When zinc metal sits in copper sulfate solution, copper plates out and the zinc dissolves: a redox reaction, as in balancing redox equations. The electrons pass straight from zinc atoms to copper ions, and the energy only warms the solution. Separate the two half-reactions, connect them with a wire, and those same electrons do useful work on the way. That is a galvanic (voltaic) cell.
🔑 Key idea
Oxidation happens at the anode, reduction at the cathode. Electrons flow through the external wire from anode to cathode. A salt bridge (or porous barrier) lets ions move between the two solutions to keep each one neutral: anions drift toward the anode, cations toward the cathode. In a galvanic cell, the anode is the negative terminal.
💡 Why it works
At the zinc electrode, $\mathrm{Zn \to Zn^{2+} + 2e^-}$ leaves electrons behind in the metal, so it becomes the negative terminal. At the copper electrode, $\mathrm{Cu^{2+} + 2e^- \to Cu}$ takes electrons from the metal. Without the salt bridge, positive charge would pile up in the zinc solution and negative charge in the copper solution, and the current would stop almost at once. Ions moving through the bridge carry the charge inside the cell; electrons cannot travel through water.
A zinc–copper cell. Electrons leave the zinc anode through the wire; in the salt bridge, anions move toward the anode and cations toward the cathode.
Cell notation lists the anode on the left and the cathode on the right: a single line for a phase boundary, a double line for the salt bridge.
$$\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$$
When neither species of a half-reaction is a metal (as for $\mathrm{Fe^{3+}/Fe^{2+}}$), an inert electrode such as platinum carries the electrons.
Worked examples
✏️ Worked example: Copper and silver
A cell has a copper electrode in $\mathrm{Cu(NO_3)_2}$ and a silver electrode in $\mathrm{AgNO_3}$. Silver ions are reduced. Identify the anode and cathode, write the overall reaction and the cell notation, and give the direction of electron flow.
Reduction at the cathode: $\mathrm{Ag^+ + e^- \to Ag}$, so silver is the cathode. Copper is oxidized at the anode: $\mathrm{Cu \to Cu^{2+} + 2e^-}$.
Notation: $$\mathrm{Cu(s)\,|\,Cu^{2+}(aq)\,\|\,Ag^+(aq)\,|\,Ag(s)}$$ Electrons flow through the wire from copper to silver.
Answer: Anode Cu, cathode Ag; electrons go Cu → Ag through the wire.
✏️ Worked example: Reading notation
For $$\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Fe^{3+}(aq), Fe^{2+}(aq)\,|\,Pt(s)}$$ write the half-reactions and say which electrode gains mass.
Left is the anode: $\mathrm{Mg \to Mg^{2+} + 2e^-}$. Right is the cathode: $\mathrm{Fe^{3+} + e^- \to Fe^{2+}}$, on inert platinum.
Magnesium loses mass; platinum does not change (both iron species stay in solution). No electrode gains mass.
Answer: Mg is oxidized; $\mathrm{Fe^{3+}}$ is reduced on Pt; neither electrode gains mass.
⚠️ Common mistake
Thinking electrons travel through the salt bridge or the solution. Electrons move only through the metal wire and electrodes; ions carry the current inside the cell. Check: follow the charge around the loop: electrons in the wire, ions in the solutions and bridge.
A memory aid: an ox and red cat: anode = oxidation, reduction = cathode.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.