Worked examples — Galvanic (voltaic) cells — anode (oxidation), cathode (reduction)

This page is the drill hall for the parent topic on galvanic cells. The parent taught you what an anode and cathode are and why electrons flow. Here we hit every kind of number a problem can throw at you — both potentials positive, both negative, mixed signs, a near-tie, a same-metal cell, a real-world battery, and an exam trap where the reaction is run backwards.

Before we start, let us re-earn the two tools, from zero, so no symbol appears unexplained.

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The scenario matrix

Every galvanic-cell number problem falls into one of these cells. The examples below are labelled with the matrix cell they cover.

#	Case class	What's tricky	Example
A	Mixed signs (one $+E°$, one $-E°$)	The textbook case	Ex 1
B	Both $E°$ positive	"Negative" number is still the anode	Ex 2
C	Both $E°$ negative	Both hate electrons; least-hungry still wins	Ex 3
D	Near-tie ($E°\approx E°$)	Tiny but real voltage; sign decides direction	Ex 4
E	Degenerate: same metal, different concentration	$E{°}_{\text{cell}}=0$, yet a real voltage exists	Ex 5
F	Faraday link: charge → mass deposited	Convert electrons to grams	Ex 6
G	Real-world word problem	Strip the story, find the two half-cells	Ex 7
H	Exam twist: reaction written backwards	Recognise a non-spontaneous / electrolytic setup	Ex 8

Reference table (all standard, vs Standard Hydrogen Electrode):

Look at this picture first — it is the mental number line every example plugs into.

How to read this figure. The vertical axis is $E°$ in volts, running from very negative at the bottom to positive at the top — it is the "hunger for electrons" scale. Each short horizontal tick is one electrode, labelled with its metal and $E°$ value (e.g. Ag sits high at $+0.80$, Mg sits far down at $-2.37$). The red arrow is $E{°}_{\text{cell}}$: it always starts at the anode (the lower electrode) and points up to the cathode (the higher electrode); its length equals the voltage, and the label spells out $\text{big}-\text{small}$ for the Zn→Cu case ($1.10\text{V}$). The words at top and bottom remind you that "higher = more hungry = cathode" and "lower = less hungry = anode." That single picture solves cases A, B, C, and D — only the positions of the two chosen electrodes change; the red arrow always runs bottom-up.

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Case A — Mixed signs

Forecast: one score is negative, one positive — guess which metal is forced to give up electrons, and roughly how many volts the gap is.

1. Compare hunger scores. $+0.34>-0.76$, so Cu is hungrier. Why this step? Bigger $E°$ = cathode; this is the whole decision.
2. Assign electrodes. Cathode = Cu ($+0.34$), anode = Zn ($-0.76$). Why? The loser (Zn) is oxidised; that is the definition of anode from the parent note.
3. Apply the formula. $E{°}_{\text{cell}}=0.34-(-0.76)=1.10\text{V}$ Why? Big $-$ small = the gap the winner out-pulls the loser by.
4. Electron direction: out of Zn, through the wire, into Cu. Why? Electrons are produced where oxidation happens (anode).

Verify: $E{°}_{\text{cell}}>0$ ✓ (spontaneous, as a real Daniel cell must be). Units: $\text{V}-\text{V}=\text{V}$ ✓. On the figure, red arrow runs Zn → Cu, length $1.10$ ✓.

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Case B — Both potentials positive

Forecast: both numbers are positive — does that mean neither wants to be the anode? Guess before reading.

