Physical

Galvanic Cells

Galvanic cells traditionally are used as sources of DC electrical power. A simple galvanic cell may contain only one electrolyte separated by a semi-porous membrane, while a more complex version involves two separate half-cells connected by a salt bridge. The salt bridge contains an inert electrolyte like potassium sulfate whose ions will diffuse into the separate half-cells to balance the building charges at the electrodes. According to the mnemonic “Red Cat An Ox”, oxidation occurs at the anode and reduction occurs at the cathode. Since the reaction at the anode is the source of electrons for the current, the anode is the negative terminal for the galvanic cell.

Let’s look at an example of a galvanic cell like the classic AA alkaline battery, in which the equations for the two half-reactions are listed below:

MnO2(s)+H2O+e−→MnOOH(s)+OH-(aq)                        ;Eo=+0.382 V

Zn(OH)2(s)+2e−→Zn(s)+2OH-(aq)                                    ;Eo=−1.221 V

Here we are given two reduction potentials for the anode and cathode. Most data tables listing cell potentials will list reduction potentials. Since they are both reduction potentials, we need to decide which one needs to be flipped. Let’s keep in mind that when we flip the reduction potential into an oxidation potential, we also need to flip the sign. Since galvanic cells have a positive EMF, we are looking to flip the equation that when added to the other EMF will give us a positive value. By flipping the zinc half-reaction, we have the two EMF values as +0.382 V and +1.221 V. To find the cell EMF, we simply add them together to give us an approximate value of 1.5 V, which we already know to be the EMF of an alkaline AA battery.

Since voltage is an intensive property, which is one that does not depend on the system size or the amount of material in the system, we do not have to multiply the EMF by any stoichiometric coefficient to cancel out the electrons in calculating EMF.