ElectrochemistryStep 1: Investigating redox reactions Practice with redox reactions Reduction tendencies of metal ions Step 2: Step 3: Step 4: Practice with nonstandard cells Feedback Form: 
Electrochemistry Tutorial: Galvanic Cells and the Nernst Equation >> Step 3: Calculating cell potentials
Electrochemistry: Galvanic Cells and the Nernst EquationStep 3: Calculating cell potentialsIn the preceding simulations you measured cell potentials of several cells made of various combinations of halfcells. It would be a monumental task to assemble a list of all possible cells and report a cell voltage for each. It is more useful to create a list of the potentials of halfcells and then calculate the potentials of the required combinations. Here, we will first learn how to use halfcell potentials, and then consider how to determine halfcell potentials from experiments. The voltage of an electrochemical cell depends on the redox reaction occurring in the cell. Above, we saw that it is often useful to break a redox reaction, such as: Zn(s) + Pb^{2+}(aq) > Zn^{2+}(aq) + Pb(s) into the following half reactions.
Given how useful halfcell potentials are (you can use them to calculate the potential of any fullcell), it is not surprising that these are well tabulated in handbooks and on the web. These tables, by convention, contain the halfcell potentials for reduction. Reduction potential, the tendency to gain electrons, represents the relative tendency for a given metal ion to undergo reduction. For the above cell, the table would contain:
The Pb reaction in the table and in the cell are in the same direction (i.e. reduction). This will always be the case for the cathode, because the table lists reduction potentials and reduction always occurs at the cathode. However, the Zn reaction is in the opposite direction. This will always be the case for the anode, because the table lists reduction potentials and oxidation occurs at the anode. Because of this, the cell potential is the difference between the cell potential of the cathode and that of the anode: Eº_{cell} = Eº_{cathode}  Eº_{anode} The only potential we can measure experimentally is that of a full cell (a reduction process must always be coupled with an oxidation process, otherwise free electrons would build up). So all measurements correspond to differences between the numbers in a table of halfcell potentials. This is analogous to the altitudes of cities. All we really care about is the difference in altitudes between cities. All tables define altitude relative to sea level, but if we instead created a table that gave the altitudes of cities relative to the top of Mt Everest (or the center of the earth), we would still get the same values for the difference in altitudes between cities. Sea level is the "reference" for altitude, meaning that sea level has an altitude of zero. For halfcell potentials, we also need to decide on a "reference". The convention among chemists is to take the following halfreaction as the standard and assign it a zero halfcell potential:
The electrode corresponding to this halfreaction is called the "standard hydrogen electrode". Here the notation Eº is used to represent hydrogen's standard electrode potential (declared to be zero at all temperatures). The measured cell voltage using the standard hydrogen electrode as one of the halfcells is, therefore, the potential of the other half reaction. The standard hydrogen electrode consists of hydrogen gas at 1 atmosphere and so is not convenient for us to work with in the laboratory. Instead, you will create your own table of reduction potentials using a tin halfcell as a standard instead of the hydrogen halfcell. Using the information from a standard reduction potentials table (see link below), the standard cell potential for the zinc/lead cell is then: Eº_{cell} = Eº_{cathode}  Eº_{anode} = 0.13  ( 0.76) = 0.63 V The following example shows how to use a table of standard reduction potentials [such as the one available online at: http://hyperphysics.phyastr.gsu.edu/hbase/tables/electpot.html] to predict the cell potentials that you measured in the simulation on the previous page:
For the reaction between Cu and Sn, we can perform the following calculation:
Since the copper halfcell is the cathode (this is where the reduction occurs) and the tin halfcell is the anode (where the oxidation occurs), our calculation would be: Eº_{cell} = Eº_{cathode}  Eº_{anode} = 0.34  ( 0.14) = 0.48 V For the reaction between Cu and Ag, we can perform the following calculation:
Since the silver halfcell is the cathode (this is where the reduction occurs) and the copper halfcell is the anode (where the oxidation occurs), our calculation would be: Eº_{cell} = Eº_{cathode}  Eº_{anode} = 0.80  0.34 = 0.46 V Note that we multiplied the Ag halfcell reaction by 2, to balance the electrons. However, we did not multiply the cell potential by 2. When we multiple a halfcell reaction by a constant, we do not need to alter the halfcell potential.

Page Last Updated: 11.07.2016 