62265893 electrochemistry
TRANSCRIPT
Chapter 18 Notes
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18Chemistry: A Molecular Approach by Nivaldo J. TroChemistry: A Molecular Approach by Nivaldo J. TroCHEM 1411 General ChemistryCHEM 1411 General Chemistry
Mr. Kevin A. BoudreauxAngelo State Universitywww.angelo.edu/faculty/kboudrea
Mr. Kevin A. BoudreauxAngelo State Universitywww.angelo.edu/faculty/kboudrea
ElectrochemistryElectrochemistry
Objectives:• Understand oxidation-reductions reactions, and be
able to balance a redox equation.• Understand the operation of galvanic and
electrolytic cells.• Understand how to calculate the potential of a cell,
and how this quantity is related to other thermodynamic quantities.
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ElectrochemistryElectrochemistry• Electrochemistry is the study of the relationship
between chemical and electrical energy.– Galvanic or voltaic cells convert chemical
energy into electrical energy (batteries).– Electrolytic cells use electrical energy to drive
chemical changes.– Many essential chemicals and materials are made
through electrochemical processes.• Electrochemistry is also deeply connected to
thermodynamics.
Chapter 18 Notes
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A Review of Oxidation and Reduction
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Oxidation and ReductionOxidation and Reduction• An oxidation-reduction (redox) reaction is a
process in which electrons are transferred from one substance to another.– Oxidation is the loss of electron(s).– Reduction is the gain of electron(s).
Oxidation: Zn → Zn2+ + 2e-
Reduction: 2H+ + 2e- → H2
– Oxidation always accompanies reduction.
Zn(s) + 2H+(aq) ⎯⎯→ Zn2+(aq) + H2(g)
oxidation
reduction
half-reactions
Chapter 18 Notes
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Oxidizing and Reducing AgentsOxidizing and Reducing Agents• The oxidizing agent is the substance that causes
oxidation to occur by accepting electrons. The oxidizing agent itself becomes reduced.
• The reducing agent is the substance that causes reduction to occur by losing electrons. The reducing agent itself becomes oxidized.
• The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent. This is important when balancing redox equations.
Zn(s) + 2H+(aq) ⎯⎯→ Zn2+(aq) + H2(g)
oxidation
reduction
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Balancing Redox ReactionsBalancing Redox Reactions• Since oxidation and reduction reactions always
occur together, balancing a redox reaction also requires balancing the number of electrons which are transferred.– A redox reaction may look balanced, but if the
charges are not the same on both sides, it isn’t:Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s)
[not balanced]• Simple redox reactions can often be balanced by
inspection, but more complicated ones require a systematic approach. There are two methods for balancing redox reactions:– the oxidation number method– the half-reaction method
Chapter 18 Notes
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The HalfThe Half--Reaction Method Reaction Method Step 1. Divide the skeleton reaction into two half-
reactions, each of which contains the oxidized and reduced forms of one of the species.
Step 2. Balance atoms and charges in each half-reaction.a. Balance atoms other than O and H.b. Balance O atoms by adding H2O.c. Balance H atoms by adding H+ ions.d. Balance charge by adding electrons (e-).
Step 3. Multiply each half-reaction by some integer to make the number of e- gained in the reduction equal the number of e- lost in the oxidation.
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The HalfThe Half--Reaction Method Reaction Method Step 4. Add the balanced half-reactions and include
states of matter. Step 4a. If the reaction occurs in basic solution, add
one OH- to both sides of the equation for every H+ ion present; combine H+ and OH-
ions to form waters, and cancel out those that appear on both sides.
Step 5. Double-check that the atoms and charges are balanced.
Chapter 18 Notes
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The Oxidation Number MethodThe Oxidation Number MethodStep 1. Assign oxidation numbers to all elements in
the reaction.Step 2. Identify the oxidized and reduced species.Step 3. Compute the number of electrons lost in the
oxidation and gained in the reduction from the oxidation number changes.
Step 4. Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients.
Step 5. Complete the balancing by inspection, adding phase labels.
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Examples: Balancing Redox ReactionsExamples: Balancing Redox Reactions1. Use the half-reaction method to balance the
following redox reactions. (sim. to Ex. 18.1-18.3)a. Cr2O7
2-(aq) + I-(aq) → Cr3+(aq) + I2(s)[acidic solution]
b. Sn(s) + NO3-(aq) → SnO2(aq) + NO(g)
[acidic solution]
c. MnO4-(aq) + C2O4
2-(aq) → MnO2(s) + CO32-(aq)
[basic solution]
Rules
Chapter 18 Notes
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Galvanic (Voltaic) Cells
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An Overview of Electrochemical CellsAn Overview of Electrochemical Cells• There are two types of electrochemical cells:• A galvanic cell (or voltaic cell) uses a spontaneous
reaction (ΔG° < 0) to generate electrical energy.– The difference in potential energy between higher
energy reactants and lower energy products is converted into electrical energy.
– The reactants do work on the surroundings.• An electrolytic cell uses electrical energy to drive a
nonspontaneous reaction (ΔG° > 0).– Lower energy reactants are converted into higher
energy products by electrical energy.– The surroundings do work on the reactants.
Chapter 18 Notes
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An Overview of Electrochemical CellsAn Overview of Electrochemical Cells
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A Zn Bar in a CuA Zn Bar in a Cu2+2+ SolutionSolution• If a bar of zinc metal is placed in a solution of Cu2+
ions, a spontaneous reaction takes place:Zn(s) → Zn2+(aq) + 2e- [oxidation]
Cu2+(aq) + 2e- → Cu(s) [reduction]——————————————————————————————————Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) [overall]
• When this is done in a beaker (Fig. 18.1), the zinc slowly dissolves as it is oxidized into soluble Zn2+
ions, and the blue color of the solution fades as Cu2+
is reduced to copper metal and deposits on the bar.• However, no electrical energy is generated because
the oxidizing and reducing agents are in physical contact; the enthalpy of the reaction escapes into the surroundings as heat.
Chapter 18 Notes
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A Zn Bar in a CuA Zn Bar in a Cu2+2+ SolutionSolution
Figure 18.1
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A Zn A Zn -- CuCu2+2+ Galvanic CellGalvanic Cell• If, however, the oxidation and reduction processes
are separated from each other in two separate half-cells (Fig. 18.2), electrical energy can be generated:– A rod of zinc metal (the anode) is immersed in a
Zn2+ solution and connected by a wire to a rod of copper metal (the cathode) in a Cu2+ solution.
– The zinc is oxidized, releasing electrons which travel through the wire and reduce Cu2+ in the other half-cell, producing and electrical current.
– A salt bridge between the half-cells maintains electrical neutrality and completes the circuit.
– The chemical energy in the reactants can then be used to light a bulb or run an electrical device.
– This type of galvanic cell is called a Daniell cell.
