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Lecture 12: Cell Potentials. Reading: Zumdahl 11.2 Outline What is a cell potential? SHE, the electrochemical zero. Using standard reduction potentials. Cell Potentials.

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lecture 12 cell potentials
Lecture 12: Cell Potentials
  • Reading: Zumdahl 11.2
  • Outline
    • What is a cell potential?
    • SHE, the electrochemical zero.
    • Using standard reduction potentials.
cell 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
Cell Potentials (cont.)
  • We can measure the magnitude of the EMF causing electron (i.e., current) flow by measuring the voltage.

e-

Anode

Cathode

1 2 cell potentials
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
1/2 Cell Potentials (cont.)
  • Consider the following galvanic cell
  • Electrons are spontaneously flowing from the Zn/Zn+2 half cell (anode) to the H2/H+ half cell (cathode)
1 2 cell potentials cont6
1/2 Cell Potentials (cont.)
  • We define the 1/2 cell potential of the hydrogen 1/2 cell as zero.

SHE

P(H2) = 1 atm

[H+] = 1 M

2H+ + 2e- H2

E°1/2(SHE) = 0 V

1 2 cell potentials cont7
1/2 Cell Potentials (cont.)
  • With our “zero” we can then measure the voltages of other 1/2 cells.
  • In our example, Zn/Zn+2 is the anode: oxidation

Zn Zn+2 + 2e-

E°Zn/Zn+2 = 0.76 V

2H+ + 2e- H2

E° SHE = 0 V

Zn + 2H+ Zn+2 + H2

E°cell = E°SHE + E°Zn/Zn+2 = 0.76 V

0

standard reduction potentials
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- Zn E° = -0.76 V

Cu+2 + 2e- Cu E° = +0.34 V

Fe+3 + e- Fe+2 E° = 0.77 V

standard potentials cont
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- Zn E° = -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
Writing Galvanic Cells

For galvanic cells, Ecell > 0

In this example:

Zn/Zn+2 is the anode

Zn Zn+2 + 2e- E° = +0.76 V

Cu/Cu+2 is the cathode

Cu+2 + 2e- Cu E° = 0.34 V

writing galvanic cells cont
Writing Galvanic Cells (cont.)

Zn Zn+2 + 2e- E° = +0.76 V

Cu+2 + 2e- Cu E° = 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

writing galvanic cells cont12
Writing Galvanic Cells (cont.)

Shorthand Notation

Zn|Zn+2||Cu+2|Cu

Anode

Cathode

Salt bridge

predicting galvanic cells
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
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

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 agent

Weaker

another example
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)

oxidation

+2 0 +3

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
Another Example (cont.)

weaker reduction

Fe(s) Fe+2(aq) + 2e-

Fe+2(aq) + 2e- Fe(s)

E° = +0.44 V

E° = -0.44 V

Fe+3(aq) + e- Fe+2(aq)

E° = +0.77 V

2 x

2Fe+3(aq) + Fe(s) 3Fe+2(aq)

E°cell = 1.21 V