1. Chapter 21 Electrochemistry 21.2 Half-Cells and Cell Potentials
21.1 Electrochemical Cells
21.2 Half-Cells and Cell Potentials
21.3 Electrolytic Cells
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2. CHEMISTRY & YOU
How can you calculate the electrical potential of a cell in a laptop battery?
Batteries provide current to power lights and many kinds of electronic devices—such as the laptop shown here.
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3. What causes the electrical potential of an electrochemical cell?
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4. What causes the electrical potential of an electrochemical cell?
The electrical potential of a voltaic cell is a measure of the cell’s ability to produce an electric current.
Electrical potential is usually measured in volts (V).
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5. The potential of an isolated half-cell cannot be measured.
Electrical Potential
The potential of an isolated half-cell cannot be measured.
You cannot measure the electrical potential of a zinc half-cell or of a copper half-cell separately.
When these two half-cells are connected to form a voltaic cell, however, the difference in potential can be measured.
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6. Electrical Potential
The electrical potential of a cell results from a competition for electrons between two half-cells.
The half-cell that has a greater tendency to acquire electrons is the one in which reduction occurs.
Oxidation occurs in the other half-cell.
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7. Electrical Potential
The tendency of a given half-reaction to occur as a reduction is called the reduction potential.
The half-cell in which reduction occurs has a greater reduction potential than the half-cell in which oxidation occurs.
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8. Electrical Potential
The difference between the reduction potentials of the two half-cells is called the cell potential.
cell potential = –
reduction potential of half-cell in which reduction occurs
reduction potential of half-cell in which oxidation occurs
or Ecell = Ered – Eoxid
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9. Electrical Potential
The standard cell potential (E0cell) is the measured cell potential when the ion concentrations in the half-cells are 1M, any gases are at a pressure of 101 kPa, and the temperature is 25°C.
E0cell = E0red – E0oxid
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10. Electrical Potential
The standard hydrogen electrode is used with other electrodes so the reduction potentials of the other cells can be measured.
The standard reduction potential of the hydrogen electrode has been assigned a value of 0.00 V.
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11. Electrical Potential
The standard hydrogen electrode consists of a platinum electrode immersed in a solution with a hydrogen-ion concentration of 1M.
The solution is at 24°C.
Hydrogen gas at a pressure of 101 kPa is bubbled around the platinum electrode.
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12. Electrical Potential
The half-cell reaction that occurs at the platinum black surface is as follows:
2H+(aq, 1M) + 2e– H2(g, 101kPa)
E0H+ = 0.00 V
The double arrows in the equation indicate that the reaction is reversible.
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13. Electrical Potential
The half-cell reaction that occurs at the platinum black surface is as follows:
2H+(aq, 1M) + 2e– H2(g, 101kPa)
Whether this half-cell reaction occurs as a reduction or as an oxidation is determined by the reduction potential of the half-cell to which the standard hydrogen electrode is connected.
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14. What is the standard reduction potential of the standard hydrogen electrode?
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15. What is the standard reduction potential of the standard hydrogen electrode?
E0H+ = 0.00 V
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16. Standard Reduction Potentials
How can you determine the standard reduction potential of a half-cell?
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17. Standard Reduction Potentials
A voltaic cell can be made by connecting a standard hydrogen half-cell to a standard zinc half-cell.
To determine the overall reaction for this cell, first identify the half-cell in which reduction takes place.
Reduction takes place at the cathode, and oxidation takes place at the anode.
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18. Standard Reduction Potentials
A voltmeter gives a reading of V when the zinc electrode is connected to the negative terminal and the hydrogen electrode is connected to the positive terminal.
The zinc is oxidized, which means that it is the anode.
Hydrogen ions are reduced, which means that the hydrogen electrode is the cathode.
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19. Standard Reduction Potentials
You can now write the half-reactions and the overall cell reaction.
Oxidation: Zn(s) → Zn2+(aq) + 2e– (at anode)
Reduction: 2H+(aq) + 2e– → H2(g) (at cathode)
Cell reaction: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
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20. Standard Reduction Potentials
You can determine the standard reduction potential of a half-cell by using a standard hydrogen electrode and the equation for standard cell potential.