1. Compare. $+0.80>+0.34$. Ag is hungrier. Why this step? The rule is relative, not about the sign. A positive score can still be the smaller one.
2. Assign. Cathode = Ag, anode = Cu. Why? Copper, though positive, is the less hungry of the two, so it is forced to give up electrons. "Positive $E°$" does not mean "cannot be an anode."
3. Formula. $E{°}_{\text{cell}}=0.80-0.34=0.46\text{V}$ Why this step? We need the size of the voltage the cell delivers; it is the gap between the winner (Ag) and loser (Cu) on the number line. Note: $E°$ values are intensive — they do not get multiplied when you scale a half-reaction, so no scaling affects this number.
4. Half-reactions — and balancing the electrons. Cathode: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$. Anode: $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$. The anode releases 2 electrons but the silver half-reaction only absorbs 1, so we multiply the silver reaction by 2 to make the electrons match: $2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$ Adding the two balanced halves (the $2e^{-}$ now cancel): $\text{Cu}+2{\text{Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$ Why this step? Electrons can never be left over — every electron the anode gives up must be caught by the cathode. Doubling silver is the only way to conserve them. Crucially, doubling a half-reaction does not double its $E°$ (voltage is a "per-electron pressure", not a total), so $E{°}_{\text{cell}}$ stays $0.46\text{V}$.

Verify: $E{°}_{\text{cell}}=0.46>0$ ✓ spontaneous. Balanced net reaction $\text{Cu}+2{\text{Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$ conserves charge ($+2$ each side) and electrons ✓.

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Case C — Both potentials negative

Forecast: two negative scores — is there even a cathode? Yes: "least reluctant" still counts as "most hungry."

1. Compare. On the number line, $-0.44$ sits above $-0.76$ (less negative = larger). Why this step? $-0.44>-0.76$. Being "less negative" makes Fe the relative winner.
2. Assign. Cathode = Fe ($-0.44$), anode = Zn ($-0.76$). Why this step? The rule "bigger $E°$ = cathode" is about relative position, not sign — so even though Fe's score is negative, it is the larger of the two and therefore eats electrons (cathode), while Zn is forced to give them up (anode).
3. Formula. $E{°}_{\text{cell}}=(-0.44)-(-0.76)=0.32\text{V}$ Why? Subtracting a more negative anode value flips it to a plus, giving a positive gap.

Verify: $0.32>0$ ✓ (a real Zn–Fe cell does drive current — this is why zinc coating protects iron, sacrificial anode idea). The two minus signs made a plus: $-0.44+0.76=0.32$ ✓.

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Case D — Near-tie: sign of the gap decides the direction

Forecast: the scores differ by only $0.01$. Guess: does the cell die, or run at a tiny voltage?

1. Compare. $-0.13>-0.14$ by a hair. Pb is (barely) hungrier. Why this step? Even a razor-thin difference still ranks the two electrodes; the comparison, not the closeness, decides who is cathode.
2. Assign. Cathode = Pb, anode = Sn. Why this step? Pb has the larger (less negative) $E°$, so by "bigger $E°$ = cathode" Pb eats electrons and Sn is forced to dissolve.
3. Formula. $E{°}_{\text{cell}}=(-0.13)-(-0.14)=+0.01\text{V}$ Why this matters: even a razor-thin positive gap means spontaneous. The sign, not the size, decides direction; the size only tells you how weak the battery is.

Verify: $+0.01>0$ ✓, so it runs (feebly). Had we mislabelled and computed $-0.01$, that would mean "runs the other way" — a signal we swapped electrodes. Cross-check: bigger $E°$ (Pb) really is the cathode ✓.

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Case E — Degenerate: same metal both sides (a concentration cell)

Here $E{°}_{\text{cell}}=0$ because both electrodes are chemically identical — yet a voltage still appears. This is the "zero/degenerate input" cell of the matrix.

Forecast: same metal, so $E{°}_{\text{cell}}=0$. Guess: is the voltage exactly zero, or not?

