1. Chapter 21 Electrochemistry: Fundamentals
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.

2. A summary of redox terminology. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
OXIDATION
One reactant loses electrons.
Zn loses electrons.
Reducing agent is oxidized.
Zn is the reducing agent and becomes oxidized.
Oxidation number increases.
The oxidation number of Zn increases from 0 to + 2.
REDUCTION
Other reactant gains electrons.
Hydrogen ion gains electrons.
Oxidizing agent is reduced.
Hydrogen ion is the oxidizing agent and becomes reduced.
Oxidation number decreases.
The oxidation number of H decreases from +1 to 0.

3. General characteristics of voltaic and electrolytic cells.
Electrochemical cell
General characteristics of voltaic and electrolytic cells.
VOLTAIC / GALVANIC CELL
ELECTROLYTIC CELL
System does work on its surroundings
Energy is released from spontaneous redox reaction
Energy is absorbed to drive a nonspontaneous redox reaction
Surroundings(power supply)
do work on system(cell)
Oxidation half-reaction
X X+ + e-
Oxidation half-reaction
A A + e-
Reduction half-reaction
Y++ e- Y
Reduction half-reaction
B++ e B
Overall (cell) reaction
X + Y X+ + Y; DG < 0
Overall (cell) reaction
A- + B A + B; DG > 0

4. A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s) Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

5. A voltaic cell using inactive electrodes.
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e-
Mn2+(aq) + 4H2O(l)
Oxidation half-reaction
2I-(aq) I2(s) + 2e-
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq) Mn2+(aq) + 5I2(s) + 8H2O(l)

6. 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

7. Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
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:
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). e-
SOLUTION:
Voltmeter
salt bridge
Oxidation half-reaction
Cr(s) Cr3+(aq) + 3e- Cr
Cr3+ Ag
Ag+ K+
NO3-
Reduction half-reaction
Ag+(aq) + e Ag(s)
Overall (cell) reaction
Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s)
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

8. Determining an unknown E0half-cell with the standard reference (hydrogen) electrode.
Oxidation half-reaction
Zn(s) Zn2+(aq) + 2e-
Overall (cell) reaction
Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
Reduction half-reaction
2H3O+(aq) + 2e H2(g) + 2H2O(l)

9. Calculating an Unknown E0half-cell from E0cell
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 E0zinc = -0.76V
PLAN:
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 = E0cathode - E0anode.
SOLUTION:
anode: Zn(s) Zn2+(aq) + 2e- E = +0.76
E0Zn as Zn2+(aq) + 2e Zn(s) is -0.76V
E0cell = E0cathode - E0anode = = E0bromine - (-0.76)
E0bromine = = 1.07 V

10. Selected Standard Electrode Potentials (298K)
Half-Reaction
E0(V)
F2(g) + 2e F-(aq)
+2.87
strength of oxidizing agent
strength of reducing agent
Cl2(g) + 2e Cl-(aq)
+1.36
MnO2(g) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
+1.23
NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
+0.96
Ag+(aq) + e Ag(s)
+0.80
Fe3+(g) + e Fe2+(aq)
+0.77
O2(g) + 2H2O(l) + 4e OH-(aq)
+0.40
Cu2+(aq) + 2e Cu(s)
+0.34
2H+(aq) + 2e H2(g)
0.00
N2(g) + 5H+(aq) + 4e N2H5+(aq)
-0.23
Fe2+(aq) + 2e Fe(s)
-0.44
2H2O(l) + 2e H2(g) + 2OH-(aq)
-0.83
Na+(aq) + e Na(s)
-2.71
Li+(aq) + e Li(s)
-3.05

11. Writing Spontaneous Redox Reactions
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 E0cell.
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
stronger oxidizing agent
weaker oxidizing agent
weaker reducing agent

12. 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 E0cell 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)
PLAN:
Put the equations together in varying combinations so as to produce (+) E0cell 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.

13. 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
(a)
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
E0 = +0.23V
Rev
E0cell = 1.19V
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l) X4
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- X3
4NO3-(aq) + 3N2H5+(aq) + H+(aq) NO(g) + 3N2(g) + 8H2O(l)
(A)
E0 = -0.96V
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-
Rev
E0 = 1.23V
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
E0cell = 0.27V
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- X2
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l) X3
2NO(g) + 3MnO2(s) + 4H+(aq) NO3-(aq) + 3Mn3+(aq) + 2H2O(l)
(B)

14. Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (3 of 4)
E0 = +0.23V
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
Rev
E0 = 1.23V
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
E0cell = 1.46V
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l) X2
N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l)
(C)
(b) Ranking oxidizing and reducing agents within each equation:
(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+

15. 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+

16. 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 (Eocell).
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.