In the zinc-hydrogen cell, zinc is oxidized and hydrogen ions are reduced.
E0cell = E0red – E0oxid
E0cell = E0H+ – E0Zn2+
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21. Standard Reduction Potentials
You can determine the standard reduction potential of a half-cell by using a standard hydrogen electrode and the equation for standard cell potential.
The cell potential (E0cell) is measured at V.
E0H+ always equals 0.00 V.
E0cell = E0H+ – E0Zn2+
+0.76 V = 0.00 V – E0Zn2+
E0Zn2+ = –0.76 V
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22. Standard Reduction Potentials
You can determine the standard reduction potential of a half-cell by using a standard hydrogen electrode and the equation for standard cell potential.
The standard reduction potential for the zinc half-cell is –0.76 V.
The value is negative because the tendency of zinc ions to be reduced to zinc metal in this cell is less than the tendency of hydrogen ions to be reduced to hydrogen gas.
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23. Reduction Potentials at 25°C with 1M Concentrations of Aqueous Species
Interpret Data
Reduction Potentials at 25°C with 1M Concentrations of Aqueous Species
Electrode
Half-reaction
E0 (V)
Least tendency to occur as a reduction
Greatest tendency to occur as a reduction
Li+/Li
Li+ + e– → Li
–3.05
Pb2+/Pb
Pb2+ + 2e– → Pb
–0.13
K+/K
K+ + e– → K
–2.93
Fe3+/Fe
Fe3+ + 3e– → Fe
–0.036
Ba2+/Ba
Ba2+ + 2e– → Ba
–2.90
H+/H2
2H+ + 2e– → H2
0.000
Ca2+/Ca
Ca2+ + 2e– → Ca
–2.87
Cu2+/Cu
Cu2+ + 2e– → Cu
+0.34
Na+/Na
Na+ + e– → Na
–2.71
Cu+/Cu
Cu+ + e– → Cu
+0.52
Al3+/Al
Al3+ + 3e– → Al
–1.66
I2/I–
I2 + 2e– → 2I–
+0.54
Zn2+/Zn
Zn2+ + 2e– → Zn
–0.76
Fe3+/Fe2+
Fe3+ + e– → Fe2+
+0.77
Cr3+/Cr
Cr3+ + 3e– → Cr
–0.74
Hg22+/Hg
Hg22++ 2e– → 2Hg
+0.79
Fe2+/Fe
Fe2+ + 2e– → Fe
–0.44
Ag+/Ag
Ag+ + e– → Ag
+0.80
Cd2+/Cd
Cd2+ + 2e– → Cd
–0.40
Hg2+/Hg
Hg2++ 2e– → Hg
+0.85
Co2+/Co
Co2+ + 2e– → Co
–0.28
Br2/Br–
Br2+ 2e– → 2Br–
+1.07
Ni2+/Ni
Ni2+ + 2e– → Ni
–0.25
Cl2/Cl–
Cl2+ 2e– → 2Cl–
1.36
Sn2+/Sn
Sn2+ + 2e– → Sn
–0.14
F2/F–
F2+ 2e– → 2F–
+2.87
Increasing tendency to occur as a reduction
(stronger oxidizing agent)
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24. CHEMISTRY & YOU
What do you need to know to calculate the electrical potential of a cell in a laptop battery?
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25. CHEMISTRY & YOU
What do you need to know to calculate the electrical potential of a cell in a laptop battery?
In order to determine the electric potential, you need to know the reactions occurring in each half-cell, and the standard reduction potentials for those half-cell reactions.
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26. In a voltaic cell in which one half-reaction is Fe3+ + 3e– → Fe, which of the following half-reactions would occur as an oxidation?
A. 2H+ + 2e– → H2
B. Al3+ + 2e– → Al
C. Br2 + 2e– → 2Br–
D. Fe3+ + e– → Fe2+
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27. In a voltaic cell in which one half-reaction is Fe3+ + 3e– → Fe, which of the following half-reactions would occur as an oxidation?