1. Standard part. $E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=0.34-0.34=0\text{V}$. Why this step? Identical metals ⇒ identical hunger scores ⇒ zero standard gap. This is the degenerate case.
2. Write the actual cell reaction so we can build $Q$ honestly. The concentrated side is reduced (cathode) and the dilute side is oxidised (anode): $\underset{\text{cathode}}{\underbrace{{\text{Cu}}^{2+}(\text{1.0 M})+2e^{-}\to \text{Cu}}}+\underset{\text{anode}}{\underbrace{\text{Cu}\to {\text{Cu}}^{2+}(\text{0.01 M})+2e^{-}}}$ $\text{net:}{\text{Cu}}^{2+}(\text{1.0 M})\to {\text{Cu}}^{2+}(\text{0.01 M})$ Why this step? You cannot form $Q$ until you know which ions are products and which are reactants; the net reaction shows the $1.0\text{M}$ ion is consumed (reactant) and the $0.01\text{M}$ ion is produced (product).
3. Correct for concentration with the Nernst tool. Because concentrations differ, $E°$ alone is not enough — reach for the Nernst Equation, which corrects $E°$ for real concentrations: $E_{\text{cell}}=E{°}_{\text{cell}}-\frac{0.0592}{n}{\log }_{10}Q$ Why this tool and not the standard formula? The standard formula only knows the metals, not the crowding of ions. Nernst is the only tool that reads concentration. Here $n=2$ (two electrons transferred), and ${\log }_{10}$ is the base-10 logarithm (the $0.0592\text{V}$ constant is tied to base-10 at $25{}^{\circ }\text{C}$).

4. Plug in. $E_{\text{cell}}=0-\frac{0.0592}{2}{\log }_{10}(0.01)=-0.0296\times (-2)=0.0592\text{V}$ Why this step? Substituting $Q=0.01$ and $n=2$ turns the "zero standard gap" into the real, nonzero working voltage.

Verify: ${\log }_{10}(0.01)=-2$, so $E=-0.0296\times (-2)=+0.0592\text{V}>0$ ✓. Direction: electrons flow from the dilute (anode) to the concentrated (cathode) electrode ✓.

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Case F — Faraday link: from charge to grams

Forecast: guess whether it is grams or milligrams. (Ampere-minutes are small — expect under a gram.)

1. Charge passed: $Q=It=2.0\text{A}\times (30\times 60\text{s})=3600\text{C}$. Why this step? Charge = current × time; this is the total "electron traffic."
2. Moles of electrons: $n_{e^{-}}=\frac{Q}{F}=\frac{3600}{96485}=0.03731\text{mol}$. Why? Divide total charge by charge-per-mole to count electron-moles.
3. Moles of Cu: cathode reaction ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$ needs 2 electrons per Cu, so $n=2$ and $n_{\text{Cu}}=n_{e^{-}}/2=0.018655\text{mol}$. Why? The half-reaction's coefficient is the exchange rate electrons ↔ atoms.
4. Mass (finish the calculation): $m=n_{\text{Cu}}\times M=0.018655\times 63.5=1.185\text{g}$ Equivalently, straight from the master formula: $m=\frac{QM}{nF}=\frac{3600\times 63.5}{2\times 96485}=1.185\text{g}$. Why this step? Multiplying moles of Cu by molar mass converts the atom count into a weighable mass — the final answer the question asked for.

Verify: $Q=3600\text{C}$; $3600/96485=0.03731$ mol e⁻; halved $=0.018655$ mol Cu; $\times 63.5=1.185\text{g}$ ✓. Master formula gives the same $1.185\text{g}$ ✓. Units: $\text{A}\cdot \text{s}=\text{C}$, $\text{C}/(\text{C/mol})=\text{mol}$ ✓. Sanity: under one-and-a-bit grams for a small current — matches the forecast ✓.

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Case G — Real-world word problem

Forecast: Mg is very negative — expect a large voltage. Guess which metal disappears over time.

1. Find the two half-cells hiding in the words. Metals: Mg, Ag. Ignore "sensor" and "engineer" — only the electrodes matter. Why this step? Word problems bury two $E°$ values in a story; the physics is unchanged.
2. Compare. $+0.80>-2.37$. Ag is the cathode; Mg is the anode. Why this step? Silver's much larger $E°$ makes it the winner that eats electrons (cathode); Mg, far lower on the number line, is forced to give them up (anode).
3. Formula. $E{°}_{\text{cell}}=0.80-(-2.37)=3.17\text{V}$ Why this step? The voltage the sensor sees is the gap between the two electrodes; big $-$ small gives it directly.
4. Which metal wears away? The anode dissolves ($\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}$), so the magnesium bar shrinks over time. Why? Oxidation is the metal atoms leaving as ions.