Chapter 18 Notes
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A Zn A Zn -- CuCu2+2+ Galvanic CellGalvanic Cell
Figure 18.2 MOV: Copper-Zinc Cell movie
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Galvanic Cells Galvanic Cells —— ElectrolytesElectrolytes• In a galvanic (voltaic) cell, the components of each
half-reaction are placed in separate containers called half-cells, consisting of an electrode dipping into an electrolyte solution; the cells are joined by a wireand a salt bridge.
• The electrolyte solutions are solutions of ions (usually dissolved in water) that are involved in the reaction or that carry the charge between the electrodes.– Unless an inert electrode is being used, the
electrolyte is usually a soluble ionic compound containing a cation of the electrode metal.
– In this example, the zinc half-cell electrolyte could be any soluble zinc compound, such as zinc nitrate, zinc chloride, zinc sulfate, etc.
Fig. 18.2
Chapter 18 Notes
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Galvanic Cells Galvanic Cells —— ElectrodesElectrodes• The electrodes are usually metal rods or sheets that
are immersed in the electrolyte solutions of each half-cell; they conduct the electricity between the cell and the surroundings. An electrode is classified by the half-reaction which takes place there:– The anode is where the oxidation half-reaction
takes place. Electrons are given up by the substance being oxidized (reducing agent) and leave the cell at the anode.
– The cathode is where the reduction half-reaction takes place. Electrons are taken up by the substance being reduced (oxidizing agent) and enter the cell at the cathode.
Fig. 18.2
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Galvanic Cells Galvanic Cells —— ElectrodesElectrodes• In this example, the oxidation half-cell consists of a
Zn metal bar acting as the anode, which is immersed in a Zn2+ electrolyte solution [Zn(NO3)2].– The Zn anode will dissolve as it is oxidized into
Zn2+, and will conduct e-’s out of its half-cell.Zn(s) → Zn2+(aq) + 2e-
• In this example, the reduction half-cell consists of a Cu metal bar acting as the cathode, which is immersed in a Cu2+ electrolyte solution [Cu(NO3)2]. – The Cu cathode will gain more Cu metal as Cu2+
is reduced, and will conduct e-’s into its half-cell.Cu2+(aq) + 2e- → Cu(s)
• By convention, the anode compartment is always drawn on the left.
Fig. 18.2
Chapter 18 Notes
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Galvanic Cells Galvanic Cells —— ElectrodesElectrodes• The relative charges on the electrodes are
determined by the source of the electrons and the direction of the electron flow.– Electrons are generated at the anode and
consumed at the cathode, therefore:– In any voltaic cell, the anode is negative and the
cathode is positive.
CPRCathodePositiveReduction
ANOAnodeNegativeOxidation
Fig. 18.2
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Galvanic Cells Galvanic Cells —— The Salt BridgeThe Salt Bridge• The salt bridge consists of a non-reactive electrolyte
suspended in a gel. It completes the circuit between the half-cells by maintaining charge neutrality in the electrolyte solutions.– As Zn is oxidized to Zn2+, anions (NO3
-) diffuse out of the salt bridge into its electrolyte solution, balancing out the charge.
– As Cu2+ is reduced to Cu, cations (K+) diffuse out of the salt bridge into its electrolyte solution.
– As electrons move left to right through the wire, anions move right to left (towards the anode) and cations move left to right (toward the cathode) through the salt bridge.
– Thus, the overall charge in each solution is maintained, and the circuit is completed.
Fig. 18.2
Chapter 18 Notes
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Galvanic Cells Galvanic Cells —— SummarySummaryFig. 18.2
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Active and Inactive ElectrodesActive and Inactive Electrodes• In this example, active electrodes are used, which
are active participants in the half-reactions. The electrodes are consumed or added to during the course of the reaction.
• Often, inactive (or inert) electrodes are used, usually made of graphite or platinum:
2I-(aq) → I2(s) + 2e- [anode]MnO4
-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)[cathode]
– In this case, the species involved can’t act as electrodes, and so something else must be used. The electrolyte solutions must contain all of the species involved in the half-reaction. (Fig 21.6)
Chapter 18 Notes
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Active and Inactive ElectrodesActive and Inactive Electrodes
Active Electrodes:
anode: Zn(s) → Zn2+(aq) + 2e-
cathode: Cu2+(aq) + 2e- → Cu(s)overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Inactive Electrodes:
anode: 2I-(aq) → I2(s) + 2e-
cathode: MnO4-(aq) + 8H+(aq) + 5e- →
Mn2+(aq) + 4H2O(l)overall: 2MnO4
-(aq) + 16H+(aq) + 10I-(aq)→ 2Mn2+(aq) + 5I2(s) + 8H2O(l)
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Active and Inactive ElectrodesActive and Inactive Electrodes• This examples uses both an active and an inactive
electrode:
anode: Fe(s) → Fe2+(aq) + 2e-
cathode: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
overall: 5Fe(s) + 2MnO4-(aq) + 16H+(aq) →
5Fe2+(s) + 2Mn2+(aq) + 8H2O(l)Figure 18.4
Chapter 18 Notes
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Cell NotationCell Notation• Cell notation is a useful shorthand description for
the components of a voltaic cell:
– The anode components are always written on the left and the cathode components on the right
– A vertical line | represents a boundary between two phases. Two or more components in the same phase are separated by commas.
– A double vertical line || represents the salt bridge.– The electron flow is left to right in this notation.
Anode Anodeelectrolyte
Cathodeelectrolyte Cathode||| |
electron flow
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Cell NotationCell Notation• In the Zn-Cu2+ voltaic cell, the cell notation is:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)– The order in which the cell notation is written
suggests the overall reaction: Zn is oxidized to Zn2+, and Cu2+ is reduced to Cu.
• In the I--MnO4- cell, where inert electrodes are used,
and the electrolyte solutions contain all of the reacting species, the cell notation is
Graphite(s) |I2(s) | I-(aq) ||H+(aq), MnO4
-(aq), Mn2+(aq) | Graphite(s)
Chapter 18 Notes
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Examples: Galvanic CellsExamples: Galvanic Cells2. Diagram, show balanced equations, and write the
cell notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.
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Examples: Galvanic CellsExamples: Galvanic Cells3. Design a galvanic cell that uses the redox reaction
Fe(s) + 2Fe3+(aq) → 3Fe2+(aq)Identify the anode and cathode half-reactions, and sketch the experimental setup. Label the anode and cathode, indicate the direction of electron and ion flow, and identify the sign of each electrode.
Chapter 18 Notes
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Examples: Galvanic CellsExamples: Galvanic Cells4. Given the following cell notations, write balanced
equations for the cell reactions, and give a brief description of the cells.
a. Pt(s)|Sn2+(aq),Sn4+(aq)||Ag+(aq)|Ag(s)
b. Cu(s)|Cu2+(aq)||Cl2(g)|Cl-(aq)|C(s)
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Examples: Galvanic CellsExamples: Galvanic Cells5. Write the shorthand notations for galvanic cells that
use the reactions shown below.
a. Fe(s) + Sn2+(aq) → Fe2+(aq) + Sn(s)
b. Pb(s) + Br2(l) → Pb2+(aq) + 2Br-(aq)
Chapter 18 Notes
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Cell Potentials
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Cell PotentialCell Potential• The driving force that causes electrons to move from
anode to the cathode is the electromotive force(emf), or cell potential (Ecell). This represents the potential energy difference between the materials in the anode and cathode compartments.
