Galvanic (voltaic) cells — anode (oxidation), cathode (reduction): Question bank

Before you start, pin these anchors so the reveals make sense:

Oxidation = losing electrons. The electrode that loses electrons feeds them into the wire.
Reduction = gaining electrons. The electrode that pulls electrons out of the wire.
Anode = where oxidation happens. Cathode = where reduction happens. (Related tools: Oxidation Numbers to track who lost/gained, Reduction Potentials to predict which is which.)

Definition The symbols this page leans on — define them before you use them

$E$ = potential (voltage), the generic case. A "push strength" for electrons, measured in volts (V). Bigger $E$ = harder shove. Use plain $E$ (no circle) whenever conditions are not specified — any concentration, any temperature.
The little circle $°$ means "standard conditions" — a fixed reference setup so everyone measures on the same footing: every dissolved ion at 1 mol/L (1 M) concentration, any gas at 1 bar, temperature 25 °C. So $E °$ reads "E-standard": the potential you'd measure specifically under those agreed conditions. The rule for the whole page: a symbol with $°$ is locked to standard conditions; the same symbol without $°$ is the general value.
$E °_{cell}$ = the standard potential of the whole cell — the net push driving electrons around the circuit, measured under standard conditions. (Its non-standard cousin $E_{cell}$ , without the circle, is what the Nernst Equation gives you once concentrations drift.)
$E °_{cathode}$ , $E °_{anode}$ = the standard reduction potentials of the two electrodes (both looked up as reductions in a table, see Reduction Potentials).
$n$ = number of electrons transferred per unit of reaction (e.g. $n = 2$ for the Zn–Cu cell, since 2 electrons move per Zn atom). Here "unit of reaction" means one mole of the reaction exactly as the balanced equation is written — for $Zn + Cu^{2 +} \to Zn^{2 +} + Cu$ that is one mole of Zn reacting, carrying $n = 2$ moles of electrons.
$F$ = Faraday's constant $\approx 96, 500$ C/mol — the charge carried by one mole of electrons. It is just the "exchange rate" between moles of electrons and coulombs of charge (see Faraday's Laws of Electrolysis).
$Δ G °$ = standard Gibbs free energy change — the maximum useful energy the reaction can release, under standard conditions. Negative $Δ G °$ = spontaneous (runs on its own); positive = needs a push.

Formula Why

$Δ G ° = - n F E °_{cell}$ — built from scratch This link between energy and voltage is the hinge of the whole topic, so let's earn it rather than quote it. We start with the generic voltage $E$ (no circle, any conditions) and only lock to $E °_{cell}$ once we impose standard conditions in Step 4. Step 1 — what electrical work is (generic $E$ ). Voltage $E$ is defined as energy per unit charge: $E = \frac{energy}{charge}$ . Rearranged, the electrical work a cell can do is $work = charge \times E$ . Nothing here assumes standard conditions yet, so we keep the plain $E$ . Step 2 — how much charge moves. Take one mole of the reaction as written in the balanced equation — that is our "unit of reaction." For that one mole, $n$ moles of electrons cross over. Each mole carries $F$ coulombs, so total charge $= n F$ . Step 3 — put them together (still generic). Maximum electrical work $= (n F) \times E$ , using the plain generic voltage. Step 4 — impose standard conditions, then connect to free energy. Now fix everything at standard conditions: $E$ becomes $E °_{cell}$ and the work becomes $Δ G °$ . $Δ G °$ is defined as the maximum useful (non-expansion) work the reaction can deliver, but with a minus sign by convention — energy released by the system counts as negative. Hence: $Δ G ° = - n F E °_{cell}$ Read it out: a positive $E °_{cell}$ makes $Δ G °$ negative → spontaneous → a working galvanic cell. That single sign flip is why "positive cell voltage = it runs on its own."

The picture below ties all these symbols to the physical cell — glance at it, then the reveals will land.