A. 2H+ + 2e– → H2
B. Al3+ + 2e– → Al
C. Br2 + 2e– → 2Br–
D. Fe3+ + e– → Fe2+
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28. Calculating Standard Cell Potentials
How can you determine if a redox reaction is spontaneous?
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29. Calculating Standard Cell Potentials
In an electrochemical cell, the half-cell reaction having the more positive (or less negative) reduction potential occurs as a reduction in the cell.
You can use the known standard reduction potentials for the half-cells to:
predict the half-cells in which reduction and oxidation will occur.
find the E0cell value without having to actually assemble the cell.
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30. Calculating Standard Cell Potentials
If the cell potential for a given redox reaction is positive, then the reaction is spontaneous as written.
If the cell potential is negative, then the reaction is nonspontaneous.
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31. Determining Reaction Spontaneity
Sample Problem 21.1
Determining Reaction Spontaneity
Show that the following redox reaction between zinc metal and silver ions is spontaneous.
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
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32. Analyze List the knowns and the unknown.
Sample Problem 21.1
Analyze List the knowns and the unknown. 1
Identify the half-reactions, and calculate the standard cell potential (E0cell = E0red – E0oxid). If E0cell is positive, the reaction is spontaneous.
KNOWNS
Cell reaction: Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
UNKNOWN
Is the reaction spontaneous?
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33. Calculate Solve for the unknown.
Sample Problem 21.1
Calculate Solve for the unknown. 2
First identify the half-reactions.
Oxidation: Zn(s) → Zn2+(aq) + 2e–
Reduction: Ag+(aq) + e– → Ag(s)
Write both half-cells as reductions with their standard reduction potentials.
Zn2+(aq) + 2e– → Zn(s) E0Zn2+ = –0.76 V
Ag+(aq) + e– → Ag(s) E0Ag+ = V
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34. Calculate Solve for the unknown.
Sample Problem 21.1
Calculate Solve for the unknown. 2
Calculate the standard cell potential.
E0cell = E0red – E0oxid = E0Ag+ – E0Zn2+
= V – (–0.76 V) = V
E0cell > 0, so the reaction is spontaneous.
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35. Evaluate Does the result make sense?
Sample Problem 21.1
Evaluate Does the result make sense? 3
Zinc is above silver in the activity series for metals.
It makes sense that zinc is oxidized in the presence of silver ions.
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36. Writing the Cell Reaction
Sample Problem 21.2
Writing the Cell Reaction
Determine the cell reaction for a voltaic cell composed of the following half-cells:
Fe3+(aq) + e– → Fe2+(aq) E0Fe3+ = V
Ni2+(aq) + 2e– → Ni(s) E0Ni2+ = –0.25 V
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37. Analyze Identify the relevant concepts.
Sample Problem 21.2
Analyze Identify the relevant concepts. 1
The half-cell with the more positive reduction potential is the one in which reduction occurs (the cathode).
The oxidation reaction occurs at the anode.
Add the half-reactions, making certain that the number of electrons lost equals the number of electrons gained.
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38. Solve Apply the concepts to this problem.
Sample Problem 21.2
Solve Apply the concepts to this problem. 2
First identify the cathode and the anode.
The Fe3+ half-cell has the more positive reduction potential, so it is the cathode.
The Ni2+ half-cell has the more negative reduction potential, so it is the anode.
Thus the voltaic cell, Fe3+, is reduced and Ni is oxidized.
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39. Solve Apply the concepts to this problem.
Sample Problem 21.2
Solve Apply the concepts to this problem. 2
Write the half-cell reactions in the direction in which they actually occur.
Oxidation: Ni(s) → Ni2+(aq) + 2e– (at anode)
Reduction: Fe3+(aq) + e– → Fe2+(aq) (at cathode)
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40. Solve Apply the concepts to this problem.
Sample Problem 21.2
Solve Apply the concepts to this problem. 2
If necessary, multiply the half-reactions by the appropriate factor(s) so that the electrons cancel when the half-reactions are added.
