{°}_{{\text{Ni}}^{2+}/\text{Ni}}=-0.25\text{V}$ Why this step? SHE is defined as $0$, so its term vanishes and isolates the unknown.

Verify: A negative $E°$ means "reduction is non-spontaneous vs hydrogen" — i.e. Ni prefers to oxidise, which is exactly what we observed (Ni dissolved). Consistent. Units: volts. ✓

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[!example] Example 2 — Case B: a cathode above SHE (reference is the anode)

Statement. Cell: $\text{Pt}|{\text{H}}_2(1\text{bar})|{\text{H}}^{+}(1\text{M})||{\text{Ag}}^{+}(1\text{M})|\text{Ag}$. Silver deposits, H₂ is consumed. $E{°}_{\text{cell}}=+0.80\text{V}$. Find $E{°}_{{\text{Ag}}^{+}/\text{Ag}}$.

Forecast: which side is the anode now — Ag or SHE?

1. Assign electrodes. Ag deposits (Ag⁺ gains e⁻) → reduction → Ag is cathode. H₂ is consumed (oxidised) → SHE is anode. Why this step? Here hydrogen is being oxidised, the opposite role to Example 1 — this is the whole point of case B.
2. Apply master formula. $+0.80=E{°}_{{\text{Ag}}^{+}/\text{Ag}}-E{°}_{{\text{H}}^{+}/{\text{H}}_2}=E{°}_{{\text{Ag}}^{+}/\text{Ag}}-0.00$ Why this step? The formula is the bridge from the measured cell reading to the single unknown; here Ag is the cathode term and SHE is the anode term, so this arrangement (not the Example 1 arrangement) is the correct one to write.
3. Solve. $E{°}_{{\text{Ag}}^{+}/\text{Ag}}=+0.80\text{V}$. Why this step? Direct read-off once SHE drops out.

Verify: Positive $E°$ ⇒ Ag⁺ reduction is more favourable than H⁺ reduction, so Ag⁺ should win over SHE — matches Ag depositing. ✓

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[!example] Example 3 — Case C: both electrodes on the same side (no SHE in sight)

Statement. Build the classic Daniell cell $\text{Zn}|{\text{Zn}}^{2+}(1\text{M})||{\text{Cu}}^{2+}(1\text{M})|\text{Cu}$. Given $E{°}_{{\text{Zn}}^{2+}/\text{Zn}}=-0.76\text{V}$, $E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=+0.34\text{V}$. Find $E{°}_{\text{cell}}$.

Forecast: SHE never appears here. Guess the number before reading on.

1. Rank on the number line. $+0.34>-0.76$, so Cu²⁺/Cu is higher → cathode; Zn²⁺/Zn is lower → anode. Why this step? Higher $E°$ = pulls electrons harder = gets reduced. This is the picture, not a rule to memorise.
2. Master formula. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=(+0.34)-(-0.76)$ Why this step? Note the double minus — the anode's own value is negative, so subtracting it adds.
3. Compute. $E{°}_{\text{cell}}=0.34+0.76=+1.10\text{V}$.

Verify: Positive ⇒ spontaneous; the real Daniell cell famously delivers ≈ 1.10 V. The reference cancelled because both potentials were already measured against the same SHE. ✓

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[!example] Example 4 — Case D: three species, who actually reacts?

Statement. Iron filings are dropped into a solution containing both ${\text{Cu}}^{2+}$ and ${\text{Ag}}^{+}$. Using $E{°}_{{\text{Fe}}^{2+}/\text{Fe}}=-0.44\text{V}$, $E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=+0.34\text{V}$, $E{°}_{{\text{Ag}}^{+}/\text{Ag}}=+0.80\text{V}$, which ion is reduced most favourably by Fe, and what is that cell potential?

Forecast: Fe is the reducing agent (lowest, gets oxidised). Which oxidiser gives the biggest gap?

