Chapter 21
Electrochemistry:
Chemical Change and Electrical Work
Key Points About Redox Reactions
•Oxidation (electron loss) always accompanies reduction (electron gain).
•The oxidizing agent is reduced, and the reducing agent is oxidized.
•The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.
A summary of redox terminology
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
OXIDATION
Zn loses electrons.
Zn is the reducing agent and becomes oxidized.
The oxidation number of Zn increases from x to +2.
REDUCTION
One reactant loses electrons.
Reducing agent is oxidized.
Oxidation number increases.
Hydrogen ion gains electrons.
Hydrogen ion is the oxidizing agent and becomes reduced.
The oxidation number of H decreases from +1 to 0.
Other reactant gains electrons.
Oxidizing agent is reduced.
Oxidation number decreases.
Half-Reaction Method for Balancing Redox Reactions
Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions.
•Each reaction is balanced for mass (atoms) and charge.
•One or both are multiplied by some integer to make the number of electrons gained and lost equal.
•The half-reactions are then recombined to give the balanced redox equation.
•The separation of half-reactions reflects actual physical separations in electrochemical cells.
Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions -
Determine the O.N.s for the species undergoing redox.
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)
+6 -1 +3 0
2. Balance atoms and charges in each half-reaction -
I - I2
2 + 7H2O(l)
Cr2O72- Cr3+
14H+(aq) + Cr2O72- Cr3+
Cr is going from +6 to +3
I is going from -1 to 0
net: +12 net: +6 Add 6e- to left.
2Cr2O72- Cr3+ + 7H2O(l)14H+(aq) +6e- +
Balancing Redox Reactions in Acidic Solution continued
+ 2e-I - I22
Cr2O72- Cr3+ + 7H2O(l)14H+(aq) +6e- +
3. Multiply each half-reaction by an integer, if necessary -
Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.
X 3
4. Add the half-reactions together -
+ 2e-I - I22
3I - I26 + 6e-
Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)14H+(aq) +
2
Cr2O72- Cr3+ + 7H2O(l)14H+ +6e- + 2
Do a final check on atoms and charges.
Balancing Redox Reactions in Basic Solution
Balance the reaction in acid and then add OH- so as to neutralize the H+ ions.
Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)14H+(aq) +
+ 14OH-(aq) + 14OH-(aq)
Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH-14H2O +
Reconcile the number of water molecules.
+ 14OH-Cr2O72- + 6 I- 2Cr3+ + 3I27H2O +
Do a final check on atoms and charges.
The redox reaction between dichromate ion and iodide ion.
Cr2O72- I-
Cr3+ + I2
Practice 1 Balancing Redox Reactions by the Half-Reaction Method
PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O4
2-(aq) MnO2(s) + CO32-(aq)
Energy is absorbed to drive a nonspontaneous redox reaction
General characteristics of voltaic and electrolytic cells
VOLTAIC CELL ELECTROLYTIC CELLEnergy is released from
spontaneous redox reaction
Reduction half-reactionY++ e- Y
Oxidation half-reactionX X+ + e-
System does work on its surroundings
Reduction half-reactionB++ e- B
Oxidation half-reactionA- A + e-
Surroundings(power supply)do work on system(cell)
Overall (cell) reactionX + Y+ X+ + Y; G < 0
Overall (cell) reactionA- + B+ A + B; G > 0
The spontaneous reaction between zinc and copper(II) ion
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A voltaic cell based on the zinc-copper reaction
Oxidation half-reactionZn(s) Zn2+(aq) + 2e-
Reduction half-reactionCu2+(aq) + 2e- Cu(s)
Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Notation for a Voltaic Cell
components of anode compartment
(oxidation half-cell)
components of cathode compartment
(reduction half-cell)
phase of lower oxidation state
phase of higher oxidation state
phase of higher oxidation state
phase of lower oxidation state
phase boundary between half-cells
Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)
graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite
inert electrode
A voltaic cell using inactive electrodes
Reduction half-reactionMnO4
-(aq) + 8H+(aq) + 5e-
Mn2+(aq) + 4H2O(l)
Oxidation half-reaction2I-(aq) I2(s) + 2e-
Overall (cell) reaction2MnO4
-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)
Practice 2 Diagramming Voltaic Cells
PROBLEM: Diagram, show balanced equations, and write the 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 AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.
Why Does a Voltaic Cell Work?
The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit.
Ecell > 0 for a spontaneous reaction
1 Volt (V) = 1 Joule (J)/ Coulomb (C)
Voltages of Some Voltaic Cells
Voltaic Cell Voltage (V)
Common alkaline battery
Lead-acid car battery (6 cells = 12V)
Calculator battery (mercury)
Electric eel (~5000 cells in 6-ft eel = 750V)
Nerve of giant squid (across cell membrane)
2.0
1.5
1.3
0.15
0.070
Determining an unknown E0half-cell with the standard
reference (hydrogen) electrode
Oxidation half-reactionZn(s) Zn2+(aq) + 2e-
Reduction half-reaction2H3O+(aq) + 2e- H2(g) + 2H2O(l)Overall (cell) reaction
Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 3 Calculating an Unknown E0half-cell from E0
cell
PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V
Calculate E0bromine given E0
zinc = 0.76V
Selected Standard Electrode Potentials (298K)
Half-Reaction E0(V)
2H+(aq) + 2e- H2(g)
F2(g) + 2e- 2F-(aq)
Cl2(g) + 2e- 2Cl-(aq)
MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)
Ag+(aq) + e- Ag(s)
Fe3+(g) + e- Fe2+(aq)
O2(g) + 2H2O(l) + 4e- 4OH-(aq)
Cu2+(aq) + 2e- Cu(s)
N2(g) + 5H+(aq) + 4e- N2H5+(aq)
Fe2+(aq) + 2e- Fe(s)
2H2O(l) + 2e- H2(g) + 2OH-(aq)
Na+(aq) + e- Na(s)
Li+(aq) + e- Li(s)
+2.87
-3.05
+1.36
+1.23
+0.96
+0.80
+0.77
+0.40
+0.34
0.00
-0.23
-0.44
-0.83
-2.71
•By convention, electrode potentials are written as reductions.