Verify: $0.80+2.37=3.17\text{V}>0$ ✓ (a genuinely strong single cell). Mg, the more-negative electrode, is the sacrificial anode ✓ — the same principle used in Battery Technologies and corrosion protection.

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Case H — Exam twist: the reaction written backwards

Forecast: guess whether the reversed reaction can be a spontaneous battery, or whether you must force it.

1. Compute $E{°}_{\text{cell}}$ for the reversed reaction as written. Reversed means Cu²⁺ is now reduced (cathode) and Ag is oxidised (anode): $E{°}_{\text{cell}}=E{°}_{\text{Cu}}-E{°}_{\text{Ag}}=0.34-0.80=-0.46\text{V}$ Why this step? We test the claimed direction by assigning the electrodes exactly as that direction demands, then computing the gap.
2. Read the sign. $E{°}_{\text{cell}}<0$. Why this step? A negative gap means the "winner" chosen is actually the loser — the reaction does not run on its own.
3. Conclusion. The reversed reaction is non-spontaneous. To make it happen you must push electrons uphill with an external power supply of more than $0.46\text{V}$ — that is an electrolytic cell, not a galvanic one. Why? Positive $E{°}_{\text{cell}}$ = spontaneous (galvanic); negative = must be forced (electrolytic). This links to Gibs Free Energy and Cell Potential via $\Delta G°=-nFE{°}_{\text{cell}}$: a negative $E{°}_{\text{cell}}$ makes $\Delta G°$ positive, and a positive $\Delta G°$ means "not spontaneous."

Verify: $0.34-0.80=-0.46\text{V}$ ✓ (negative confirms non-spontaneous). The forward direction $\text{Cu}+2{\text{Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$ gives $+0.46\text{V}$ (Ex 2) — the true spontaneous direction ✓.

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Recall

Recall Which electrode is decided by the

sign of $E°$? Neither — it's decided by comparison. Bigger $E°$ = cathode ::: A positive $E°$ metal can still be the anode if its partner is even more positive (Ex 2).

Recall Both electrodes have negative

$E°$. Is there still a cathode? Yes ::: the less negative one is the relative "winner" and becomes the cathode (Ex 3).

Recall Same metal, both sides, different concentrations. What is

$E{°}_{\text{cell}}$, and is the voltage zero? $E{°}_{\text{cell}}=0$, but the actual voltage is not zero ::: the Nernst equation gives a small nonzero $E$ from the concentration difference (Ex 5).

Recall In the Nernst equation, what is

$Q$ and which log is it? $Q$ = products over reactants (each to its coefficient), pure solids count as 1; the log is base-10 (${\log }_{10}$) ::: for the Cu concentration cell $Q=[\text{dilute}]/[\text{concentrated}]=0.01$ (Ex 5).

Recall You compute

$E{°}_{\text{cell}}=-0.46\text{V}$. What does the negative sign tell you? The reaction as written is non-spontaneous ::: either you swapped anode/cathode, or it needs an external power supply (electrolytic cell) (Ex 8).

Recall

$2.0\text{A}$ for $30\text{min}$ deposits how much Cu? Which formula? About $1.185\text{g}$ ::: via the master formula $m=QM/(nF)$ with $Q=It=3600\text{C}$, $n=2$ (Ex 6).

Recall Iron appears twice in the table. Which

$E°$ do I use? Match the species present ::: solid Fe electrode → ${\text{Fe}}^{2+}/\text{Fe}$ ($-0.44$); only Fe³⁺/Fe²⁺ ions on inert electrode → $+0.77$.