Figure — Galvanic (voltaic) cells — anode (oxidation), cathode (reduction)

True or false — justify

True or false: In a galvanic cell the anode is the positive terminal.

False. In a galvanic cell the anode supplies electrons to the wire, so it is electron-rich and labelled negative; the cathode pulls electrons in and is the positive terminal.

True or false: The electrode called "anode" is always negative, in every kind of cell.

False. "Anode" only ever means oxidation happens here — that is the fixed rule. Its sign flips depending on cell type: negative in a galvanic cell, positive in an electrolytic cell where a power supply forces oxidation there.

True or false: Electrons travel through the salt bridge from anode to cathode.

False. Electrons never enter the salt bridge — they go through the external metal wire. The salt bridge carries ions, which is how the circuit is completed inside the solution.

True or false: A positive

E °_{cell}

guarantees the reaction is spontaneous as written.

True. By

Δ G ° = - n F E °_{cell}

, a positive cell potential forces the Gibbs free energy change negative — the definition of spontaneity.

True or false: The metal with the more positive standard reduction potential becomes the anode.

False. More positive

E °

means it "wants" electrons more, so it gets reduced — that makes it the cathode. The more negative

E °

metal is the anode.

True or false: Making the electrodes bigger increases the cell voltage.

False.

E °_{cell}

is an intensive quantity — it depends on the identity of the materials and their concentrations, not their size. Bigger electrodes give more total charge (longer life), not more volts.

True or false: In the external wire, conventional current and electron flow point the same way.

False. Electrons flow anode → cathode; conventional current is defined as the flow of positive charge, so it points the opposite way, cathode → anode (outside the cell).

True or false: If you swap which beaker you call "left," the anode becomes the cathode.

False. The roles are fixed by chemistry (which reaction is spontaneous), not by your drawing. Naming or repositioning does nothing; the metal that oxidizes is the anode no matter where you put it.

Spot the error

Spot the error: "

E °_{cell} = E °_{anode} - E °_{cathode}

The subtraction is backwards. The correct formula is

E °_{cell} = E °_{cathode} - E °_{anode}

(both as reduction potentials); reversing it just gives you the wrong sign.

Spot the error: "The anode's

E °

must be flipped to

+ 0.76

V before subtracting because oxidation is the reverse reaction."

No flipping is needed in the formula

E °_{cathode} - E °_{anode}

. You plug in tabulated reduction potentials for both; the subtraction already accounts for the anode running in reverse. Flipping and subtracting double-counts the sign.

Spot the error: "Cu²⁺ ions travel through the wire to reach the zinc."

Ions cannot travel through a metal wire — only electrons move there. Cu²⁺ stays in its own beaker and gets reduced at the copper surface; ion movement to balance charge happens through the salt bridge.

Spot the error: "The salt bridge's job is to let Zn and Cu²⁺ mix so they can react."

That is the opposite of its job. If they mixed directly, electrons would transfer without going through the wire — no usable current. The salt bridge only lets inert spectator ions move to keep each beaker neutral.

Spot the error: "Anions in the salt bridge move toward the cathode."

Anions move toward the anode. The anode beaker fills with positive Zn²⁺ ions, so negative anions migrate there to cancel that build-up; cations head to the cathode instead.

Spot the error: "For the Daniel cell, doubling the reaction (2Zn + 2Cu²⁺ → …) doubles

E °_{cell}

to 2.20 V."

Wrong. Multiplying a half-reaction by a factor multiplies

Δ G °

and

n

together, but

E ° = - Δ G °/ (n F)

divides them back out — so

E °

stays unchanged at 1.10 V. Potential is per-electron.

Why questions

Why does the anode carry a negative sign even though "oxidation loses electrons" sounds like it should end up positive?

Because the electrode metal becomes electron-rich the instant before those electrons leave: oxidation deposits electrons onto the metal, which then flow out. That electron surplus at the terminal is what "negative" labels — not the charge of the ions it releases.

Why do we compare reduction potentials to decide the anode, instead of oxidation potentials?