1. Fe is the anode. It sits lowest ($-0.44$), so Fe → Fe²⁺ + 2e⁻ (oxidation). Why? Lowest species on the line is forced to give up electrons.
2. Compare each oxidiser's gap to Fe.
   • With Cu²⁺: $E{°}_{\text{cell}}=0.34-(-0.44)=+0.78\text{V}$
   • With Ag⁺: $E{°}_{\text{cell}}=0.80-(-0.44)=+1.24\text{V}$ Why this step? Bigger positive gap = stronger driving force, so it happens preferentially.
3. Conclude. Ag⁺ (gap +1.24 V) is reduced most favourably; both are spontaneous, but Ag⁺ dominates.

Verify: Both gaps positive ⇒ Fe reduces both ions (correct — iron does displace both). Ag⁺ having the larger gap matches Ag⁺ being higher on the series. ✓

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[!example] Example 5 — Case E: the non-spontaneous direction

Statement. Can $\text{Cu}$ metal reduce ${\text{Zn}}^{2+}$ ions (i.e. run the Daniell reaction backwards)? Same values as Example 3.

Forecast: we already know Zn→Cu²⁺ works; guess the sign of the reverse.

1. Propose the reverse reaction. Cu → Cu²⁺ (oxidation, anode) and Zn²⁺ → Zn (reduction, cathode).
2. Apply master formula for THIS assignment. $E{°}_{\text{cell}}=E{°}_{{\text{Zn}}^{2+}/\text{Zn}}-E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=-0.76-0.34=-1.10\text{V}$ Why this step? We deliberately put the lower species as cathode to test the reverse — the formula reports the penalty.
3. Read the sign. $E{°}_{\text{cell}}=-1.10\text{V}<0$ ⇒ non-spontaneous.

Verify: Exactly the negative of Example 3 — reversing a reaction flips the sign of $E{°}_{\text{cell}}$, as it must. So Cu will not reduce Zn²⁺. ✓

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[!example] Example 6 — Case F: the limiting ends of the series

Statement. Fluorine gas ($E{°}_{{\text{F}}_2/{\text{F}}^{-}}=+2.87\text{V}$) is bubbled over lithium metal ($E{°}_{{\text{Li}}^{+}/\text{Li}}=-3.04\text{V}$). Find $E{°}_{\text{cell}}$ and comment on why this is the "biggest possible" driving force among common species.

Forecast: these are the two extremes of the table. Guess how large the voltage gets.

1. Assign by position. F₂ is the highest common $E°$ → strongest oxidiser → cathode. Li is the lowest → strongest reducer → anode.
2. Master formula. $E{°}_{\text{cell}}=2.87-(-3.04)=+5.91\text{V}$
3. Interpret the limit. No common pair gives a bigger gap, because F₂ sits at the top and Li near the bottom of the electrochemical series.

Verify: Enormous positive value ⇒ violently spontaneous (fluorine + lithium is indeed explosive). The magnitude ~5.9 V is why lithium-based cells reach high voltages. ✓

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[!example] Example 7 — Case G (degenerate): the concentration-cell zero

Statement. A cell has copper on both sides, both at $1\text{M}$ Cu²⁺: $\text{Cu}|{\text{Cu}}^{2+}(1\text{M})||{\text{Cu}}^{2+}(1\text{M})|\text{Cu}$. What is $E{°}_{\text{cell}}$?

Forecast: same electrode, same concentration — will it push electrons at all?

1. Identify the cathode and anode. Both are the same half-reaction $C{u}^{2+}+2e^{-}\to Cu$ with $E°=+0.34\text{V}$.
2. Master formula. $E{°}_{\text{cell}}=0.34-0.34=0.00\text{V}$ Why this step? Subtracting a value from itself is the degenerate case — the maths must give exactly zero.
3. Interpret. $E{°}_{\text{cell}}=0$ ⇒ at standard conditions there is no driving force; the cell is at equilibrium.

Verify: Zero output is the correct degenerate answer. (If the concentrations differed, a small voltage would appear — that's the Nernst equation, the next topic.) Units check: V − V = V. ✓

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[!example] Example 8 — Case H: real-world word problem (galvanised iron)

Statement. A steel (iron) fence is coated with zinc to stop it rusting ("galvanising"). Using $E{°}_{{\text{Fe}}^{2+}/\text{Fe}}=-0.44\text{V}$ and $E{°}_{{\text{Zn}}^{2+}/\text{Zn}}=-0.76\text{V}$, explain with a number why zinc protects iron, and find the driving-force potential for zinc dissolving in preference to iron.

