When handling chemical batteries, a major property of their use is E Cell. This E Cell, which is a measure of the electric potential contained within, indicates the overall voltage of the battery. Based on the voltages of the half-reactions at the cathode and anode, E cathode and E anode, the total E Cell can be calculated.
Be sure to balance the two half-reactions before calculating E Cell if there are unequal moles of electrons transferred between reactions. If the reactants are not kept at standard conditions (1.0 M), you may wish to use the Nernst equation to convert the E Cell to an adjusted value.
If working with an actual battery instead of data alone, be sure to wear proper safety equipment. Take necessary precautions to avoid electric shock, such as keeping the circuit away from ionized water.
Determine which chemicals are involved in the battery. Do so by identifying the metals present at the anode and cathode.
Write out half-reactions for what is occurring at each side of the battery. For example, in a Copper/Zinc battery the reactions might be:
Cu2+ + 2e- ' Cu Zn ' Zn2+ + 2e-
Now switch the reactions such that both are in the form of reduction. Once finished, all electrons should be on the left side of the equations. In the example given, this would look like:
Cu2+ + 2e- ' Cu Zn2+ + 2e- ' Zn
Consult the Table of Standard Reduction Potentials and record the values that match up with the written half-reactions. In the Cu/Zn example, this would be:
Cu2+ + 2e- ' Cu 0.34 Volts Zn2+ + 2e- ' Zn -0.76 Volts
Look at the values obtained for E anode and E cathode (from the Table of Standard Reduction Potentials). To determine which metal is at the anode and which metal is at the cathode, switch the reaction around that has the greatest negative voltage. In the example, this looks like:
Cu2+ + 2e- ' Cu E cathode: 0.34 Volts Zn ' Zn2+ + 2e- E anode: 0.76 Volts
Now add the two voltages together to obtain E Cell. In a Cu/Zn battery, this is (0.34V) + (0.76V) = 1.10 Volts.
- Flickr user "makelessnoise"