Purely by convention — tables list every half-reaction as a reduction (measured against the Standard Hydrogen Electrode). Comparing on one common scale lets us subtract cleanly; the metal with the lower reduction potential is simply the one more willing to run in reverse.

Why is the cell potential a difference of two potentials rather than just the cathode's value?

Voltage is always a potential difference — a "height gap" for electrons. A single electrode's potential is meaningless in isolation; only the drop from the high-potential cathode to the low-potential anode drives current.

Why does the formula

Δ G ° = - n F E °_{cell}

carry a minus sign?

Because

Δ G °

counts energy from the system's point of view: energy released is negative. A spontaneous cell (positive

E °_{cell}

) gives out energy, so its

Δ G °

must come out negative — the minus sign enforces that bookkeeping.

Why does adding the two half-reactions make the electrons cancel?

Because the same number of electrons produced at the anode are consumed at the cathode — they are a real, conserved species passing between the two, so they appear on opposite sides and cancel in the net equation.

Why does a very small positive

E °_{cell}

(like the Pb–Sn cell at

+ 0.01

V) still count as spontaneous?

Any positive value, however tiny, means

Δ G ° < 0

, so the reaction can proceed and generate current. It will just be a very weak driving force — low voltage, easily reversed by small concentration changes (see Nernst Equation).

Why can't we get useful electricity by putting two identical zinc electrodes in identical zinc solutions?

Both half-cells have the same reduction potential, so

E °_{cell} = 0

. With no potential difference there is no push on the electrons — no net direction, no current.

Edge cases

Edge case: What happens to

E °_{cell}

if you calculate it for a cell and get a negative number?

It means the reaction is non-spontaneous as written (

Δ G ° > 0

) — this is not a galvanic cell in that direction. To make it run you must supply energy, which is an electrolytic cell; or simply swap which electrode is anode/cathode to get the spontaneous

+ E °

Edge case: Two metals have exactly equal standard reduction potentials. Which is the anode?

Neither — with

Δ E ° = 0

there is no thermodynamic preference, so no net reaction and no defined anode. The system sits at equilibrium producing zero voltage.

Edge case: What if the salt bridge dries out or is removed while the cell is running?

Charge quickly builds up — the anode beaker goes positive, the cathode beaker negative — and this counter-voltage halts electron flow almost immediately. The reaction stops even though the chemistry is still favourable.

Edge case: Under standard conditions all ions are 1 M. What tool do you reach for once concentrations change?

The Nernst Equation. It corrects

E °_{cell}

for non-standard concentrations; e.g. raising the Cu²⁺ concentration pushes reduction forward and raises the cell voltage.

Edge case: As a galvanic cell discharges over time, what happens to its voltage?

It falls toward zero. Reactant ions deplete and product ions accumulate, shrinking the concentration-driven push (via the Nernst relation) until

E_{cell} = 0

— the cell is "dead" and at equilibrium.

Edge case: If the mass of copper deposited depends only on charge passed, does raising the voltage deposit more copper for the same charge?

No. Mass deposited depends only on charge (moles of electrons), governed by Faraday's Laws of Electrolysis. Voltage sets how forcefully/fast, but a fixed number of coulombs always plates the same fixed mass.

Recall Quick self-test before you leave

In a galvanic cell, is the anode positive or negative? ::: Negative — it is the source of electrons. Which direction do electrons flow in the external wire? ::: From anode to cathode. What moves through the salt bridge? ::: Ions (anions toward anode, cations toward cathode), never electrons. Correct sign convention for cell potential? ::: $E °_{cell} = E °_{cathode} - E °_{anode}$ , using reduction potentials for both. What does the little $°$ superscript mean? ::: Standard conditions — 1 M ions, 1 bar gases, 25 °C. Why does positive $E °_{cell}$ mean spontaneous? ::: Because $Δ G ° = - n F E °_{cell}$ , so positive $E °$ forces $Δ G °$ negative.