Ni(s) →Ni2+(aq) + 2e–
2[Fe3+(aq) + e– → Fe2+(aq)]
Multiply the Fe3+ half-cell equation by 2 so that the electrons are present in equal numbers on both sides of the equation.
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41. Solve Apply the concepts to this problem.
Sample Problem 21.2
Solve Apply the concepts to this problem. 2
Add the half-reactions.
Ni(s) → Ni2+(aq) + 2e–
2Fe3+(aq) + 2e– → 2Fe2+(aq)
Ni(s) + 2Fe3+(aq) → Ni2+(aq) + 2Fe2+(aq)
The electrons lost by the species that is oxidized must be equal to the electrons gained by the species that is reduced.
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42. Calculating the Standard Cell Potential
Sample Problem 21.3
Calculating the Standard Cell Potential
Calculate the standard cell potential for the voltaic cell described in Sample Problem The half-reactions are as follows:
Fe3+(aq) + e– → Fe2+(aq) E0Fe3+ = V
Ni2+(aq) + 2e– → Ni(s) E0Ni2+ = –0.25 V
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43. Analyze List the knowns and the unknown.
Sample Problem 21.3
Analyze List the knowns and the unknown. 1
Use the equation E0cell = E0red – E0oxid to calculate the standard cell potential.
UNKNOWN
KNOWNS
E0cell = ?
E0Fe3+ = V
E0Ni2+ = –0.25 V
anode: Ni2+ half-cell
cathode: Fe3+ half-cell
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44. Calculate Solve for the unknown.
Sample Problem 21.3
Calculate Solve for the unknown. 2
First write the equation for the standard cell potential.
E0cell = E0red – E0oxid = E0Fe3+ – E0Ni2+
Substitute the values for the standard reduction potentials and solve the equation.
E0cell = V – (–0.25 V) = V
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45. Evaluate Does the result make sense?
Sample Problem 21.3
Evaluate Does the result make sense? 3
The reduction potential of the reduction is positive, and the reduction potential of the oxidation is negative.
Therefore, E0cell must be positive.
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46. Fe3+(aq) + 3e– → Fe(s) E0cell = –0.036 V
Determine the cell reaction for a voltaic cell composed of the following half-cells.
Fe3+(aq) + 3e– → Fe(s) E0cell = –0.036 V
Al3+(aq) + 3e– → Al(s) E0cell = –1.66 V
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47. Fe3+(aq) + 3e– → Fe(s) E0cell = –0.036 V
Determine the cell reaction for a voltaic cell composed of the following half-cells.
Fe3+(aq) + 3e– → Fe(s) E0cell = –0.036 V
Al3+(aq) + 3e– → Al(s) E0cell = –1.66 V
Oxidation: Al(s) → Al3+(aq) + 3e– (at anode)
Reduction: Fe3+(aq) + 3e– → Fe(s) (at cathode)
Al(s) + Fe3+(aq) → Al3+(aq) + Fe(s)
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48. Key Concepts
The electrical potential of a cell results from a competition for electrons between two half-cells.
You can determine the standard reduction potential of a half-cell by using a standard hydrogen electrode and the equation for standard cell potential.
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49. Key Concepts & Key Equation
If the cell potential for a given redox reaction is positive, then the reaction is spontaneous as written. If the cell potential is negative, then the reaction is nonspontaneous.
E0cell = E0red – E0oxid
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50. Glossary Terms
electrical potential: the ability of a voltaic cell to produce an electric current
reduction potential: a measure of the tendency of a given half-reaction to occur as a reduction (gain of electrons) in an electrochemical cell
cell potential: the difference between the reduction potentials of two half-cells
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51. Glossary Terms
standard cell potential (E0cell): the measured cell potential when the ion concentrations in the half-cells are 1.00M at 1 atm of pressure and 25°C
standard hydrogen electrode: an arbitrary reference electrode (half-cell) used with another electrode (half-cell) to measure the standard reduction potential of that cell; the standard reduction potential of the hydrogen electrode is assigned a value of 0.00 V
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52. END OF 21.2
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