• Electrical units in the SI system:– electrical charge: coulomb, C
– electrical potential: volt, V
– electrical current: ampere, A
C1J 1 V 1 =
s1C 1 A 1 =
Chapter 18 Notes
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The Standard Hydrogen Electrode (SHE)The Standard Hydrogen Electrode (SHE)• Since it is not possible to measure the potential of a
half-cell without connecting it to another cell, a standard reference half-cell has been defined to have a potential of 0 V. All other cell potentials are combined with the reference half-cell, and measured relative to that value.
MOV: Zinc-Hydrogen Cell movieFigure 18.6
• The standard hydrogen electrode (SHE) consists of a platinum electrode in a 1 MH+(aq) solution at 25°C, through which H2(g) at 1 atm is bubbled:
2H+(aq, 1 M) + 2e- → H2(g, 1 atm); E° = 0 VH2(g, 1 atm) → 2H+(aq, 1 M) + 2e-; E° = 0 V
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Using the SHE to Obtain a Reduction PotentialUsing the SHE to Obtain a Reduction Potential• If we connect a Zn/Zn2+ half-cell to a SHE half-cell
(Fig. 18.7), the cell potential is found to be +0.76 V, with the zinc half-cell acting as the anode:
Zn(s) → Zn2+(aq) + 2e-; E° = ? V2H+(aq) + 2e- → H2(g); E° = 0 V
_______________________________________________________________________________________________________
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g); E° = +0.76 V
Figure 18.7
Chapter 18 Notes
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Using the SHE to Obtain a Reduction PotentialUsing the SHE to Obtain a Reduction Potential• The cell potential is obtained from the formula:
E°cell = E°red + E°ox
• Since E°cell = +0.76 V, and E°red = 0 V (SHE); E°oxfor the zinc cell is 0.76 V.
• If the Zn/Zn2+ half-cell instead runs as a reduction, we would write the reaction backwards and reverse the sign on the cell potential. Its standardreduction potential would be -0.76 V:
Zn2+(aq) + 2e- → Zn(s); E° = -0.76 V• By convention, tables of half-cell potentials are
written as reductions; the reaction for the cell which occurs as an oxidation is reversed.
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Using the SHE to Obtain a Reduction PotentialUsing the SHE to Obtain a Reduction Potential• If we connect a Cu/Cu2+ half-cell to a SHE half-cell,
the overall cell potential is found to be +0.34 V, with the copper half-cell acting as the cathode:
H2(g) → 2H+(aq) + 2e-; E° = 0 VCu2+(aq) + 2e- → Cu(s); E° = ? V
_______________________________________________________________________________________________________
H2(g) + Cu2+(aq) → 2H+(aq) + Cu(s); E° = +0.34 V
• Since E°cell = +0.34 V, and E°ox = 0 V (SHE); E°redfor the copper cell is +0.34 V:
Cu2+(aq) + 2e- → Cu(s); E° = +0.34 V
Chapter 18 Notes
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Standard Reduction PotentialsStandard Reduction Potentials• In the following charts, the standard reduction
potentials (or standard electrode potentials) for a number of half-cells are listed. In all cases, the half-reactions are written as reductions. In other words, the potential listed is the potential that the half-cell has if the half-cell reactions occurs as a reduction.
• Oxidizing agents (which get reduced) and electrons are on the left side of each half-reaction; reducing agents (which get oxidized) are on the right.
• The half-reactions are listed by decreasing standard reduction potential. Thus, the strongest oxidizing agents are located in the upper left (F2, H2O2, MnO4
-, etc.) and the strongest reducing agents are found in the lower right (Li, Na, Mg, etc.).– This list is essentially where the activity series for single displacement
reactions is derived.
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Standard Reduction PotentialsStandard Reduction Potentials
Chapter 18 Notes
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Standard Reduction PotentialsStandard Reduction Potentials
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Standard Reduction PotentialsStandard Reduction Potentials
more standard reduction potentials are listed in Appendix IID, p. A-15
Chapter 18 Notes
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Standard Reduction PotentialsStandard Reduction PotentialsStandard Reduction Potentials at 25°C
Half-Reaction E° (V) F2(g) + 2e- h 2F-(aq) +2.87 O3(g) + 2H+(aq) + 2e- h O2(g) + 2H2O(l) +2.07 Co3+(aq) + e- h Co2+(aq) +1.82 H2O2(aq) + 2H+(aq) + 2e- h 2H2O(l) +1.77 PbO2(s) + SO4
2-(aq) + 4H+(aq) + 2e- h PbSO4(s) + 2H2O(l) +1.69 2HOCl(aq) + 2H+(aq) + 2e- h Cl2(g) + 2H2O(l) +1.63 Ce4+(aq) + e- h Ce3+(aq) +1.61 MnO4
-(aq) + 8H+(aq) + 5e- h Mn2+(aq) + 4H2O(l) +1.51 Au3+(aq) + 3e- h Au(s) +1.50 BrO3
-(aq) + 6H+(aq) + 6e- h Br-(aq) + 3H2O +1.44 Cl2(g) + 2e- h 2Cl-(aq) +1.36 Cr2O7
2-(aq) + 14H+(aq) + 6e- h 2Cr3+(aq) + 7H2O(l) +1.33 MnO2(s) + 4H+(aq) + 2e- h Mn2+(aq) + 2H2O(l) +1.23 O2(g) + 4H+(aq) + 4e- h 2H2O +1.23 Br2(g) + 2e- h 2Br-(aq) +1.07 NO3
-(aq) + 4H+(aq) + 3e- h NO(g) + 2H2O +0.96 2Hg2+(aq) + 2e- h Hg2
2+(aq) +0.92 Hg2
2+(aq) + 2e- h Hg(l) +0.85 Ag+(aq) + e- h Ag(s) +0.80 Fe3+(aq) + e- h Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e— h H2O2(aq) +0.70 MnO4
-(aq) + 2H2O(l) + 3e- h MnO2(s) + 4OH-(aq) +0.59 I2(g) + 2e- h 2I-(aq) +0.53 NiO2(s) + 2H2O + 2e- h Ni(OH)2(s) + 2OH-(aq) +0.49 O2(g) + 2H2O(l) + 4e- h 4OH-(aq) +0.40 Cu2+(aq) + 2e- h Cu(s) +0.34 AgCl(s) + e- h Ag(s) + Cl-(aq) +0.22 SO4
2- + 4H+(aq) + 2e- h SO2(g) + 2H2O(l) +0.20 Cu2+(aq) + e- h Cu+(aq) +0.15 Sn4+(aq) + 2e- h Sn2+(aq) +0.15 2H+(aq) + 2e- h H2(g) 0.00
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Standard Reduction PotentialsStandard Reduction PotentialsStandard Reduction Potentials at 25°C
