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

Core Concept

Why does this happen? Different metals have different tendencies to oxidize. Zinc would rather be Zn²⁺ and lose electrons; copper would rather stay as Cu metal or gain electrons to become Cu from Cu²⁺. When you connect them, electrons flow through the wire from the metal that oxidizes more easily (Zn) to the one that reduces more easily (Cu²⁺ → Cu).

The Two Electrodes

Memory aid: "An Ox" — Anode is where Oxidation happens.

Memory aid: "Red Cat" — Reduction happens at the Cathode.

Visual Structure

Key components:

1. Two half-cells: Each contains a metal electrode dipped in a solution of its own ions
2. Salt bridge: Allows ion flow to maintain electrical neutrality (prevents charge buildup)
3. External wire: Path for electron flow from anode to cathode
4. Voltmeter (optional): Measures the cell potential (voltage)

First-Principles Derivation: Why Electrons Flow

Step 1: The Oxidation-Reduction Driving Force

Every metal has a standard reduction potential ($E°$), which measures its tendency to gain electrons (be reduced). This is measured against the standard hydrogen electrode (SHE).

For the classic Daniel cell:

• Zn²⁺ + 2e⁻ → Zn has $E°_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$
• Cu²⁺ + 2e⁻ → Cu has $E°_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V}$

What does this mean? Copper has a POSITIVE reduction potential—it "wants" to gain electrons. Zinc has a NEGATIVE reduction potential—it "wants" to LOSE electrons (get oxidized).

Step 2: Writing the Half-Reactions

At the anode (Zn): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$

Why this step? Zinc metal oxidizes because its reduction potential is more negative. The reverse reaction (oxidation) is spontaneous. The electrons produced flow out through the wire.

At the cathode (Cu): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Why this step? Copper ions in solution gain the electrons that arrive from the zinc anode via the wire. Copper plates out on the cathode.

Step 3: The Net Cell Reaction

Add the half-reactions (electrons cancel): $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Why is this spontaneous? The total free energy change is negative when the more easily oxidized metal (Zn) gives electrons to the more easily reduced ion (Cu²⁺).

Step 4: Calculating Cell Potential

Derivation from first principles:

The cell potential represents the maximum electrical work the cell can do per mole of electrons transferred. It's the difference in "electron pressure" between the two electrodes.

• At the cathode, reduction happens: $E°_{\text{cathode}}$ is the potential for Cu²⁺ + 2e⁻ → Cu = +0.34 V
• At the anode, oxidation happens (reverse of reduction): $E°_{\text{anode}}$ is the potential for Zn²⁺ + 2e⁻ → Zn = -0.76 V

The cell pulls electrons from the anode and pushes them to the cathode. The "push" is:

$E°_{\text{cell}} = 0.34 - (-0.76) = 1.10 \text{ V}$

Why subtract? You're finding the potential difference. The anode is at lower potential (more negative), the cathode at higher potential. Current flows from high to low potential in the external circuit.

Important: A positive $E°_{\text{cell}}$ means the reaction is spontaneous. If $E°_{\text{cell}} < 0$, you'd need to supply energy (electrolytic cell, not galvanic).

The Role of the Salt Bridge

As Zn oxidizes, the anode solution gains Zn²⁺ ions (becomes positively charged). As Cu²⁺ reduces, the cathode solution loses Cu²⁺ ions (becomes negatively charged). Without a way to balance charge, the reaction would stop immediately.

The salt bridge contains an inert electrolyte (like KNO₃ or Na₂SO₄) that allows ions to migrate:

• Anions (NO₃⁻, SO₄²⁻) move toward the anode to neutralize the buildup of Zn²⁺
• Cations (K⁺, Na⁺) move toward the cathode to replace the Cu²⁺ being removed

Why not just connect the solutions? That would allow direct mixing and reaction between Zn and Cu²⁺ without electron flow through the external circuit—no electricity generated.

Worked Examples

Setup:

• Anode: Zn electrode in 1 M ZnSO₄
• Cathode: Cu electrode in 1 M CuSO₄
• Salt bridge: KNO₃

Find: Cell potential, direction of electron flow, mass of Cu deposited when 2moles of electrons flow.

Solution:

Step 1: Write half-reactions

• Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
• Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Why these? Zn has more negative $E°$, so it oxidizes.

Step 2: Calculate $E°_{\text{cell}}$ $E°_{\text{cell}} = E°_{\text{Cu}^{2+}/\text{Cu}} - E°_{\text{Zn}^{2+}/\text{Zn}} = 0.34 - (-0.76) = 1.10 \text{ V}$

Why positive? The reaction is spontaneous in this direction.

Step 3: Electron flow direction Electrons flow from Zn (anode) → Cu (cathode) through the external wire.

Why this direction? Electrons are produced at the anode and consumed at the cathode.

Step 4: Mass of Cu deposited From the cathode half-reaction: 2 moles e⁻ → 1 mole Cu

If 2 moles e⁻ flow: $n_{\text{Cu}} = 1$ mole

$m_{\text{Cu}} = 1 \text{ mol} \times 63.5 \text{ g/mol} = 63.5 \text{ g}$

Why this calculation? The stoichiometry of the half-reaction directly relates electrons to copper atoms.

Given: A cell made from Ag|Ag⁺ (E° = +0.80 V) and Ni|Ni²⁺ (E° = -0.25 V)

Find: Which is the anode? Which is the cathode? Cell voltage?

