Lecture 12: Cell Potentials Reading: Zumdahl 11.2 Outline –What is a cell potential? –SHE, the electrochemical zero. –Using standard reduction potentials.
Cell Potentials In our galvanic cell, we had a species being oxidized at the anode, a species being reduced at the cathode, and electrons flowing from anode to cathode. The force on the electrons causing them to full is referred to as the electromotive force (EMF). The unit used to quantify this force is the volt (V) 1 volt = 1 Joule/Coulomb of charge V = J/C
Cell Potentials (cont.) We can measure the magnitude of the EMF causing electron (i.e., current) flow by measuring the voltage. AnodeCathode e-e-
1/2 Cell Potentials What we seek is a way to predict what the voltage will be between two 1/2 cells without having to measure every possible combination. To accomplish this, what we need to is to know what the inherent potential for each 1/2 cell is. The above statement requires that we have a reference to use in comparing 1/2 cells. That reference is the standard hydrogen electrode (SHE)
1/2 Cell Potentials (cont.) Consider the following galvanic cell Electrons are spontaneously flowing from the Zn/Zn +2 half cell (anode) to the H 2 /H + half cell (cathode)
1/2 Cell Potentials (cont.) We define the 1/2 cell potential of the hydrogen 1/2 cell as zero. SHE P(H 2 ) = 1 atm [H + ] = 1 M 2H + + 2e - H 2 E° 1/2 (SHE) = 0 V
1/2 Cell Potentials (cont.) With our “zero” we can then measure the voltages of other 1/2 cells. Zn Zn +2 + 2e - E° SHE = 0 V In our example, Zn/Zn +2 is the anode: oxidation 2H + + 2e - H 2 Zn + 2H + Zn +2 + H 2 E° cell = E° SHE + E° Zn/Zn+2 = 0.76 V 0 E° Zn/Zn+2 = 0.76 V
Standard Reduction Potentials Standard Reduction Potentials: The 1/2 cell potentials that are determined by reference to the SHE. These potentials are always defined with respect to reduction. Zn +2 + 2e - ZnE° = -0.76 V Cu +2 + 2e - CuE° = +0.34 V Fe +3 + e - Fe +2 E° = 0.77 V
Standard Potentials (cont.) If in constructing an electrochemical cell, you need to write the reaction as a oxidation instead of a reduction, the sign of the 1/2 cell potential changes. Zn +2 + 2e - ZnE° = -0.76 V Zn Zn +2 + 2e- E° = +0.76 V 1/2 cell potentials are intensive variables. As such, you do NOT multiply them by any coefficients when balancing reactions.
Writing Galvanic Cells For galvanic cells, E cell > 0 In this example: Zn/Zn +2 is the anode Cu/Cu +2 is the cathode Zn Zn +2 + 2e- E° = +0.76 V Cu +2 + 2e - CuE° = 0.34 V
Writing Galvanic Cells (cont.) Zn Zn +2 + 2e- E° = +0.76 V Cu +2 + 2e - CuE° = 0.34 V Cu +2 + Zn Cu + Zn +2 E° cell = 1.10 V Notice, we “reverse” the potential for the anode. E° cell = E° cathode - E° anode
Predicting Galvanic Cells Given two 1/2 cell reactions, how can one construct a galvanic cell? Need to compare the reduction potentials of the two half cells. Turn the reaction for the weaker reduction (smaller E° 1/2 ) and turn it into an oxidation. This reaction will be the anode, the other the cathode.
Predicting Galvanic Cells (cont.) Example. Describe a galvanic cell based on the following: Ag + + e - Ag E° 1/2 = 0.80 V Fe +3 + e - ----> Fe +2 E° 1/2 = 0.77 V We ake r Fe +2 ----> Fe +3 + e - E° 1/2 = -0.77 V Ag + + Fe +2 Ag + Fe +3 E° cell = 0.03 V E° cell > 0….cell is galvanic reducing agentreducing agent
Another Example For the following reaction, identify the two half cells, and use these half cells to construct a galvanic cell 3Fe +2 (aq) Fe(s) + 2Fe +3 (aq) +2 0 +3 oxidation reduction Fe +2 (aq) + 2e- Fe(s) E° = -0.44 V Fe +3 (aq) + e- Fe +2 (aq)E° = +0.77 V
Another Example (cont.) Fe +2 (aq) + 2e- Fe(s) E° = -0.44 V Fe +3 (aq) + e- Fe +2 (aq)E° = +0.77 V weaker reduction Fe(s)Fe +2 (aq) + 2e-E° = +0.44 V 2 x 2Fe +3 (aq) + Fe(s) 3Fe +2 (aq)E° cell = 1.21 V