Half-Reaction E° (V) 2H+(aq) + 2e- h H2(g) 0.00 Pb2+(aq) + 2e- h Pb(s) -0.13 Sn2+(aq) + 2e- h Sn(s) -0.14 N2(g) + 5H+(aq) + 4e- h N2H5
+(aq) -0.23 Ni2+(aq) + 2e- h Ni(s) -0.25 Co2+(aq) + 2e- h Co(s) -0.28 PbSO4(s) + 2e- h Pb(s) + SO4
2-(aq) -0.31 Cd2+(aq) + 2e- h Cd(s) -0.40 Fe2+(aq) + 2e- h Fe(s) -0.44 Cr3+(aq) + 3e- h Cr(s) -0.74 Zn2+(aq) + 2e- h Zn(s) -0.76 2H2O + 2e- h H2(g) + 2OH-(aq) -0.83 Mn2+(aq) + 2e- h Mn(s) -1.18 Al3+(aq) + 3e- h Al(s) -1.66 Mg2+(aq) + 2e- h Mg(s) -2.37 Na+(aq) + e- h Na(s) -2.71 Ca2+(aq) + 2e- h Ca(s) -2.87 Sr2+(aq) + 2e- h Sr(s) -2.89 Ba2+(aq) + 2e- h Ba(s) -2.90 K+(aq) + e- h K(s) -2.93 Li+(aq) + e- h Li(s) -3.05
Chapter 18 Notes
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Using Standard Reduction PotentialsUsing Standard Reduction Potentials• An overall redox reaction is the sum of the half-
reactions associated with each half-cell process:– one occurs as a reduction (at the cathode), written
in the forward direction.– the other occurs as an oxidation (at the anode),
written in the reverse direction.– The potential of the cell (E°cell) is the difference
between the standard half-cell potentials of the cathode and the anode:
E°cell = E°cathode + (- E°anode)E°cell = E°red + (- E°ox)
• NOTE: Do NOT multiply E°half-cell values by the coefficient; it is an intensive property, since it is the ratio of J/C.
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Using Standard Reduction PotentialsUsing Standard Reduction Potentials• For example, to calculate the cell potential (E°cell)
for the oxidation of Zn(s) by Ag+(aq):2Ag+(aq) + Zn(s) → 2Ag(s) + Zn2+(aq)
consider the relevant half-reactions:
reduction (cathode):2 [Ag+(aq) + e- → Ag(s)] E° = 0.80 V
oxidation (anode):Zn(s) → Zn2+(aq) + 2e- -E° = -(-0.76 V)
E°cell = E°cathode + (- E°anode)E°cell = 0.80 V + 0.76 V = 1.56 V
Chapter 18 Notes
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Relative Strengths of Oxidizing & Reducing AgentsRelative Strengths of Oxidizing & Reducing AgentsCu2+(aq) + 2e- → Cu(s) E°cell = +0.34 V2H+(aq) + 2e- → H2(g) E°cell = 0.00 VZn2+(aq) + 2e- → Zn(s) E°cell = -0.76 V
• The more positive the E°cell value, the more the reaction tends to occur as written. Thus, Cu2+ gains two electrons more easily than H+, and H+ gains electrons more easily than Zn2+.– Therefore, Cu2+ is the strongest oxidizing agent,
and Zn2+ is the weakest oxidizing agent.– Since reversing the reaction reverses the sign of
E°cell, Zn is the strongest reducing agent and Cu is the weakest reducing agent.
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Writing and Predicting Spontaneous ReactionsWriting and Predicting Spontaneous Reactions• Table 18.1 can be used as a guide to writing
spontaneous redox reactions, which is useful when trying to construct electrochemical cells.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)stronger reducing
agent
weaker reducing
agent
stronger oxidizing
agent
weaker oxidizing
agent
• The members of a redox couple (such as Zn and Zn2+) differ by one or more electrons.
• A spontaneous reaction (E°cell > 0) will occur between an oxidizing agent and a reducing agent that lies below it on Table 18.1.
Chapter 18 Notes
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Writing and Predicting Spontaneous ReactionsWriting and Predicting Spontaneous Reactions• In other words, if the overall reaction is split into
half reactions, the half-reaction with the more positive reduction potential takes place as a reduction, while the other one runs in reverse as an oxidation.
• To determine whether a given reaction is spontaneous, look at the oxidation (anode) and reduction (cathode) processes, and calculate the cell potential. If the cell potential is positive, the reaction is spontaneous; if negative it is nonspontaneous (or spontaneous in the reverse direction).
E°cell = E°cathode + -E°anode
E°cell = E°red + -E°ox
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Metals Dissolving in AcidsMetals Dissolving in Acids• Some metals, such as Zn, dissolve in acidic solutions:
Zn2+(aq) + 2e- → Zn(s); E° = -0.76 V2H+(aq) + 2e- → H2(g); E° = 0 V
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g); E° = +0.76 V• Cu, however, does not dissolve in most acids:
Cu2+(aq) + 2e- → Cu(s); E° = +0.34 VCu(s) + 2H+(aq) → Cu2+(aq) + H2(g); E° = -0.34 V
• An important exception is nitric acid, which oxidizes metals through the redox reactionNO3
-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O(l); Ered
° = 0.96 V
Chapter 18 Notes
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials6. Arrange the following oxidizing agents in order of
increasing strength under standard-state conditions: Br2(l), Fe3+(aq), Cr2O7
2-(aq).
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials7. Arrange the following reducing agents in order of
increasing strength under standard-state conditions: Al(s), Na(s), Zn(s).
Chapter 18 Notes
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials8. Use the following half-reactions to predict the
spontaneous reaction that occurs when Br-, SO42-,
H2SO3, and Br2 are mixed in an acidic solution.Br2(aq) + 2e- → 2Br-(aq); E° = +1.07 VSO4
2-(aq) + 4H+(aq) + 2e- → H2SO3(aq) + H2O
E° = +0.17 V
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials9. What spontaneous reaction occurs if Cl2 and Br2
are added to a solution that contains both Cl- and Br-?
Chapter 18 Notes
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials10. Predict from Table 18.1 whether Pb2+(aq) can
oxidize Al(s) or Cu(s) under standard-state conditions. Calculate E° for each reaction at 25°C.
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials11. What is the overall cell reaction and the standard
cell potential of a galvanic cell employing the following half-reactions?