Forecast: which metal will be sacrificed — the coating or the fence?

1. Translate to competing anodes. Two metals could be oxidised by the same oxygen cathode: Fe ($-0.44$) or Zn ($-0.76$). Whichever is lower on the line oxidises preferentially. Why this step? Corrosion = oxidation; the series ranks willingness to be oxidised, so the lower metal is sacrificed.
2. Compare positions. Zn ($-0.76$) is lower than Fe ($-0.44$), so Zn is the anode and corrodes instead of the iron. This is "sacrificial protection." Why this step? Lowest species = strongest reducer = the one actually consumed, so the iron is spared.
3. Quantify how much more willing Zn is by pairing the two metals directly. Treat Zn as the anode (oxidised) and Fe²⁺/Fe as the reference cathode of the comparison, then apply the master formula: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=E{°}_{{\text{Fe}}^{2+}/\text{Fe}}-E{°}_{{\text{Zn}}^{2+}/\text{Zn}}=-0.44-(-0.76)=+0.32\text{V}$ Why this step? A positive gap in this metal-versus-metal comparison confirms, with an actual voltage, that Zn oxidises in preference to Fe — the "how much" behind the "which."

Verify: $E{°}_{\text{cell}}=+0.32\text{V}>0$ ⇒ Zn spontaneously oxidises in preference to Fe, so the iron is spared until the zinc is used up. Matches real galvanising behaviour — with oxygen (not Fe²⁺) doing the actual electron-accepting in the environment. ✓

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[!example] Example 9 — Case I (exam twist): does doubling a half-reaction double $E°$?

Statement. A student writes $2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$ and claims its potential is $2\times 0.80=+1.60\text{V}$. For the cell $\text{Cu}|{\text{Cu}}^{2+}||{\text{Ag}}^{+}|\text{Ag}$ (with $E{°}_{{\text{Cu}}^{2+}/\text{Cu}}=+0.34\text{V}$, $E{°}_{{\text{Ag}}^{+}/\text{Ag}}=+0.80\text{V}$), is $E{°}_{\text{cell}}=1.60-0.34$? Find the correct value, and the reaction's $\Delta G°$.

Forecast: does multiplying atoms multiply volts? (Trap alert.)

1. State the principle. $E°$ is an intensive property — a per-electron pulling strength. Scaling the equation does not scale $E°$. So $E{°}_{{\text{Ag}}^{+}/\text{Ag}}$ stays $+0.80\text{V}$. Why this step? Voltage is like temperature (intensive): two cups of hot water aren't hotter. Only $\Delta G$ (extensive, via $n$) scales.
2. Correct cell potential. Ag⁺/Ag is the cathode (higher), Cu²⁺/Cu the anode: $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=0.80-0.34=+0.46\text{V}$ Why this step? Master formula with the unscaled $E°$ values — the student's $1.60$ is the trap.
3. Where $n$ does matter — the $\Delta G°$ link. Balancing the whole reaction $\text{Cu}+2{\text{Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$ transfers $n=2$ electrons, so: $\Delta G°=-nFE{°}_{\text{cell}}=-2\times 96485\times 0.46\approx -8.877\times 10^4\text{J}=-88.8\text{kJ}$ Why this step? This is the promised payoff of the $\Delta G°=-nFE{°}_{\text{cell}}$ relation from the definition block: the energy (extensive) uses $n$; the voltage (intensive) does not. See cell potentials for how $n$ enters a full cell.

Verify: $E{°}_{\text{cell}}=+0.46\text{V}$ (not $1.26\text{V}$ from the wrong "doubled" claim). $\Delta G°\approx -88.8\text{kJ}<0$ ⇒ spontaneous, consistent with the positive $E{°}_{\text{cell}}$. ✓

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[!recall]- Quick self-test (reveal after guessing)