Chapter 21 Electrochemistry: Fundamentals

1
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.

2
A summary of redox terminology.
OXIDATION
Zn loses electrons.
One reactant loses electrons.
Zn is the reducing agent and becomes oxidized.
Reducing agent is oxidized.
The oxidation number of Zn increases from 0 to
2.
Oxidation number increases.
REDUCTION
Hydrogen ion gains electrons.
Other reactant gains electrons.
Hydrogen ion is the oxidizing agent and becomes
reduced.
Oxidizing agent is reduced.
The oxidation number of H decreases from 1 to 0.
Oxidation number decreases.
3
Electrochemical cell
General characteristics of voltaic and
electrolytic cells.
VOLTAIC / GALVANIC CELL
ELECTROLYTIC CELL
Energy is absorbed to drive a nonspontaneous
redox reaction
Energy is released from spontaneous redox reaction
System does work on its surroundings
Surroundings(power supply) do work on system(cell)
4
A voltaic cell based on the zinc-copper reaction.
5
A voltaic cell using inactive electrodes.
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 lower oxidation state
phase of higher oxidation state
phase of higher oxidation state
phase boundary between half-cells
Examples
Zn(s) Zn2(aq) Cu2(aq) Cu (s)
graphite I-(aq) I2(s) H(aq), MnO4-(aq) ,
Mn2(aq) graphite
7
Diagramming Voltaic Cells
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).
SOLUTION
Voltmeter
salt bridge
Cr(s) Cr3(aq) Ag(aq) Ag(s)
8
Determining an unknown E0half-cell with the
standard reference (hydrogen) electrode.
9
Calculating an Unknown E0half-cell from E0cell
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
E0cell E0cathode - E0anode 1.83 E0bromine
- (-0.76)
E0bromine 1.86 - 0.76 1.07 V
10
Selected Standard Electrode Potentials (298K)
Half-Reaction
E0(V)
2.87
1.36
1.23
0.96
0.80
0.77
0.40
0.34
0.00
2H(aq) 2e- H2(g)
-0.23
-0.44
-0.83
-2.71
-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.

stronger reducing agent
weaker oxidizing agent
stronger oxidizing agent
weaker reducing agent
12
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
(b) Rank the relative strengths of the oxidizing
and reducing agents
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
(a)
E0cell 1.19V
X4
X3
E0cell 0.27V
X3
14
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (3 of 4)
E0cell 1.46V
X2
(b) Ranking oxidizing and reducing agents within
each equation
(A) oxidizing agents NO3- gt N2
reducing agents N2H5 gt NO
(B) oxidizing agents MnO2 gt NO3-
reducing agents NO gt Mn2
(C) oxidizing agents MnO2 gt N2
reducing agents N2H5 gt 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 gt NO3- gt N2
Reducing agents N2H5 gt NO gt 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
Relative Reactivities (Activities) of Metals
Li K Ba Ca Na
can displace H from water
1. Metals that can displace H from acid
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
The interrelationship of DG0, E0, and K.
Reaction at standard-state conditions
DG0
K
E0cell
DG0
lt 0
spontaneous
gt 0
gt 1
at equilibrium
0
0
1
nonspontaneous
gt 0
lt 0
lt 1
DG0 -nFEocell
DG0 -RT lnK
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
Nernst equation

• When Q lt 1 and thus reactant gt product, lnQ lt
  0, so Ecell gt E0cell

• When Q 1 and thus reactant product, lnQ
  0, so Ecell E0cell

• When Q gt1 and thus reactant lt product, lnQ gt
  0, so Ecell lt E0cell

20
Calculating K and DG0 from E0cell
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
E0 -0.13V Anode
E0 0.80V Cathode
E0cell E0cathode E0anode 0.93V
2X
E0cell - (RT/n F) ln K
DG0 -nFE0cell
-(2)(96.5kJ/molV)(0.93V)
log K
DG0 -1.8x102kJ
K 2.6x1031
21
Using the Nernst Equation to Calculate Ecell
PLAN
Find E0cell and Q in order to use the Nernst
equation.
SOLUTION
Determining E0cell
Q 4.8x10-4
Ecell 0.76 - (0.0592/2)log(4.8x10-4)
0.86V
22
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
Zn
Zn2
Cathode
Anode
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
DG wmax charge x (-Ecell)
DG - n F Ecell
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/Vmol
E0cell - (0.05916/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
Nernst equation

• When Q lt 1 and thus reactant gt product, lnQ
  lt 0, so Ecell gt E0cell forward reaction

• When Q 1 and thus reactant product, lnQ
  0, so Ecell E0cell Equilibrium

• When Q gt1 and thus reactant lt product, lnQ
  gt 0, so Ecell lt E0cell Reverse reaction

25
Sample Problem
Using the Nernst Equation to Calculate Ecell
2H(aq) Zn (s) H2(g) Zn2 (aq)
Calculate Ecell at 298 K.
PLAN
Find E0cell and Q in order to use the Nernst
equation.
Ecell E0cell -
ln Q
SOLUTION
Determining E0cell
E0 0.00V cathode
E0 -0.76V anode
Ecell0 E0c-E0a 0.00-(-0.76)V 0.76 V
Q 4.8x10-4
Ecell 0.76 - (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.