chapter 21 electrochemistry: fundamentals key points about redox reactions 1.oxidation (electron...
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Chapter 21 Electrochemistry: Fundamentals
Key Points About Redox Reactions
1. Oxidation (electron loss) always accompanies reduction (electron gain).
2. The oxidizing agent is reduced, and the reducing agent is oxidized.
3. 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 0 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.
Energy is absorbed to drive a nonspontaneous redox reaction
General characteristics of voltaic and electrolytic cells.
VOLTAIC / GALVANIC 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
Electrochemical cell
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)
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)
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
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.
PLAN:
SOLUTION:
Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).
Voltmeter
Oxidation half-reactionCr(s) Cr3+(aq) + 3e-
Reduction half-reactionAg+(aq) + e- Ag(s)
Overall (cell) reactionCr(s) + Ag+(aq) Cr3+(aq) + Ag(s)
Cr
Cr3+
Ag
Ag+
K+
NO3-
salt bridge
e-
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
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) reactionZn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
Calculating an Unknown E0half-cell from E0
cell
PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal:
PLAN:
SOLUTION:
Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V
Calculate E0bromine given E0
zinc = -0.76V
The reaction is spontaneous as written since the E0cell is (+). Zinc is
being oxidized and is the anode. Therefore the E0bromine can be found
using E0cell = E0
cathode - E0anode.
anode: Zn(s) Zn2+(aq) + 2e- E = +0.76
E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V
E0cell = E0
cathode - E0anode = 1.83 = E0
bromine - (-0.76)
E0bromine = 1.86 - 0.76 = 1.07 V
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
strength
of reducin
g agent
stre
ngt
h o
f ox
idiz
ing
agen
t
• 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
Writing Spontaneous Redox Reactions
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0
cell for each.
PLAN:
(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)
Put the equations together in varying combinations so as to produce (+) E0
cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0.
In ranking the strengths, compare the combinations in terms of E0cell.
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
continued (2 of 4)
SOLUTION: (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = 0.96V
(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
X4
X3
E0cell = 1.19V
(a)
E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23VRev
E0 = -0.96V(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-Rev
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)
X2
X3
E0cell = 0.27V
4NO3-(aq) + 3N2H5
+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l)(A)
2NO(g) + 3MnO2(s) + 4H+(aq) 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l)(B)
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
continued (3 of 4)
E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)
E0 = +0.23V(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-Rev
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) X2
E0cell = 1.46V
(b) Ranking oxidizing and reducing agents within each equation:
N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l)(C)
(A): oxidizing agents: NO3- > N2 reducing agents: N2H5
+ > NO
(B): oxidizing agents: MnO2 > NO3- reducing agents: NO > Mn2+
(C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
continued (4 of 4)
A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of
Oxidizing agents: MnO2 > NO3- > N2
Reducing agents: N2H5+ > NO > Mn2+
Summary
• A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge.
• The salt bridge provides ions to maintain the charge balance when the cell operates.
• Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell.
• The output of a cell is called cell potential (Ecell) and is measured in volts.
• When all substances are in standard states, the cell potential is the standard cell potential (Eo
cell).
• Ecell equals Ecathode minus Eanode, Ecell = Ecathode - Eanode.
• Conventionally, the half cell potential refers to its reduction half-reaction.
• Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking
the oxidizing agent or reducing agent.
• Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones.
• Spontaneous reaction is indicated negative ∆G and positive ∆E,
∆G = - nF∆E.
• We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.
