Electrochemistry
Oxidation-Reduction
 REDOX reactions are at the heart of all batteries.
 Take a simple single replacement reaction like:
 Zn(s) + CuCl2(aq)  ZnCl2(aq) + Cu(s)
 Reducing this to a net ionic equation and then to half-
reactions reveals the transfer of the electrons.
 Oxidation = loss of electron(s).
 Reduction = gain of electron(s).
Oxidation-Reduction
 In Chapter 4, the
activity series was
used to predict the
outcome of a single
replacement
reaction.
 LEP #1
Oxidation-Reduction
 The substance being oxidized can also be referred to as
the reducing agent and the substance being reduced as
the oxidizing agent.
 2 Zn(s) + O2(g)  2 ZnO(s)
 Co(s) + NiCl2(aq)  CoCl2(aq) + Ni(s)
 LEP #2
Balancing REDOX Reactions
 Reactions are split into half-reactions –
 Oxidation half-reaction
 Reduction half-reaction
 These are then balanced separately.
 Electrons are added to show the net change in oxidation
number of the REDOX half-reactions.
 Each half reaction is then multiplied by the LCM to cancel
out the electrons.
 LEP #3
Voltaic Cells
 In a spontaneous
reaction, electrons are
transferred from the
more active metal to the
less active one.
 If we can put these in
separate compartments,
then the electron transfer
can be harnessed.
Voltaic Cells
 A voltaic cell is one that
produces a positive
voltage.
 A complete pathway
must be present for the
electrons to travel.
Voltaic Cells
 A standard cell looks
like this.
 Oxidation takes place
at the anode.
 Reduction takes place
at the cathode.
 The flow of electrons
is from the anode to
the cathode.
Voltaic Cells
 In a
molecular
view of the
reaction…
Electromotive force (emf)
 Water flows
spontaneously from
higher to lower levels.
 Electrons do the same
– they flow from
higher electron
pressure to lower
electron pressure.
Emf
 This difference in pressure is called the emf – aka the voltage. Note: 1 V
=1J/1C
 Emf’s are relative to a position, so you must have a point of reference.
 S.H.E. = standard hydrogen electrode
 Thus, 2H+(aq) + 2eH2(g) ; Eocell = 0.00V
Emf
 Cell voltages are for 1 M solutions, 1 atm, 25oC
 For a Zn / Cu cell (LEP#4):
 Zn+2(aq) + 2e-  Zn(s) ; Eo = -0.76V
 Cu+2(aq) + 2e-  Cu(s) ; Eo = +0.34V
 Because Zn metal is oxidized, that equation is reversed
and its sign is changed.
 Or: Eocell = Ecathode - Eanode
 LEP #5
Oxidizing / Reducing Agents
 If placed in order of
most positive to most
negative voltages
(handout), then top
left has strongest
oxidizing agent and
bottom right has
strongest reducing
agent.
Free Energy and REDOX
 Can use reduction potentials to predict whether a
REDOX reaction is spontaneous.
 Eo = +, then spontaneous.
 Eo = -, then non-spontaneous.
 I2(s) + 5 Cu+2(aq) + 6 H2O(l)  2 IO3-(aq) + 5 Cu(s) + 12 H+(aq)
 H2SO3(aq) + 2 Mn(s) + 4 H+(aq)  S(s) + 2 Mn+2(aq) + 3 H2O(l)
Free Energy
 The relationship between Free Energy (DG) and cell emf is:
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DGo = -nFE0 ; where n is the net number of electrons
transferred (balanced reaction) and F is the Faraday
constant 96,485 C/mol.
From Chapter 19, we also know that: DGo = -RT lnK.
Thus, -nFEo = -RT lnK.
And, ln K = nFEo / RT.
The constants can be combined and the original equations
used log instead of the natural log.
Log K = nEo / (0.0592).
LEP #6
Nernst Equation
 What if concentrations are not 1 M?
 What happens to a battery in use over time?
 Recall from Chapter 19:
 DG = DGo + RT lnQ
 Since DG = -nFE, we can replace this for both DG and
DGo.
Nernst Equation
 Suppose [Cu+2] = 0.10M and [Zn+2] = 0.10M, will we
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need to use the Nernst Equation?
Net: Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)
What if the reaction replaced the Cu with Ag?
Net: Zn(s) + 2 Ag+1(aq)  Zn+2(aq) + 2 Ag(s)
LEP #7 and #8
Concentration Cells
 All voltaic cells so far involved two different metals.
 Can we produce a voltage when the same metal is used
for both half-cells?
 Yes!
Concentration Cells
 Since both half-reactions are the same, but opposite
reactions, then Eo = 0.00 V (always!).
 Ni+2(aq) + 2e-  Ni(s) ; Eored = -0.28V
 Ni(s)  Ni+2(aq) + 2e- ; Eoox = +0.28V
 LEP #9
Importance of Concentration Cells
 Pacemaker cells in the heart
are a concentration cell.
 Potassium ion concentrations
inside and outside of the cell
are different – ICF = 135 mM
and ECF = 4 mM.
 Generates an electrical pulse of
about 94 mV.
 This can be measured with an
EKG.
Electrolysis
 Can use electrical energy to force a non-spontaneous
reaction to occur.
 This is called and electrolytic cell.
 Two types of electrolysis:
 Aqueous – the ions are present in a solution of water and
requires only the energy for the electolysis.
 Molten Salt – the ionic compound is heated until it
liquefies. This requires energy (lots) before the
electrolysis can take place.
Electrolysis
 Suppose we have a solution of NaCl(aq) and try to
electrolyze this solution.
 At the cathode, reduction of the metal is possible:
 Na+(aq) + 1e-  Na(s) ; Eo = -2.71 V
 However, there is a second competing reaction:
 2 H2O(l) + 2e-  H2(g) + 2 OH-(aq) ; Eo = -0.83 V
 Only the easier one will occur!!!
Electrolysis
 Likewise, at the anode where oxidation occurs, the
oxidation of the non-metal is possible:
 2 Cl-(aq)  Cl2(g) + 2e- ; Eo = -1.36 V
 Once again, there is a competing reaction:
 2 H2O(l)  O2(g) + 4 H+(aq) + 4e- ; Eo = -1.23 V
 As before, only the easier one will occur!!!
 Therefore, electrolysis on NaCl(aq) will yield:
Electrolysis
 The ONLY way to produce sodium metal is with
molten salt electrolysis.
Predicting the Products
 We will use our electrode potentials to predict the
products of an aqueous electrolysis.
 Key reactions to highlight are:
 Cathode:
 2 H2O(l) + 2e-  H2(g) + 2 OH-(aq) ; Eo = -0.83 V
 Anode:
 2 H2O(l)  O2(g) + 4 H+(aq) + 4e- ; Eo = -1.23 V
 LEP #11 and #12
Applications of Electrolysis
 Besides producing certain metals there are other uses
of electrolysis like electroplating and purifying certain
metals.
Quantitative Electrolysis
 Current is measured in Amperes or amps for short.
 1 Amp = 1 C / 1 s
 wmax = -nFE
 Problems are set up using standard dimensional
analysis.
 Will require a conversion between moles of electrons
and moles of metal reduced.
 LEP #13 and #14