Describe the operation of galvanic cells and use standard electrode potentials to calculate cell potential and predict spontaneity.

What this dot point is asking

You must describe how a galvanic cell works and calculate its EMF from standard electrode potentials.

How a galvanic cell works

A galvanic (voltaic) cell has two half-cells, each a metal electrode in a solution of its ions, connected by an external wire and a salt bridge.

• At the anode, oxidation occurs (electrons released). In a galvanic cell the anode is the negative electrode.
• At the cathode, reduction occurs (electrons gained). The cathode is the positive electrode.
• Electrons flow through the external circuit from anode to cathode.
• The salt bridge allows ions to move between the half-cells to keep each solution electrically neutral, completing the circuit.

Standard electrode potentials

Each half-reaction has a standard reduction potential $E^\circ$, measured relative to the standard hydrogen electrode ($0.00\ \text{V}$) under standard conditions ($25\ ^\circ\text{C}$, $1\ \text{mol L}^{-1}$, $100\ \text{kPa}$). A more positive $E^\circ$ means a stronger tendency to be reduced (a stronger oxidising agent).

Calculating cell potential

The half-cell with the more positive $E^\circ$ is reduced (the cathode); the other is oxidised (the anode). The standard cell EMF is:

A positive $E^\circ_{cell}$ means the cell reaction is spontaneous.

Conventions

Cell diagrams are written anode | anode solution || cathode solution | cathode, with the double bar representing the salt bridge. Electron flow in the external wire is always anode → cathode.