Electrochemistry
Chapter 20
Brown-LeMay
Review of Redox Reactions
• Oxidation - refers to the loss of electrons by
a molecule, atom or ion - LEO goes
• Reduction - refers to the gain of electrons
by an molecule, atom or ion – GER
• Chemical reactions in which the oxidation
state of one or more substances changes
are called oxidation-reduction reactions
(or redox reactions)
Zn(s)+
+
2H
(aq)
2+
Zn
(aq)+
H2(g)
• Zn = 0, Zn2+ = +2 (LEO) reducing agent
H+ = +1, H2 = 0 (GER) oxidizing agent
• Thus, the oxidation number of both the
Zn(s) and H+(aq) change during the course
of the reaction, and so, this must be a redox
reaction
• Review balancing redox-equations
Balancing Redox by Half Reactions
• Half reactions are a convenient way of
separating oxidation and reduction
reactions.
• Balance the titration of acidic solution of
Na2C2O4 (colorless) with KMnO4(deep purple)
• 1st Write the incomplete ½ reactions
• MnO4(aq)  Mn2+(aq) is reduced (pale pink)
• C2O4(aq)  CO2(g) is oxidized
• MnO4-(aq)  Mn2+(aq) + 4H2O
C2O42-(aq)  2CO2(g)
• Then bal H by adding H+
8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
C2O42-(aq)  2CO2(g)
• finish by balancing the e’s
• For the permanganate 7+ left and 2+ on the right
5e- + 8H+ + MnO4(aq)  Mn2+(aq) + 4H2O
• On the oxalate 2- on the right and o on the left
C2O42-(aq)  2CO2(g) + 2e-
• 2(5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O)
5(C2O42-(aq)  2CO2(g) + 2e-)
10e- + 16H+ + MnO4- (aq)  2Mn2+(aq) + 8H2O
5C2O42-(aq)  10CO2(g) +10e16H++2 MnO4- +5C2O42-  2Mn2+(aq)+8H2O
+ 10CO2
Balancing Eq in Basic Solution
• The same method is used but OH- is added
to neutralize the H+ used.
• The equation must again be simplified by
canceling the terms on both sides of the
equation.
Voltaic Cells
• Spontaneous redox
reactions may be use
to perform electrical
work
• Voltaic or galvanic
cells are devices that
electron transfer
occurs in an external
circuit.
Zn(s) +
2+
Cu
(aq)

2+
Zn
(aq)
+ Cu(s)
• Zn0 is spontaneously oxidized to Zn2+
• Cu2+ is spontaneously reduced to Cu0
• Oxidization half reaction at the anode
Zn(s)  Zn2+(aq) + 2e• Reduction half reaction at the cathode
Cu2+(aq) + 2e-  Cu(s)
• As oxidization occurs, Zn is converted to
Zn2+ and 2e-.
• The electrons flow toward the cathode,
where they are used in the reduction
reaction
• We expect the Zn electrode to lose mass
• Electrons flow from the anode to cathode so
the anode is negative and the cathode is
positive
• Electrons cannot flow through the solution;
they have to be transported though an
external wire.
• Anions and cations move through a porous
barrier or salt bridge
• Cations move into the cathodic
compartment to neutralize the excess
negatively charged ions (cathode: Cu2+ + 2e Cu, so the counter ion of Cu is in excess)
• A build up of excess charge is avoided by
movement of cations and anions through the
salt bridge
• The anions move into the anodic
compartment to neutralize the excess Zn2+
ions formed by the oxidation.