Cr3+(aq) + 3e- h Cr(s); E°Cr3+ = -0.74 VMnO4
-(aq) + 8H+(aq) + 5e- h Mn2+(aq) + 4H2O
E°MnO4- = +1.51 V
Chapter 18 Notes
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Examples: Using Reduction PotentialsExamples: Using Reduction Potentials12. Which of the following reactions occur
spontaneously in the forward direction? (sim. to Ex. 18.5)
Br2(aq) + Cl2(g) + 2H2O → 2Br-(aq) + 2HOCl(aq) + 2H+(aq)
3Zn(s) + 2Cr3+(aq) → 3Zn2+(aq) + 2Cr(s)
58
A Shocking Dental Galvanic CellA Shocking Dental Galvanic Cell
• Biting on aluminum foil produces pain if you have fillings in your teeth. Al acts as the anode (-1.66 V), saliva as the electrolyte, and the silver/tin/mercury alloy filling as an inert cathode. O2 is reduced to water (1.23 V), and the short circuit between the foil in contact with the filling creates a current that jolts the nerve in the tooth.
Chapter 18 Notes
59
Cell Potentials and Thermodynamics
60
Cell Potential and Free EnergyCell Potential and Free Energy• In the examples we’ve seen of spontaneous cell
reactions, the cell potentials have been positive. There must then be some connection between the cell potential and the free energy change.
• The cell potential (Ecell) and the free energy change (∆G) are related by the equation
∆G = - n F Ecell
n = the number of moles of electrons transferred in the reaction.
F = the faraday, F , the charge on one mole of electrons (named for Michael Faraday):
1 F = 96,485 C / mol e-
1 F = 96,485 J V-1 mol e-1
Chapter 18 Notes
61
Cell Potential and Free EnergyCell Potential and Free Energy• This gives us a connection between cell potential
and spontaneity:
• The more positive the Ecell, the more work the cell can do, and the farther the reaction proceeds to the right as written.
• If Ecell = 0, the reaction has reached equilibrium, and the cell can do no more work (“dead battery”).
at equilibrium00nonspontaneous-+spontaneous+-SpontaneityEcell∆G
62
Standard Cell PotentialsStandard Cell Potentials• Since Ecell is affected by concentration, pressure, and
temperature, a standard cell potential, E°cell is defined, in which all components are at standard thermodynamic conditions (25°C, 1 atm, pure solids for electrodes, ions at 1.00 M).
• When all components are at standard conditions,
ΔG° = - nFE°cell
Chapter 18 Notes
63
Examples: Cell Potential and Free EnergyExamples: Cell Potential and Free Energy13. Calculate the standard free-energy change at 25°C
for the following reaction. (sim. to Ex. 18.6)Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
64
Examples: Cell Potential and Free EnergyExamples: Cell Potential and Free Energy14. Calculate the standard cell potential at 25°C for the
following reaction, which has a standard free-energy change of -266.3 kJ.
Al(s) + Cr3+(aq) → Al3+(aq) + Cr(s)
Chapter 18 Notes
65
The Relationship Between The Relationship Between EE°°cellcell, , ΔΔGG°°, and K, and K• When all components are in their standard states,
ΔG° = - nFE°cell
• Using the relationship between K and ΔG°,ΔG° = - RT ln K
- RT ln K = - nFE°cell
KnFRTE ln o
cell =
Kn
E log V 0.0592 ocell =
V 0.0592 log
ocellnEK =
66
Examples: Calculating K and Examples: Calculating K and ΔΔGG°° from Efrom E°°15. Lead can displace silver from solution:
Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s)As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and ΔG° at 25°C for this reaction. (sim. to Ex. 18.7)
Chapter 18 Notes
67
The Nernst EquationThe Nernst Equation• Most electrochemical cells do not start out with all
of their components in their standard states or at their standard concentrations, and even if they did, would not remain that way for long.
• In order to obtain Ecell, the cell potential under nonstandard conditions, we start with the expression for ΔG under nonstandard conditions:
∆G = ∆G° + RT ln Q-nFEcell = -nFE°cell + RT ln Q
• Solving for Ecell gives the Nernst equation (W. H. Nernst, 1889)
QnFRTEE ln - o
cellcell =
68
The Nernst EquationThe Nernst Equation
• Remember when calculating the value of Q, solids (even solid electrodes) do not appear in the mass-action expression:
Cd(s) + 2Ag+(aq) → Cd2+(aq) + 2Ag(s)Q = [Cd2+] / [Ag+]2
C)25(at log V 0.0592 - oocellcell Q
nEE =
Ecell = E°cell
Ecell < E°cell
Ecell > E°cell
log Q = 0log Q > 0log Q < 0
[product] = [reactant]If Q = 1[product] > [reactant]If Q > 1[product] < [reactant]If Q < 1
Chapter 18 Notes
69
Examples: Using the Nernst EquationExamples: Using the Nernst Equation16. Consider a galvanic cell that uses the reaction
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)Calculate Ecell at 25°C when [H+] = 1.0 M, [Zn2+] = 0.0010 M, and PH2 = 0.10 atm. (sim. to Ex. 18.8)
70
Examples: Using the Nernst EquationExamples: Using the Nernst Equation17. A galvanic cell was constructed by connecting a
nickel electrode that was dipped into 1.20 M NiSO4solution to a chromium electrode that was dipping into a solution containing Cr3+ at an unknown concentration. The potential of the cell was measured to be 0.552 V, with the chromium serving as the anode. The standard cell potential for this system was determined to be 0.487 V. What was the concentration of Cr3+ in the solution of unknown concentration?
Chapter 18 Notes
71
Concentration CellsConcentration Cells• Because cell potentials change with concentration,
two of the same half-cells at different concentrations can be joined to produce a current:– In cell A, two Cu/Cu2+ half-cells are connected,
with all of the components at standard conditions. There is no voltage difference, and no current.
Figure 18.12
A B
72
Concentration CellsConcentration Cells– In cell B, the left half-cell has a concentration
0.010 M and the right half-cell is at 2.0 M. This concentration difference causes electrons to flow from the half-cell with the dilute Cu2+ to the half-cell with the concentrated Cu2+, and produces a voltage of 0.068 V.
– The flow of electrons increases the concentration of Cu2+ in the dilute cell (as Cu gets oxidized) and decreases the concentration of Cu2+ in the concentrated cell (as Cu2+ is reduced to Cu).
– Nerve cells use a concentration gradient to cause Na+ and K+ ions to diffuse into or out of a nerve cell, creating a spike in the electrochemical potential across the membrane, which travels down the nerve cell.
