Sending the electrons through a wire
Drop a strip of zinc into copper(II) sulfate solution (lesson 2.7.2) and red copper quickly coats the zinc while the solution warms slightly. Zn hands its electrons straight to the Cu²⁺ touching its surface, and the energy released becomes heat (lesson 2.10). If Zn and Cu²⁺ are kept apart, the electrons can only travel from the zinc to the copper side through a wire. That flow of electrons is an electric current, and it can light a bulb. A device that turns chemical energy into electrical energy this way is a galvanic cell原电池 (also called a voltaic cell). Phone batteries and dry cells are galvanic cells.
The simplest cell
Put a zinc strip and a copper strip into dilute sulfuric acid, connect them with a wire, and add an ammeter in between. The meter deflects, the zinc slowly dissolves, and bubbles of H₂ form on the copper.
- At the zinc: Zn → Zn²⁺ + 2e⁻. Zinc loses electrons, which travel through the wire to the copper.
- At the copper: 2H⁺ + 2e⁻ → H₂. H⁺ ions in the solution pick up electrons at the copper surface. The copper itself doesn’t react; it just passes the electrons on.
The electrode where oxidation happens is the anode负极（原电池）; electrons leave the cell there, so in a galvanic cell it is the negative terminal. The electrode where reduction happens is the cathode正极（原电池）, the positive terminal. A memory aid: An Ox, Red Cat (anode–oxidation, reduction–cathode).
Two beakers and a salt bridge
In a one-beaker cell, the zinc also reacts directly with the H⁺ (or Cu²⁺) in solution, so some energy is lost as heat and the current is unsteady. The fix is to separate the two half-reactions:
- Left beaker: a zinc strip in ZnSO₄ solution.
- Right beaker: a copper strip in CuSO₄ solution.
- A salt bridge盐桥 joins the two beakers: a U-tube of gel soaked in a salt solution such as KCl.
Now Zn can give its electrons only to the wire, and Cu²⁺ can get electrons only from the copper strip:
- Anode (zinc), oxidation: Zn → Zn²⁺ + 2e⁻.
- Cathode (copper), reduction: Cu²⁺ + 2e⁻ → Cu.
- Overall: Zn + Cu²⁺ → Zn²⁺ + Cu.
The two electrode reactions are exactly the half-reactions of lesson 2.13.3.
How charge flows
- In the wire: electrons flow from the anode to the cathode. Conventional current is defined in the opposite direction, from cathode to anode.
- In the solutions: electrons can’t travel through a solution; ions carry the charge instead. Zn²⁺ keeps forming in the left beaker, adding positive charge, while Cu²⁺ is used up in the right beaker, leaving extra SO₄²⁻.
- In the salt bridge: K⁺ moves toward the cathode side (CuSO₄) and Cl⁻ toward the anode side (ZnSO₄), keeping both solutions neutral and completing the circuit. Remove the bridge and the current stops at once.
What a galvanic cell needs
- Two electrodes of different reactivity; the more active metal is the anode.
- Electrodes dipping into electrolyte solutions.
- A closed circuit: a wire between the electrodes and a salt bridge between the solutions.
- A spontaneous redox reaction.
To decide which metal is the anode, use the activity series from lesson 2.7.2: the metal higher on the list gives up electrons more easily. AP Unit 9 makes this quantitative with standard reduction potentials and calculates the cell’s voltage.
Pick two metals below to build a two-beaker cell. Watch the electrons in the wire and the ions in the salt bridge. Then try two identical metals, or take the salt bridge out:
How Chinese textbooks differ
- Chinese textbooks call a galvanic cell 原电池 and name its electrodes 负极 (negative electrode, the anode) and 正极 (positive electrode, the cathode), rather than anode and cathode.
- For electrolytic cells (like the electrolysis of water in lesson 1.7.3), Chinese textbooks use 阳极 and 阴极. The 阳极 is connected to the positive terminal and is where oxidation happens, so it is also the anode. The anode is always where oxidation happens, but it is negative in a galvanic cell and positive in an electrolytic cell.
- Chinese textbooks write electrode reactions as Zn − 2e⁻ = Zn²⁺. US textbooks write Zn → Zn²⁺ + 2e⁻ and use cell notation, Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s), with the anode on the left.
Example 1: an iron–copper cell
An iron strip in FeSO₄ solution and a copper strip in CuSO₄ solution are connected by a wire and a KCl salt bridge. Identify the anode and cathode, write the half-reactions and the overall reaction, and describe how electrons and the ions in the bridge move.
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Anode and cathode. Fe is more active than Cu and gives up electrons more easily, so iron is the anode and copper is the cathode.
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Anode reaction (oxidation). Iron loses electrons:
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Cathode reaction (reduction). Cu²⁺ from the solution gains electrons:
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Overall. Add the half-reactions; the electrons cancel:
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Electron flow. Electrons flow through the wire from the anode (iron) to the cathode (copper). They never enter the solution.
