U9.7 · Galvanic (Voltaic) and Electrolytic CellsPractice · 原电池与电解池

In a galvanic cell the side with the lower E° is oxidized and is the anode (negative): Mg → Mg²⁺ + 2e⁻. Fe²⁺ is reduced at the iron electrode, the cathode (positive). E°cell = −0.45 − (−2.37) = 1.92 V.

Iron has the lower E° and is the anode: Fe → Fe²⁺ + 2e⁻, so iron dissolves and loses mass. Copper is the cathode: Cu²⁺ + 2e⁻ → Cu deposits copper on it, so it gains mass, and the blue CuSO₄ solution fades.

The anode side keeps producing cations and builds up positive charge, so NO₃⁻ anions move there; on the cathode side cations are being reduced, so K⁺ moves in to replace them. Electrons travel only through the wire, never the salt bridge, and K⁺ and NO₃⁻ don’t react at the electrodes.

In any cell, oxidation happens at the anode and reduction at the cathode. In an electrolytic cell the positive terminal pulls electrons away from the anode, so the anode is the one wired to the positive terminal. Electrolysis uses electrical energy to drive a nonspontaneous reaction (ΔG > 0); it’s the galvanic cell that has ΔG < 0 and does work.

By convention the anode is written on the left and the cathode on the right; | marks a phase boundary (solid and solution) and ‖ the salt bridge. So Mg is oxidized on the left and Fe²⁺ is reduced to Fe on the right, consistent with comparing E° values.