Chapter 18 Notes
73
The Nernst Equation and pHThe Nernst Equation and pH• pH meters are electronic devices that are used for
measuring pH (duuuh!). They are based on a cell which is similar to the SHE half-cell, which dips into a solution of unknown pH, and a calomel reference electrode:
Pt(s) | H2(1 atm) | H+(? M) || Reference cathode• The potential for the hydrogen half-reaction is
obtained from the Nernst equation:H2(g) → 2H+(aq) + 2e-
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
→→ ++
222
H
2o
H H H H
][H log V 0.0592 - Pn
EE
74
The Nernst Equation and pHThe Nernst Equation and pH• Since E° = 0 V, n = 2, and PH2 = 1 atm, this
simplifies to:
• The overall cell potential is
• and therefore
V)(pH) (0.0592 ])[H V)(-log (0.0592
][H log (2)2
V 0.0592 -
][H log 2
V 0.0592 - 2 H H2
==
=
=
+
+
+→ +E
refcell V)(pH) (0.0592 EE +=
V 0.0592 pH refcell EE −
=
Chapter 18 Notes
75
The pH MeterThe pH Meter• In practice, a glass electrode containing a silver wire
coated with silver chloride immersed in a solution of HCl is used as the test electrode; the solution in the cell is separated from the test solution by a thin glass membrane (Figure 18.6).
• The reference electrode is a calomel (Hg2Cl2) electrode in contact with liquid mercury and KCl(aq):
Hg2Cl2(s) + 2e- → 2Hg(l) + 2Cl-(aq); E° = +0.28 V• A typical pH meter measures the cell potential, and
electronically converts the potential to the pH value:
V 0.0592V .280 pH cell −=
E
76
The pH MeterThe pH Meter
Chapter 18 Notes
77
Batteries
78
Electrochemical Processes in BatteriesElectrochemical Processes in Batteries• A battery is a self-contained group of voltaic cells
arranged in series (plus-to-minus-to-plus) so that their individual voltages are added together. The term is also sometimes used to refer to single voltaic cells.– A primary battery cannot be recharged, and
must be disposed of when it is “dead.”– A secondary, or rechargeable, battery can be
restored to use by supplying electricity to reverse the cell reactions and regenerate the reactants.
– A fuel cell, or flow battery, is not self-contained; the reactants (usually a combustible fuel and oxygen) enter the cell and the products leave, generating electricity through controlled oxidation of the fuel.
Chapter 18 Notes
79
Dry Cell (Leclanché cell) (primary)A common household battery
Anode: Zn(s) → Zn2+(aq) + 2e-
Cathode: 2MnO2(s) + 2NH4+(aq) + 2e- → Mn2O3(s) + 2NH3(aq) + H2O(l) etc.
Ecell = 1.5 V
80
Alkaline Battery (primary)A modified form of the Leclanché cell, using a basic (alkaline) electrolyte;
corrodes more slowly, and produces higher power and a more stable current
Anode: Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode: 2MnO2(s) + 2H2O(l) + 2e- → 2Mn(OH)2(s) + 2OH-(aq)
Ecell = 1.5 V
Chapter 18 Notes
81
Lead-Acid Battery (rechargeable)Six cells in series, each of which delivers about 2 V.
The electrolyte solution is ~4.5 M H2SO4.The reactions can be run in reverse, thus recharging the battery,
by using current from the alternator.
Anode: Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e-
Cathode: PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O(l)
82
Mercury and Silver (Button) Batteries (primary)
Anode: Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode (mercury): HgO(s) + H2O(l) + 2e- → Hg(l) + 2OH-(aq)Cathode (silver): Ag2O(s) + H2O(l) + 2e- → 2Ag(s) + 2OH-(aq)
Ecell = 1.3 V (mercury); 1.6 V (silver)
Chapter 18 Notes
83
Nickel-Cadmium (Nicad) Battery (rechargeable)
Anode: Cd(s) + 2OH-(aq) → Cd(OH)2(s) + 2e-
Cathode: 2NiO(OH)(s) + 2H2O(l) + 2e- → 2Ni(OH)2(s) + 2OH-(aq)
Ecell = 1.4 V
84
Nickel-Metal-Hydride (NiMH) Battery (rechargeable)Replaces the toxic cadmium anode with a metal matrix impregnated with hydrogen atoms in a strongly basic electrolyte. At the anode, hydrogen
atoms are released as water molecules; supplying external current reverses the reactions and recharges the battery.
Anode: MH(s) + OH-(aq) → M(s) + H2O(l) + e-
Cathode: NiO(OH)(s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)
Ecell = 1.2 V
Chapter 18 Notes
85
Lithium Solid-State Battery (rechargeable)Lightweight batteries which produce high voltages; common in computers, digital
camereas, cell phones, etc.Anode: Li(s) → Li+(in solid electrolyte) + e-
Cathode: MnO2(s) + Li+ + e- → LiMnO2(s)
Ecell = 3 V
86
Fuel CellsDiffers from an ordinary battery in that the reactants are not contained within the
cell, but are continually supplied from outside. In a hydrogen-oxygen fuel cell, the reactants are H2 and O2 gases; there are no pollutants formed, but the electrode
materials are expensive and short-lived.
Anode: 2H2(g) + 4OH-(aq) → 4H2O(l) + 4e-
Cathode: O2(g) + 2H2O(l) + 4e- → 4OH-(aq)Overall: 2H2(g) + O2(g) → 2H2O(l)
Ecell = 1.2 V
Chapter 18 Notes
87
Experimental Systems:
Aluminum-Air BatteryAn aluminum anode is oxidized and oxygen from flowing moist air is reduced
at an inactive porous graphite cathode. An aqueous NaOH electrolyte circulates through the cell.
Anode: 4[Al(s) + 4OH-(aq) → Al(OH)4-(aq) + 3e-]
Cathode: 3[O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
Ecell = 2.7 V
Sodium-Sulfur BatteryMolten Na is the anode, and
molten S8 is the cathode.
Anode: 2Na(l) → 2Na+(l) + 2e-
Cathode: n/8S8(l) + 2e- → nS2-(l)
Ecell = 2.1 V
88
Chapter 18 Notes
89
Corrosion
90
The Corrosion of IronThe Corrosion of Iron• Corrosion is the oxidative deterioration of a metal,
such as the rusting of iron.• Iron does not rust because of a simple reaction
between iron metal and oxygen gas:4Fe + 3O2(g) → 2Fe2O3(s)
– Water must be present for rusting to occur; in oxygen-free water or dry air, iron does not rust.
– Rusting is accelerated by electrolytes and acids.– Rusting also involves pitting of the metal surface,
but the rust is deposited at a different location from the pits.
– Iron is oxidized in one region, and oxygen is reduced in another region.
Chapter 18 Notes
91
The Corrosion of IronThe Corrosion of Iron
• The surface of the iron and the droplet of water form a tiny galvanic cell.– Portions of the iron surface act as an anode,
where iron is oxidized. Other portions act as cathodes, where oxygen is reduced to water.
– The water droplet itself acts as an electrolyte.
Figure 18.27
92
The Corrosion of IronThe Corrosion of Ironanode: 2Fe(s) → 2Fe2+(aq) + 4e-; E° = 0.45 V
cathode: O2(g) + 4H+(aq) + 4e- → 2H2O(l); E° = 1.23 V
• When Fe2+ ions migrate away from the pitted anode region, they come in contact with O2 dissolved in the water droplet and are oxidized to Fe3+ ions:
4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(s) + 2H2O
• The Fe3+ ions form a very insoluble hydrated oxide, which is deposited as red-brown rust:
2Fe3+(aq) + 4H2O(g) → 4Fe2O3•H2O(s) + 6H+(aq)
ACT: Corrosion Activity
Chapter 18 Notes
93
The Corrosion of IronThe Corrosion of Iron• Ionic compounds dissolved in the water accelerate
the rusting process by increasing the conductivity of the electrolyte solution; this is a problem in areas where salt is used to melt ice on the roads, and is a big problem in ocean-going vessels.
