| M8 | Lesson #3 | Voltage of Electrochemical Cells |
Topics
Electrochemical cells (batteries)
Calculating voltage
Comparing electrochemical and electrolytic cells
Applications of electrochemistry
Calculating the voltage of electrochemical cells
Hydrogen Half- cell
Chemists in order to measure the voltage of any two half cells that were hook together set up some standard conditions and a half cell that they used as a reference. The standard conditions where that the aqueous solutions would have a concentration of 1.0 mol/l. The reference cell chosen was the hydrogen half-cell ( Pt (s) | H2 (g) , H+ (aq) ) which consists of a platinum electrode over which hydrogen is bubbled at SATP, and it was assigned a reduction potential of zero. Chemists then proceeded to hook up all other half cells to the hydrogen cell and measure the reduction potential (voltages) created. A negative reduction potential meant that the oxidizing agent in the other half cell attracted electrons less strongly than hydrogen. A positive reduction potential meant that the oxidizing agent attracted electrons more strongly. The diagram below show a zinc half cell connected to the hydrogen half cell.
Tables listing standard reduction potentials are found in any good chemistry textbook. In these tables the half reactions are all written as reductions. The values for E0 can now be used to determine the voltage that any combination of two half cell would create.
Calculating Voltages
Important
ProcedureTo calculate voltages for any two electrochemical cell we can do the following:
a) Locate the two half cells reactions in the table of standard reduction potentials.
b) The half reaction that has the higher reduction potential will reduce and can be written as you find it in the table.
c) The half reaction that has the lower reduction potential must be reversed and written as an oxidation.
d) The Eo value of the higher half reaction is recorded unchanged from the table.
e) The sign of the Eo value of the lower half reaction is changed and its value recorded.
f) The two half reactions are balanced for the number of electrons exchanged but the value of each Eo remains unchanged.
g) The two half reactions are then added together and so are the Eo values. ( This value will always be positive for an electrochemical cell).Examples and Assignments
Using the table of standard reductions provided write the equation for the reaction between the following two half cells, and determine its voltage.
1) Gold is higher the table and is written as found in table.
2) The copper is lower so it's reaction is written as an oxidation (sign of Eo reversed ).
3) Equation is balanced for the exchange of electrons ( each needs six).
2 Au 3+ (aq) + 6e - ------> 2 Au (s) 3 Cu (s) ------>3 Cu 2+ (aq) + 6e- *** Note that the size of the values for the Eo of the reactions are not changed ***
4) Add the two half reactions and the E0 values.
2 Au 3+ (aq) + 3 Cu (s) ---> 2 Au (s) + 3 Cu 2+ (aq) E = + 1.50 V + ( - 0.34 V ) = 1.16 V
Note on Inert electrodes
In many cases the electrode in the half cell is only a site for reduction or oxidation. It does not actually take part in the electrochemical reaction. These inert electrode are usually made of either carbon or platinum.
Key IdeaWhen writing the half reactions and calculating the voltages of cells containing inert electrons, we ignore the inert electrodes.
Example of calculating voltage when an inert electrode is present in one of the cells
What is the voltage of the following battery.Cu2+ (aq) | Cu (s) || ClO4 - (aq) ; H+ (aq) | C(s)
Note that carbon C(s) is an inert electrode and therefore will be ignored in calculating the voltage and net reaction.
1) The ClO4 - (aq) H+ (aq) cell is higher in the table and is written as a reduction
ClO4 - (aq) + 8H + (aq) + 8e- Cl- (aq) + 4 H2O(l) Eo = +1.39 V 2) The Cu2+| Cu(s) half cell is lower in the table and therefore is written as a oxidation (reversed) and the sign of the E0 is reversed.
3) The reactions are balanced for # of electrons (multiply the second half reaction by 4), and added up. The voltages are added up as well
Eo = +1.39 V + (-0.34 V) = 1.05 V
Assignment: Structure of electrochemical cells
Assignment: Voltages of Electrochemical cells