Relative Reactivities (Activities) of Metals
1. Metals that can displace H from acid
2. Metals that cannot displace H from acid
3. Metals that can displace H from water
4. Metals that can displace other metals from solution
can displace Hfrom water
LiKBaCaNa
stre
ngt
h as
red
ucin
g ag
ents
can displace Hfrom steam
MgAlMnZnCrFeCd
H2
cannot displace H from any source
CuHg AgAu
can displace Hfrom acid
CoNiSnPb
G0
E0cell K
G0 K
Reaction at standard-state
conditionsE0cell
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 298 K)
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
Nernst equation
Calculating K and G0 from E0cell
PLAN:
SOLUTION:
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 298 K for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0
cell. Substitute into the equations found on slide
E0 = -0.13V Anode
E0 = 0.80V Cathode
E0cell = E0
cathode – E0anode = 0.93V
Ag+(aq) + e- Ag(s)
Pb2+(aq) + 2e- Pb(s)
Ag+(aq) + e- Ag(s)
E0cell = log K
0.592V
n
log K =
K = 2.6x1031
n x E0cell
0.592V(2)(0.93V)
0.592V=
G0 = -nFE0cell
= -(2)(96.5kJ/mol*V)(0.93V)
G0 = -1.8x102kJ
2X
E0cell = - (RT/n F) ln K
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:
PLAN:
SOLUTION:
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2
Calculate Ecell at 298 K.
Find E0cell and Q in order to use the Nernst equation.
Determining E0cell :
E0 = 0.00V2H+(aq) + 2e- H2(g)
E0 = -0.76VZn2+(aq) + 2e- Zn(s)
Zn(s) Zn2+(aq) + 2e- E0 = +0.76V
Q = P x [Zn2+]
H2
[H+]2
Q = 4.8x10-4
Q = (0.30)(0.010)
(2.5)2
Ecell = E0cell -
0.0592V
nlog Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Diagramming Voltaic Cells
PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Zn bar in a Zn(NO3)2 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Zn electrode is negative relative to the Ag electrode.
PLAN:
SOLUTION:
Identify the redox reactions
Write each half-reaction.
Associate the (-)(Zn) pole with the anode (oxidation) and the (+) (Ag) pole with the cathode (reduction).
Voltmeter
Oxidation half-reactionZn2+(aq) + 2e- Zn(s)
Reduction half-reactionAg+(aq) + e- Ag(s)
Overall (cell) reactionZn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Zn
Zn2+
Ag
Ag+
K+
NO3-
salt bridge
e-
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
CathodeAnode
Free Energy and Electrical Work
G -Ecell
-Ecell =-wmax
charge
charge = n F
n = # mols e-
F = Faraday constant
F = 96,485 C/mol
1V = 1J/C
F = 9.65x104J/V*mol
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
If there is no current flows,
the potential represents the
maximum work the cell can do.
If there is no current flows, no energy is lost
to heat the cell component.
All components are at standard state.
E0cell = - (0.05916/n) log K at RT
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 forward reaction
• When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reverse reaction
• When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell Equilibrium
Ecell = E0cell -
log Q0.0592
n
Nernst equation
Cell operates with all components at standard states. Most cells are starting at Non-standard state.
Sample Problem 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:
PLAN:
SOLUTION:
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2
Calculate Ecell at 298 K.
Find E0cell and Q in order to use the Nernst equation.
Determining E0cell :
E0 = 0.00V cathode2H+(aq) + 2e- H2(g)
E0 = -0.76V anodeZn2+(aq) + 2e- Zn(s)Q =
P x [Zn2+]H2
[H+]2
Q = 4.8x10-4
Q = (0.30)(0.010)
(2.5)2
Ecell = E0cell -
0.0592
nlog Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Ecell = E0cell - ln Q
RT
nF
Ecell0 = E0
c-E0a = 0.00-(-0.76)V = 0.76 V
2H+(aq) + Zn (s) H2(g) + Zn2+ (aq)
Ecell = E0cell -
log Q0.0592
n
Summary
• A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge.
• The salt bridge provides ions to maintain the charge balance when the cell operates.
• Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell.
• The output of a cell is called cell potential (Ecell) and is measured in volts.
• When all substances are in standard states, the cell potential is the standard cell potential (Eo
cell).
• Ecell equals Ecathode minus Eanode, Ecell = Ecathode - Eanode.
• Conventionally, the half cell potential refers to its reduction half-reaction.
• Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking
the oxidizing agent or reducing agent.
• Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones.
• Spontaneous reaction is indicated negative ∆G and positive ∆E,
∆G = -nF∆E.
• We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.