• Rusting also occurs faster at lower pH.• Many other metals, such as Al, Mg, Cr, Ti, and Zn,
form hard, impenetrable metal oxide coatings when they oxidize, which prevent any further corrosion. Iron(III) oxide is too porous to shield the underlying metal from further oxidation.
94
Preventing CorrosionPreventing Corrosion• The corrosion of iron can be prevented or minimized
by shielding the metal surface from oxygen and moisture with paint, or by coating the iron with a more durable metal such as Cr, Sn, or Zn.
• Galvanized iron is produced by dipping iron into molten zinc. When oxidation occurs, Zn is oxidized instead of Fe, since Zn has a lower reduction potential. (Zn also reduces Fe2+ back to Fe.)– The zinc oxide does not flake off the metal
surface, and forms a protective coating.
Fe2+ + 2e- h Fe(s); E° = -0.44Zn2+ + 2e- h Zn(s); E° = -0.76
Chapter 18 Notes
95
Preventing CorrosionPreventing Corrosion• Protecting a metal from corrosion by connecting it to
a more easily oxidized metal is called cathodicprotection.– It is not always necessary to coat the entire metal
surface, as long as there is electrical contact with the second metal.
– Underground iron pipelines or propellers on boats can be protected by placing them in contact with Mg, which is more easily oxidized, and acts as a sacrificial anode.
96Figure 21.16
Preventing CorrosionPreventing Corrosion• If iron is in contact with a less active metal, such as
copper, the iron anodic action is improved, and will corrode more rapidly (since Fe2+ is lost only at the anode).– When iron plumbing is connected directly to
copper plumbing with no electrical insulation between them, the iron pipe corrodes rapidly.
Chapter 18 Notes
97
Electrolytic Cells
98
Galvanic vs. Electrolytic CellsGalvanic vs. Electrolytic Cells• Galvanic cells use spontaneous redox reactions (E >
0; ΔG < 0) to generate electricity.– Chemical energy is converted to electrical energy
as the reaction proceeds toward equilibrium.• Electrolytic cells use electric current from an
external source to drive a nonspontaneous redox reaction (E < 0; ΔG > 0).– Electrical energy in converted to chemical energy
as the reaction proceeds away from equilibrium.
Dead Battery00EquilibriumElectrolytic+-NonspontaneousGalvanic-+SpontaneousCell TypeΔGEcellReaction Type
Chapter 18 Notes
99
Galvanic vs. Electrolytic CellsGalvanic vs. Electrolytic Cells
Figure 18.22
100
Electrolytic Cells and ElectrolysisElectrolytic Cells and Electrolysis• The process of using an electric current to bring
about chemical change is called electrolysis. Electrolysis can be performed on pure substances(molten salts, water), mixtures of molten salts, or aqueous solutions of salts.– In an electrolytic cell, oxidation occurs at the
anode, and reduction occurs at the cathode.– The cathode is negative and the anode is positive.
(Current is supplied to the cathode by the battery, and electrons are pulled out of the anode by the battery).
• A rechargeable battery functions as a galvanic cell when it is operating (i.e., supplying current), but as an electrolytic cell when it is being recharged.
Chapter 18 Notes
101
Electrolysis of Molten SaltsElectrolysis of Molten Salts• If a pure molten salt is subjected to electrolysis, the
cation will be reduced at the cathode and the anion will be oxidized at the anode.
Anode: 2Cl-(l) → Cl2(g) + 2e-
Cathode: Na+(l) + e- → Na(s)Overall: 2Na+(l) + 2Cl-(l) → 2Na(s) + Cl2(g)
• The electrolyte is the molten salt itself.– Na+ is attracted to the negative electrode, and Cl-
is attracted to the positive electrode.– The moving charges allow current to flow
through the molten salt, completing the circuit.
102
Electrolysis of Molten SaltsElectrolysis of Molten Salts
Figure 18.23
Chapter 18 Notes
103
Electrolysis of Electrolysis of Mixed Molten SaltsMixed Molten Salts• When there are two or more molten salts present in
an electrolytic cell:– the more easily oxidized species (stronger
reducing agent) reacts at the anode (oxidation).– the more easily reduced species (stronger
oxidizing agent) reacts at the cathode (reduction).• In determining which potential oxidations and
reductions take place, we cannot use the table of reduction potentials, since these are determined for aqueous ions. Instead, we have to rely on our knowledge of periodic atomic trends to predict which species will gain or lose electrons more easily.
104
Examples: Electrolysis of Molten SaltsExamples: Electrolysis of Molten Salts18. In the electrolysis of molten MgBr2, what products
form at the anode and cathode?
Chapter 18 Notes
105
Examples: Electrolysis of Molten SaltsExamples: Electrolysis of Molten Salts19. A chemical engineer melts a naturally occurring
mixture of NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction. (sim. to Ex. 18.9a)
106
Examples: Electrolysis of Molten SaltsExamples: Electrolysis of Molten Salts20. Metallic potassium was first prepared in 1807 by
Humphrey Davy by the electrolysis of molten potassium hydroxide. For this cell, label the anode and cathode, and show the direction of ion flow, and write balanced equations for the anode, cathode, and overall cell reactions.
Chapter 18 Notes
107
Electrolysis of WaterElectrolysis of Water• The electrolysis of water is usually carried out in the
presence of a nonreacting salt, such as Na2SO4, since pure water does not carry electrical currents well.
Anode (oxidation):2H2O(l) → O2(g) + 4H+(aq) + 4e- -E = -0.82 V
Cathode (reduction):2H2O(l) + 2e- → H2(g) + 2OH-(aq) E = -0.41 V
Overall:2H2O(l) → 2H2(g) + O2(g) E = -1.23 V
• Since the cell is not at standard conditions, the potentials are calculated from the Nernst equation.
108
Electrolysis of WaterElectrolysis of Water
Figure 18.20
Chapter 18 Notes
109
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions• When predicting the electrolysis products of an
aqueous ionic solution, it is necessary to consider that water is also present, and may be reduced or oxidized instead of the ionic compound.
• For example, in an ionic solution of NaI, there are two processes that can potentially take place at the anode and cathode:
Anode:Oxidation of I-:
2I-(aq) → I2(s) + 2e- -E° = -0.54 VOxidation of water:
2H2O(l) → O2(g) + 4H+(aq) + 4e- -E = -0.82 VOxidation of I- takes place.