•When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential.
•The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0
cell.
•When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents.
Example: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
stronger reducing agent
weaker oxidizing agent
stronger oxidizing agent
weaker reducing agent
Practice 4 Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
PROBLEM: (a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B, and C), and calculate E0
cell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
E0 = 0.96V(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)
E0 = -0.23V(2) N2(g) + 5H+(aq) + 4e- N2H5+(aq)
E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)
Free Energy and Electrical Work
G -Ecell
-Ecell =-wmax
charge
charge = n F
n = #mols e-
F = Faraday constant
F = 96,485 C/mol e-
1V = 1J/C
F = 9.65x104J/V*mol e-
G = wmax = charge x (-Ecell)
G = -n F Ecell
In the standard state -
G0 = -n F E0cell
G0 = - RT ln K
E0cell = - (RT/n F) ln K
G0
E0cell K
G0 KReaction at
standard-state conditions
E0cell
The interrelationship of G0, E0, and K
< 0 spontaneous
at equilibrium
nonspontaneous
0
> 0
> 0
0
< 0
> 1
1
< 1
G0 = -RT lnKG0 = -nFEo
cell
E0cell = -RT lnK
nF
By substituting standard state values into E0
cell, we get
E0cell = (0.0592V/n) log K (at 250C)
Practice 5 Calculating K and G0 from E0cell
PROBLEM: Lead can displace silver from solution:
As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and G0 at 250C for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
Practice 6 Using the Nernst Equation to Calculate Ecell
PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2
Calculate Ecell at 250C.
The Effect of Concentration on Cell Potential
G = G0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell - ln Q
RT
nF
•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell
Ecell = E0cell -
log Q0.0592
n
The relation between Ecell and log Q for the zinc-copper cell
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A concentration cell based on the Cu/Cu2+ half-reaction
Overall (cell) reactionCu2+(aq,1.0M) Cu2+(aq, 0.1M)
Oxidation half-reactionCu(s) Cu2+(aq, 0.1M) + 2e-
Reduction half-reactionCu2+(aq, 1.0M) + 2e- Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 7 Calculating the Potential of a Concentration Cell
PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3. What is the cell potential at 298K? Which electrode has a positive charge?
The corrosion of iron
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Enhanced corrosion at sea
The effect of metal-metal contact on the corrosion of iron
faster corrosion cathodic protection
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The use of sacrificial anodes to prevent iron corrosion
The tin-copper reaction as the basis of a voltaic and an electrolytic cell
Oxidation half-reactionSn(s) Sn2+(aq) + 2e-
Reduction half-reactionCu2+(aq) + 2e- Cu(s)
Oxidation half-reactionCu(s) Cu2+(aq) + 2e-
Reduction half-reactionSn2+(aq) + 2e- Sn(s)
Overall (cell) reactionSn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
Overall (cell) reactionSn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
voltaic cell electrolytic cell
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The processes occurring during the discharge and recharge of a lead-acid battery
VOLTAIC(discharge)
ELECTROLYTIC(recharge)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Comparison of Voltaic and Electrolytic Cells
Cell Type G Ecell
Electrode
Name Process Sign
Voltaic
Voltaic
Electrolytic
Electrolytic
< 0
< 0
> 0
> 0
> 0
> 0
< 0
< 0
Anode
Anode
Cathode
Cathode
Oxidation
Oxidation
Reduction
Reduction
-
-
+
+
Practice 8 Predicting the Electrolysis Products of a Molten Salt Mixture
PROBLEM: 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.
The electrolysis of water
Oxidation half-reaction2H2O(l) 4H+(aq) + O2(g) + 4e-
Reduction half-reaction2H2O(l) + 4e- 2H2(g) + 2OH-(aq)
Overall (cell) reaction2H2O(l) H2(g) + O2(g)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 9 Predicting the Electrolysis Products of Aqueous Ionic Solutions
PROBLEM: What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) AgNO3; (c) MgSO4?
A summary diagram for the stoichiometry of electrolysis
MASS (g)of substance oxidized or
reduced
MASS (g)of substance oxidized or
reduced
AMOUNT (MOL)of electrons transferred
AMOUNT (MOL)of electrons transferred
AMOUNT (MOL)of substance oxidized or
reduced
AMOUNT (MOL)of substance oxidized or
reduced
CHARGE (C)CHARGE (C)
CURRENT (A)CURRENT (A)
balanced half-reaction
Faraday constant (C/mol e-)
M(g/mol)
time(s)
Practice 10 Applying the Relationship Among Current, Time, and Amount of Substance
PROBLEM: A technician is plating a faucet with 0.86g of Cr from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed?
Top Related