A galvanic cell, also known as a voltaic cell, is a device designed to harness the energy released during a spontaneous chemical reaction, converting that chemical energy directly into electrical energy. This conversion relies on the movement of electrons generated by a reduction-oxidation (redox) reaction. Correctly identifying the two component electrodes—the anode and the cathode—is necessary to predict the direction of electron flow and the overall voltage produced by the cell. The methods for this determination range from fundamental chemical definitions to quantitative comparisons of electrical potential.
Defining Electrodes by Chemical Process
The fundamental way to distinguish between the two electrodes is by the specific chemical process occurring at their surfaces. The anode is defined as the electrode where oxidation takes place, which involves a species losing electrons. Conversely, the cathode is the electrode where reduction occurs, which is the process of a chemical species gaining electrons.
The mnemonic “An Ox and Red Cat” (Anode is Oxidation, Reduction is Cathode) is often used to recall these definitions. The half-reaction at the anode generates electrons, which then travel through the external wire to be consumed by the half-reaction at the cathode. The movement of ions within the cell plays a role in maintaining electrical neutrality. To prevent a charge buildup that would quickly halt the reaction, a salt bridge connects the two half-cells, allowing spectator ions to flow and balance the accumulating charges.
Using Standard Reduction Potentials
The Standard Reduction Potential (\(E^\circ\)) provides the quantitative method for predicting which electrode will be which. This potential is a measure, in volts, of a chemical species’ inherent tendency to gain electrons and undergo reduction under standard conditions. These values are always listed for reduction half-reactions. The governing rule for a spontaneous galvanic cell is that the half-reaction with the higher, or more positive, \(E^\circ\) value will proceed as the reduction half-reaction, making this half-cell the cathode.
The half-reaction with the lower, or more negative, \(E^\circ\) value must then proceed as the oxidation half-reaction, making it the anode. To represent the oxidation at the anode, the reduction half-reaction must be chemically reversed, and its potential sign must also be flipped. The overall cell voltage will be the difference between the cathode’s potential and the anode’s potential, and must always result in a positive value for a spontaneous galvanic cell. Consider a cell that pairs silver (\(E^\circ = +0.80\text{ V}\)) and copper (\(E^\circ = +0.34\text{ V}\)). Since silver has the higher standard reduction potential, the silver half-cell will be the cathode, and it will undergo reduction. The copper half-cell, having the lower potential, must be the anode, and its reaction will be reversed to oxidation, changing its contribution to \(-0.34\text{ V}\).
Polarity, Electron Flow, and Practical Mnemonics
The chemical identity of the anode and cathode translates directly into the cell’s electrical polarity and the direction of electron travel. In a galvanic cell, the anode is designated as the negative electrode. This is because the oxidation reaction at the anode generates electrons, causing a buildup of negative charge on the electrode surface before the electrons depart into the external circuit.
Therefore, electrons always travel through the external wire from the negative anode to the positive cathode. The cathode is consequently designated as the positive electrode because it is the receiving terminal that attracts the incoming electrons to facilitate the reduction of its chemical species. A helpful way to remember this polarity is the phrase “Anode is Negative” in a galvanic cell. This distinction is important because the polarity is reversed in an electrolytic cell, though the chemical definitions of oxidation and reduction remain the same. The salt bridge completes the circuit and sustains the flow by managing the charge imbalance created by the electron flow and half-reactions. As electrons leave the anode, the solution in that half-cell accumulates positive charge, which is mitigated by negative ions entering from the salt bridge.