110
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt SolutionsCathode:
Reduction of Na+: Na+(aq) + e- → Na(s) E° = -2.71 V
Reduction of water:2H2O(l) + 2e- → H2(g) + 2OH-(aq) E = -0.41 V
Reduction of water takes place.
Overall Reaction:2I-(aq) + 2H2O(l) → I2(s) + H2(g) + 2OH-(aq)
Chapter 18 Notes
111
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions
Figure 18.24
112
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions• If electrolysis were performed on aqueous KNO3:Anode:
Oxidation of NO3-: can’t be done
Oxidation of water:2H2O(l) → O2(g) + 4H+(aq) + 4e- -E = -0.82 V
Oxidation of H2O takes place.Cathode:
Reduction of K+: K+(aq) + e- → K(s) E° = -2.93 V
Reduction of water:2H2O(l) + 2e- → H2(g) + 2OH-(aq) E = -0.41 V
Reduction of water takes place.Overall Reaction:
2H2O(l) → 2H2(g) + O2(g)
Chapter 18 Notes
113
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions
114
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions• Sometimes unexpected products are obtained,
however. When gases are produced at electrodes, an additional voltage called the overvoltage is required to overcome the kinetic factors (such as high Eact).– E.g., in the electrolysis of NaCl, H2 forms at the
cathode and Cl2 forms at the anode, even through a comparison of electrode potentials would seem to indicate that O2 should form:
Oxidation of water:2H2O(l) → O2(g) + 4H+(aq) + 4e- -E = -0.82 V
(~ -1.4 V with overvoltage)Oxidation of Cl-:
2Cl-(aq) → Cl2(g) + 2e- -E° = -1.36 V
Chapter 18 Notes
115
Electrolysis of Aqueous Salt SolutionsElectrolysis of Aqueous Salt Solutions• In general, cations of less active metals are reduced
to the neutral metal, such as Au, Ag, Cu, Cr, Pt, and Cd.
• More active metals are not reduced, such as those of Group I, Group II, and Al. Water is reduced to H2and OH- instead.
• Anions that are oxidized include the halides (except F-.)
• Anions that are not oxidized include F-, SO42-, NO3
-, and PO4
3-. Water is oxidized to O2 and H+ instead.
116
Examples: Predicting Electrolysis ProductsExamples: Predicting Electrolysis Products21. Predict the half-cell reactions that occur when
aqueous solutions of the following salts are electrolyzed in a cell with inert electrodes. What is the overall cell reaction in each case? (sim. to Ex. 18.9a)a. KBrb. AgNO3
c. MgSO4
d. CuSO4
Chapter 18 Notes
117
The Stoichiometry of ElectrolysisThe Stoichiometry of Electrolysis• Faraday’s law of electrolysis — the amount of
substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell. (Michael Faraday, 1830s)
• In order to calculate how much charge is needed to produce a given amount of material:– Balance the half-reaction to find the number of
moles of electrons needed per mole of product.– Use the Faraday constant (F = 9.65 × 104 C/mol
e-) to find the corresponding charge.– Use the molar mass the find the charge needed for
a given mass of product.
118
The Stoichiometry of ElectrolysisThe Stoichiometry of Electrolysis• The amount of charge flowing through the cell is
determined from the current, measured in amperes (A):
1 A = 1 C / s• Thus, current multiplied by time gives the charge:
A × s = C/s × s = C
Chapter 18 Notes
119
Examples: Stoichiometry and ElectrolysisExamples: Stoichiometry and Electrolysis22. How many minutes will it take for a current of 10.0
A to deposit 3.00 g of gold from a solution of AuCl3?
120
Examples: Stoichiometry and ElectrolysisExamples: Stoichiometry and Electrolysis23. How many grams of copper are deposited on the
cathode of an electrolytic cell if an electric current of 2.00 A is run through a solution of CuSO4 for a period of 20.0 min? (sim. to Ex. 18.10)
Chapter 18 Notes
121
Examples: Stoichiometry and ElectrolysisExamples: Stoichiometry and Electrolysis24. What current must be supplied to deposit 3.00 g of
gold from a solution of AuCl3 in 20.0 min?
122
Chapter 18 Notes
123
Industrial Applications of
Electrolysis
124
Manufacture of SodiumManufacture of Sodium
Sodium metal is produced in a Downs cell from a mixture of sodium chloride and calcium chloride; the cell is designed to keep the sodium metal separate from the chlorine gas, or they will recombine (violently!).
Chapter 18 Notes
125
Manufacture of Chlorine and Sodium HydroxideManufacture of Chlorine and Sodium Hydroxide
The production of Cl2 gas and NaOH by electrolysis is the basis of the chlor-alkali industry, which generates sales of about $4 billion in the US alone. Saturated NaCl(aq) is oxidized in the anode compartment to produce Cl2; water is reduced in the cathode compartment to OH-; Na+ from the anode compartment moves by osmosis through a cation-permeable membrane to the cathode compartment, maintaining neutrality and thereby producing sodium hydroxide.
126
Manufacture of AluminumManufacture of Aluminum
Aluminum is the third most abundant element in the crust (8.3%), usually in the form of the ore bauxite (Al2O3). Al used to be extremely valuable because of the difficulty in extracting Al metal from the ore. Now, Al is produced using the Hall-Héroult process by dissolving bauxite in molten cryolite(Na3AlF6), and subjecting it to electrolysis. Al3+ is reduced to Al, and the liquid metal is drained off, cooled, and rolled intoplates, sheets, or foil. This process made the production of Almuch more cost-effective, but still requires large amounts of electricity (n=3: 1 mol e- produces 9.0 g Al).
Chapter 18 Notes
127
ElectroplatingElectroplatingIn electroplating, a 0.03 to 0.05 mm coating of one metal is deposited on another metal. In this illustration, silver dissolves from a solid Ag bar (the anode) and is deposited as solid silver on a fork (the cathode).Electroplating can be used to provide a variety of protective or decorative metal surfaces, such as chrome (Cr), galvanized metal (Zn), gold, silver, etc.
128
Refining of CopperRefining of Copper
Copper can be obtained from ore in 99% purity, but this is not good enough to use in wire. Impure copper is used as the anode,and a sheet of very pure copper is used as the cathode; Cu dissolves at the anode, and plates out as pure copper at the cathode. Other easily oxidized metal ions, such as Zn2+ and Fe2+
remain in solution, and less easily oxidized metals fall to the bottom of the cell as anode mud. The anode mud contains such “impurities” as silver, gold, and platinum.
Chapter 18 Notes
129
Equation SummaryEquation Summary• Cell Potentials:
E°cell = E°red + (-E°ox)• Thermodynamics:
ΔG° = - nFE°cell Kn
E log V 0.0592 ocell =
130
Equation SummaryEquation Summary• The Nernst equation:
• Electrolysis:1 A = 1 C / s 1 V = 1 J / C
1 mol e- = 9.65×104 C (F, Faraday constant)
C)25(at log V 0.0592 - oocellcell Q
nEE =