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Ions in the bridge. The anode side keeps gaining Fe²⁺, so Cl⁻ moves toward the FeSO₄ solution to balance the charge; the cathode side keeps losing Cu²⁺, so K⁺ moves toward the CuSO₄ solution. In short, cations move toward the cathode and anions toward the anode.
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What you see. The iron strip slowly thins, red copper builds up on the copper strip, and the blue CuSO₄ solution fades.
Example 2: where do the bubbles on the copper come from?
In the one-beaker zinc–copper cell in dilute sulfuric acid, bubbles form on the copper but the copper’s mass doesn’t change. What is the gas? Why doesn’t the copper dissolve? When 2 electrons pass through the circuit, how many Zn atoms dissolve and how many H₂ molecules form?
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The gas. H₂: H⁺ in the solution gains electrons at the copper:
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Why copper doesn’t dissolve. Copper is below hydrogen in the activity series and doesn’t react with dilute sulfuric acid. Here it is the cathode, just handing electrons from the wire to H⁺. The zinc at the anode is what loses electrons.
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Electron bookkeeping. Both electrodes transfer the same number of electrons. At the anode Zn → Zn²⁺ + 2e⁻, and at the cathode 2H⁺ + 2e⁻ → H₂, so for 2 electrons 1 Zn atom dissolves and 1 H₂ molecule forms.
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Scale up to moles. If 0.20 mol of electrons pass, 0.10 mol of Zn dissolves (0.10 × 65.38 = 6.5 g) and 0.10 mol of H₂ forms (0.10 × 22.4 = 2.24 L at STP).
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A common mistake. Thinking the bubbles come from copper reacting with acid. The copper is only where the transfer happens: the H₂ comes from H⁺ in the solution, and the electrons come from the zinc.
- The electrode where oxidation happens
At the anode, Zn loses electrons and is oxidized.
- The electrode where reduction happens
At the cathode, Cu²⁺ gains electrons and is reduced.
- Electrons leave the cell here
The anode gives up electrons, which flow through the wire to the cathode.
- Copper metal forms here
Cu²⁺ + 2e⁻ → Cu happens at the cathode.
- The electrode that loses mass
Zinc at the anode keeps dissolving as Zn²⁺.
- Electrons travel through it from one electrode to the other
Electrons move only through the wire (and the electrodes).
- Ions travel through it between the beakers
K⁺ and Cl⁻ move through the salt bridge, keeping both solutions neutral.
- Cations in solution move toward it
Cu²⁺ is used up on the cathode side, so cations (such as K⁺ from the bridge) move toward the cathode.
- Remove it and the current stops
Without the salt bridge the two solutions aren’t connected and the circuit is open.
Common mistakes
Mistake 1: “Electrons flow from the anode through the wire to the cathode, then back through the solution.” Electrons travel only through the wire and the electrodes. In the solutions and the salt bridge, ions carry the charge.
Mistake 2: “Cations in the salt bridge move toward the anode.” Cations (Cu²⁺) are used up on the cathode side and Zn²⁺ builds up on the anode side. To keep both sides neutral, cations in the bridge move toward the cathode and anions toward the anode.
Mistake 3: “Any two metals in any liquid, joined by a wire, make a battery.” Two identical metals have no spontaneous redox reaction between them, and zinc and copper in ethanol (a nonelectrolyte) can’t complete the circuit. All four conditions are needed.
Quick check
In a zinc–copper two-beaker cell (ZnSO₄ and CuSO₄ joined by a salt bridge), which statement is correct?
Zn is more active than Cu and loses electrons, so it is the anode, where oxidation happens. Electrons flow from zinc to copper, and Cl⁻ moves toward the ZnSO₄ (anode) side.
While a zinc–copper two-beaker cell is running, the salt bridge is removed. What does the ammeter do?
The salt bridge connects the two solutions to close the circuit. Without it the circuit is open and the current stops.
Summary
- A galvanic cell turns chemical energy into electrical energy: in a spontaneous redox reaction, the electrons lost by the reducing agent travel through a wire to the oxidizing agent.
- The anode is the more active metal, where oxidation happens (negative in a galvanic cell); the cathode is where reduction happens (positive). In the zinc–copper cell, Zn → Zn²⁺ + 2e⁻ at the anode and Cu²⁺ + 2e⁻ → Cu at the cathode.
- Electrons flow through the wire from anode to cathode; in the salt bridge, cations move toward the cathode and anions toward the anode. A cell needs two different electrodes, electrolyte solutions, a closed circuit, and a spontaneous redox reaction.
On to AP (for later)
AP topic 9.7 asks you to read or draw a galvanic cell diagram, labeling the anode, the cathode, the direction of electron flow, and the movement of ions in the salt bridge. Topic 9.8 uses standard reduction potentials to calculate E°cell and decide whether a reaction is spontaneous. This lesson stays qualitative; the numbers come in AP Unit 9.