17. cannot displace H from any source
Relative Reactivities (Activities) of Metals Li K Ba Ca Na
strength as reducing agents
1. Metals that can displace H from acid
can displace H
from water Mg Al Mn Zn Cr Fe Cd
2. Metals that cannot displace H from acid
can displace H
from steam
3. Metals that can displace H from water Co Ni Sn Pb
can displace H
from acid
4. Metals that can displace other metals from solution H2 Cu Hg Ag Au
cannot displace H from any source

18. Reaction at standard-state conditions
The interrelationship of DG0, E0, and K.
Reaction at standard-state conditions
DG0 K
E0cell
DG0
< 0
> 1
> 0
spontaneous 1
at equilibrium
> 0
< 1
< 0
nonspontaneous
DG0 = -nFEocell
DG0 = -RT lnK
By substituting standard state values into E0cell, we get
E0cell = (0.0592V/n) log K (at 298 K)
E0cell K
E0cell = -RT lnK nF

19. The Effect of Concentration on Cell Potential
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell -
ln Q RT nF
Nernst equation
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 Q
0.0592 n

20. Calculating K and DG0 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 DG0 at 298 K for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
PLAN:
Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide
SOLUTION:
Pb2+(aq) + 2e Pb(s)
E0 = -0.13V Anode
Ag+(aq) + e Ag(s)
E0 = 0.80V Cathode 2X
Ag+(aq) + e Ag(s)
E0cell = E0cathode – E0anode = 0.93V
E0cell =
log K
0.592V n
E0cell = - (RT/n F) ln K
DG0 = -nFE0cell
n x E0cell
0.592V
(2)(0.93V)
0.592V =
= -(2)(96.5kJ/mol*V)(0.93V)
log K =
DG0 = -1.8x102kJ
K = 2.6x1031

21. 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 = 0.30atm H2
Calculate Ecell at 298 K.
PLAN:
Find E0cell and Q in order to use the Nernst equation.
SOLUTION:
Determining E0cell :
Q =
P x [Zn2+] H2
[H+]2
E0 = 0.00V
2H+(aq) + 2e H2(g)
E0 = -0.76V
Zn2+(aq) + 2e Zn(s)
Q =
(0.30)(0.010)
(2.5)2
Zn(s) Zn2+(aq) + 2e-
E0 = +0.76V
Ecell = E0cell -
0.0592V n
log Q
Q = 4.8x10-4
Ecell = (0.0592/2)log(4.8x10-4) =
0.86V

22. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
Diagramming Voltaic Cells
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.
PROBLEM:
Identify the redox reactions
Write each half-reaction.
Associate the (-)(Zn) pole with the anode (oxidation) and the (+) (Ag) pole with the cathode (reduction).
PLAN:
SOLUTION: e-
Voltmeter
Oxidation half-reaction
Zn2+(aq) + 2e Zn(s)
salt bridge Ag
Ag+ Zn K+
Reduction half-reaction
Ag+(aq) + e Ag(s)
NO3-
Zn2+
Anode
Cathode
Overall (cell) reaction
Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)

23. Free Energy and Electrical Work
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.
DG a -Ecell
-Ecell =
-wmax
charge
DG = wmax = charge x (-Ecell)
DG = - n F Ecell
charge = n F
n = # mols e-
F = Faraday constant
In the standard state
All components are at standard state.
DG0 = - n F E0cell
F = 96,485 C/mol
DG0 = - RT ln K
1V = 1J/C
E0cell = - (RT/n F) ln K
F = 9.65x104J/V*mol
E0cell = - ( /n) log K at RT

24. The Effect of Concentration on Cell Potential
Cell operates with all components at standard states. Most cells are starting at Non-standard state.
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell -
ln Q RT nF
Nernst equation
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 Equilibrium
When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reverse reaction
Ecell = E0cell -
log Q
0.0592 n

25. RT nF Ecell = E0cell - ln Q 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:
2H+(aq) + Zn (s) H2(g) + Zn2+ (aq)
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm H2
Calculate Ecell at 298 K.
PLAN:
Find E0cell and Q in order to use the Nernst equation. RT nF
Ecell = E0cell -
log Q
0.0592 n
Ecell = E0cell -
ln Q
SOLUTION:
Determining E0cell :
2H+(aq) + 2e H2(g)
E0 = 0.00V cathode
Q =
P x [Zn2+] H2
[H+]2
Zn2+(aq) + 2e Zn(s)
E0 = -0.76V anode
Ecell0 = E0c-E0a = 0.00-(-0.76)V = 0.76 V
Q =
(0.30)(0.010)
(2.5)2
Ecell = E0cell -
0.0592 n
log Q
Q = 4.8x10-4
Ecell = (0.0592/2)log(4.8x10-4) =
0.86V

26. 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 (Eocell).
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.