• Molecular View
- “Rules” of voltaic cells
* at the anode electrons are products
oxidization occurs
* at the cathode electrons are reactants
reduction occurs
Cell EMF “the driving force”
• Reactions are spontaneous because the
cathode has a lower electrical potential
energy than the anode
• Potential difference: difference in
electrical potential measured in volts
• One volt is the potential difference required
to impart one joule (J) of energy to a charge
of one coulomb (C)
• 1V = 1 J/C
Cell EMF “the driving force”
• Electromotive force (emf) is the force
required to push electrons though the
external circuit
• Cell potential: Ecell is the emf of a cell
• Ecell > 0 for a spontaneous reaction
• For 1 molar, 1 atm for gases at
250C(standard conditions), the standard emf
(standard cell potential) = E0cell
Standard Reduction Potentials
& Cal. Cell Potentials
• Standard Reduction Potentials E0red are
measured relative to a standard
• The emf of a cell is
E0cell = E0red(cathode) – E0red(anode)
• The standard hydrogen electrode is used as
the standard (standard hydrogen electrode)
(SHE)
• 2H+(aq,1M)+2e-  H2(g,1atm) E0cell = 0 V
The standard hydrogen electrode
• The (SHE) is assigned a potential of zero
• Consider the following half reaction
Zn(s)  Zn2+(aq) + 2e• We can measure E0cell relative to the SHE
• In the cell the SHE is the cathode
• It cons of a Pt electrode in a tube – 1M H+sol
• H2 is bubbled through the tube
• E0cell = E0red(cathode)-E0red(anode)
• 0.76V = 0V-E0red(anode)
• Therefore E0red(anode) = -0.76V
The standard hydrogen electrode
• Standard electrode potentials are written as
reduction reactions
• Zn2+(aq) (aq,1 M)+ 2e-  Zn(s) E0 = -0.76V
• Since the reduction potential is negative in
the presence of the SHE the reduction of
Zn2+ is non-spontaneous
• However the oxidization of Zn2+ is
spontaneous with the SHE
Standard reduction potential
• The standard reduction potential is an
intensive property
• Therefore, changing the stoichiometric
coefficient does not affect E0red
• 2Zn2+(aq) + 4e-  2Zn(s) E0red= -0.76 V
• E0red > 0 are
spontaneous relative to
the SHE
• E0red < 0 are nonspontaneous relative to
the SHE
• The larger the
difference between
E0red values the larger
the E0cell
• The more positive the
E0cell value the greater
the driving force for
reduction
Oxidizing and Reducing Agents
• Consider the table of standard reduction
potentials
• We use the table to determine the relative
strength of reducing and oxidizing agents
• The more positive the E0red the stronger the
oxidizing agent (written as the reactant)
• The more negative the E0red the stronger the
reducing agent (written as the product)
Oxidizing and Reducing Agents
• We can use tables to predict if one reactant
can spontaneously oxidize or reduce another
• Example F2 can oxidize H2 or Li
Ni2+ can oxidize Al(s)
Li can reduce F2
Spontaneity of Redox Reactions
E0cell = E0red(red process) – E0red(oxid process)
• Consider the reaction
• Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)
• The standard cell potential is
• E0cell = E0red (Ag+/Ag) – E0red(Ni2+/Ni)
• E0cell = (0.80 V) - (-0.28)
• E0cell = 1.08 V the value indicates the
reaction is spontaneous
EMF and free energy change
• Delta G = -nFE
• where delta G is the change in free energy
• n = the number of moles of electrons
transferred
• F = Faraday’s constant
• E = emf of the cell
EMF and free energy change
• F = 96,500 C/mole- = 96,500 J/(V)(mole-)
• Since n and F are positive, if Delta G < 0, then E >
0 and the reaction will be spontaneous.
• Effect of concentration on cell EMF
the cell is function until E=0 at which point
equilibrium has been reached and the cell is
“dead”
The point at which E=0 is determined by the
concentrations of the species involved in the redox
reaction
Walter Nernst (Nobel Prize 1920)
• Nernst Equation
• G = G0 + RTlnQ
• -nFE = -nFE0 + RTlnQ
• Solve the equation for
E give the Nernst Eq
• E = E0 - RT/nF lnQ
• Or for base 10 log
E = E0- 2.3RT/nF logQ
• The nernst eq at 298K =
E = E0 – 0.0592/n log Q
• Consider if you may
• Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
• If [Cu2+] = 5.0M and [Zn2+] = 0.05 M
• Ecell= 1.10 V – 0.0592/2 log 0.05/5 =1.16 V
• Cell emf and chemical equilibrium
• Log K = nE0/0.0592
thus if we know cell emf, we can calc the
equilibrium constant