Solution:

Step 1: Compare reduction potentials

• Ag⁺ + e⁻ → Ag: E° = +0.80 V (more positive—wants to be reduced)
• Ni²⁺ + 2e⁻ → Ni: E° = -0.25 V (more negative—wants to be oxidized)

Why compare? The more positive reduction potential indicates the stronger oxidizing agent (gets reduced). The more negative one gets oxidized.

Step 2: Assign electrodes

• Cathode (reduction): Ag⁺ + e⁻ → Ag (E° = +0.80 V)
• Anode (oxidation): Ni → Ni²⁺ + 2e⁻ (E° = -0.25 V)

Why? Silver WANTS to gain electrons more than nickel, so silver is reduced (cathode).

Step 3: Calculate cell potential $E°_{\text{cell}} = 0.80 - (-0.25) = 1.05 \text{ V}$

Why positive? Confirms spontaneous reaction in galvanic cell.

Question: Will a cell made from Pb|Pb²⁺ (E° = -0.13 V) and Sn|Sn²⁺ (E° = -0.14 V) generate electricity? Which is the anode?

Solution:

Step 1: Identify which oxidizes more easily Sn has E° = -0.14 V (more negative) → oxidizes more easily Pb has E° = -0.13 V → reduces more easily (relatively)

Step 2: Assign electrodes

• Anode: Sn → Sn²⁺ + 2e⁻
• Cathode: Pb²⁺ + 2e⁻ → Pb

Step 3: Calculate $E°_{\text{cell}} = (-0.13) - (-0.14) = +0.01 \text{ V}$

Why very small? The two metals have VERY similar reduction potentials. The cell will work but produce only a tiny voltage.

Answer: Yes, it will generate electricity (positive $E°$), but very weakly. Sn is the anode.

Common Mistakes & Misconceptions

Why this feels right: In everyday battery labels, we think of the terminal where current "comes from" as positive.

The truth: In a galvanic cell, the anode is negative and the cathode is positive. Electrons flow OUT of the anode (negative) and INTO the cathode (positive) through the external circuit. Conventional current (opposite to electron flow) goes from + to -.

The fix: Remember: electrons are negative. They're repelled from the negative terminal (anode) and attracted to the positive terminal (cathode).

Note: This is OPPOSITE in electrolytic cells (where you supply energy). Don't confuse them.

Why this feels right: It's tempting to always do |larger - smaller| to get a positive number.

The truth: The formula is ALWAYS $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$, where these are reduction potentials as written. You must identify which electrode is the cathode (higher/more positive reduction potential) first.

The fix:

1. Find which metal/ion pair has more positive E° → that's the cathode (reduction)
2. The other is the anode (oxidation)
3. Subtract: cathode E° minus anode E°

Example: For Zn|Cu cell: $E°_{\text{cell}} = 0.34 - (-0.76)$, NOT $0.34 - 0.76$.

Why this feels right: It's connecting the two solutions, and we need to complete a circuit.

The truth: The salt bridge conducts ions (charged atoms/molecules), NOT electrons. Electrons flow through the metal wire. The salt bridge maintains electrical neutrality by allowing anions and cations to migrate between half-cells.

The fix: Electrons are conducted by metals (wire connecting electrodes). Ions are conducted by electrolytes (salt bridge, solutions).

Why this feels right: It's easy to mix up which ions go where.

The truth:

• Anions (negative ions) from the salt bridge move toward the anode (where Zn²⁺ is building up, creating positive charge)
• Cations (positive ions) from the salt bridge move toward the cathode (where Cu²⁺ is being depleted, creating negative charge)

The fix: Opposite charges attract. Positive buildup at anode attracts negative ions from salt bridge. Negative buildup at cathode attracts positive ions.

Active Recall Practice

When you put a piece of zinc in a solution with zinc ions, and a piece of copper in a solution with copper ions, and then connect them with a wire, something cool happens. The zinc starts giving away electrons (because it's generous), and those electrons travel through the wire to the copper side, where the copper ions grab them (because copper is greedy).

This flow of electrons through the wire IS electricity! That's how a battery works. The zinc piece gets smaller because it's turning into zinc ions, and copper builds up on the copper piece because copper ions are turning back into solid copper metal.

We need a salt bridge (think of it as a hallway for charged particles) because as zinc loses electrons, the solution gets too positive, and as copper gains electrons, the solution gets too negative. The salt bridge lets some negative and positive particles (not electrons, but ions dissolved in water) move between the two sides to keep everything balanced, so the reaction can keep going.

The side where zinc loses electrons (gives them away) is called the anode, and the side where copper gains electrons (takes them) is called the cathode. Electrons flow from anode to cathode through the wire, creating electric current that can power a light bulb or phone!

Flashcards

#flashcards/chemistry

What is a galvanic (voltaic) cell? :: A device that converts chemical energy into electrical energy through spontaneous redox reactions, consisting of two half-cells connected by a wire and salt bridge.

What does a negative E°_cell indicate? :: The reaction is non-spontaneous in the forward direction; you would need to supply energy to make it occur (electrolytic cell).

In a galvanic cell, is the anode positive or negative? :: Negative (it's the source of electrons)

Connections

• Reduction Potentials — The E° values that determine which electrode is anode/cathode
• Nernst Equation — Calculating cell potential under non-standard conditions
• Electrolytic Cells — The opposite: using electricity to drive non-spontaneous reactions
• Faraday's Laws of Electrolysis — Quantitative relationship between charge and mass
• Standard Hydrogen Electrode — The reference for all reduction potentials
• Oxidation Numbers — Tracking electron transfer in redox reactions
• Gibs Free Energy and Cell Potential — ΔG° = -nFE°_cell relationship
• Battery Technologies — Practical applications of galvanic cells

Last updated: 2026-06-30

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