Download E - Humble ISD

Electrochemistry
Applications of Redox
Review
 Oxidation reduction reactions involve a
transfer of electrons.
 OIL- RIG
 Oxidation Involves Loss
 Reduction Involves Gain
 LEO-GER
 Lose Electrons Oxidation
 Gain Electrons Reduction
Applications
 Moving electrons is electric current.
 8H++MnO4-+ 5Fe+2 +5e



 Mn+2 + 5Fe+3 +4H2O
Helps to break the reactions into half reactions.
8H++MnO4-+5e-  Mn+2 +4H2O
5(Fe+2  Fe+3 + e- )
In the same mixture it happens without doing
useful work, but if separate
 Connected this way the reaction starts
 Stops immediately because charge builds up.
H+
MnO4-
Fe+2
Galvanic Cell
Salt
Bridge
allows
current
to flow
H+
MnO4-
Fe+2
 Electricity travels in a complete circuit
 Instead of a salt bridge
H+
MnO4-
Fe+2
Porous
Disk
H+
MnO4-
Fe+2
e-
e-
e-
e-
Anode
e-
Reducing
Agent
Cathode
e-
Oxidizing
Agent
Cell Potential
 Oxidizing agent pushes the electron.
 Reducing agent pulls the electron.
 The push or pull (“driving force”) is called
the cell potential Ecell
 Also called the electromotive force (emf)
 Unit is the volt(V)
 = 1 joule of work/coulomb of charge
 Measured with a voltmeter
0.76
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
Standard Hydrogen Electrode
 This is the
reference all other
oxidations are
compared to
Eº = 0
H+
Cl º indicates standard
states of 25ºC,
1 atm, 1 M 1 M HCl
solutions.
H2 in
Cell Potential
 Zn(s) + Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.
 Eº cell = EºZn Zn+2 + Eº Cu+2 Cu
 We can look up reduction potentials in a
table.
 One of the reactions must be reversed,
so change it sign.
Cell Potential
 Determine the cell potential for a
galvanic cell based on the redox
reaction.
 Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)
 Fe+3(aq) + e- Fe+2(aq)
Eº = 0.77 V
 Cu+2(aq)+2e- Cu(s)
Eº = 0.34 V
 Cu(s) Cu+2(aq)+2eEº = -0.34 V
 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V
 0.77 V + - 0.34V = 0.43V
Homework
 17.27 and 17.28
Line Notation

solidAqueousAqueoussolid
 Anode on the leftCathode on the right
 Single line different phases.
 Double line porous disk or salt bridge.
 If all the substances on one side are
aqueous, a platinum electrode is
indicated.
 For the last reaction
 Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)
Galvanic Cell


1)
2)
3)
4)
The reaction always runs
spontaneously in the direction that
produced a positive cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the componentselectrodes and ions
Practice
 Completely describe the galvanic cell based on





the following half-reactions under standard
conditions.
MnO4- + 8 H+ +5e-  Mn+2 + 4H2O
Eº=1.51
Fe+2 +2e-  Fe(s)
Eº=0.44V
Reverse the smaller valued reaction
Fe(s) Fe+2 +2eEº= - 0.44V
Solution
 Balance the cell reaction:
 Balance the electrons to balance the
equation:
The First reaction requires 5 e-, the second
one supplies 2 e-, so 10 must be
exchanged.
 2(MnO4- + 8 H+ +5e-  Mn+2 + 4H2O)
 5(Fe(s) Fe+2 +2e-)
 Add the reactions together

Solution Continued
 2MnO4-(aq) + 16 H+ (aq)+5Fe (s) 
5 Fe +2 (aq) + 2 Mn+2 (aq) + 8H2O (l)
 Determine the Cell Potential:

2(1.51) + 5(-0.44) = 3.02 – 2.2 = 0.8
 Fe(s) l Fe+2(aq) ll MnO4-(aq), Mn+2(aq) l Pt(s)
More Homework
 17.29 and 17.30
Potential, Work and DG
 emf = potential (V) = work (J) / Charge(C)
 E = work done by system / charge
 E = -w/q
 Charge is measured in coulombs.
 -w = qE
 Faraday = 96,485 C/mol e-
 q = nF = moles of e- x charge/mole e w = -qE = -nFE = DG
Potential, Work and DG
 DGº = -nFE º
 if E º < 0, then DGº > 0 nonspontaneous
 if E º > 0, then DGº < 0 spontaneous
 In fact, reverse is spontaneous.
 Calculate DGº for the following reaction:
 Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)
 Fe+2(aq) + 2e-Fe(s)
 Cu+2(aq)+2e- Cu(s)
Eº = -0.44 V
Eº = 0.34 V
Practice
 Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)
 Fe(s) Fe+2(aq) + 2e-
 Cu+2(aq)+2e- Cu(s)
Eº = 0.44 V
Eº = 0.34 V
 0.44 V + 0.34 V = 0.78V
 DGº = -nFE º
 DGº = -(2 mol e-)96,485 C/mol e-(0.78J/C)
 DGº = -1.5 x105 J
Practice Predicting Spontaneity
 Will 1M HNO3 dissolve gold metal to form a 1M




Au3+ solution?
HNO3 half reaction:
NO3- + 4H+ + 3e-  NO + 2H2O E=0.96V
Au3+ Half Reaction:
Au  Au3+ + 3 e- E = -1.50V
 Au(s) + NO3-(aq) + 4H+(aq)  Au3+(s) + NO(g) + 2H2O(l)
 Ecell = 0.96V – 1.50V = -0.54V
 Since Ecell is negative, it will NOT occur under
standard conditions.
Homework
 17.37, 17.39, 17.49
Cell Potential and
Concentration
 Qualitatively - Can predict direction of change in
E from LeChâtelier.
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
 Predict if Ecell will be greater or less than Eºcell if
[Al+3] = 1.5 M and [Mn+2] = 1.0 M
LESS
 if [Al+3] = 1.0 M and [Mn+2] = 1.5M GREATER
 if [Al+3] = 1.5 M and [Mn+2] = 1.5 M SLIGHTLY
GREATER
Homework
 17.55
The Nernst Equation
 DG = DGº +RTln(Q)
 -nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q)
nF
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Eº = 0.48 V
 Always have to figure out n by balancing.
 If concentration can gives voltage, then from
voltage we can tell concentration.
The Nernst Equation
 As reactions proceed concentrations of products
increase and reactants decrease.
 Reach equilibrium where Q = K and
0 = Eº - RTln(K)
nF
Eº = RTln(K)
nF
 nFEº = ln(K)
RT
Ecell = 0
Example Using Nernst
 Describe the cell based on the following:
 VO2+ + 2H+ + e-  VO22+ + H2O
Eo = 1.00V
 Zn2+ + 2e-  Zn
Eo = -0.76V
 T = 25 C
 [VO2+] = 2.0 M
[H+] = 0.50 M
 [VO22+] = 0.01 M
[Zn2+] = 0.1 M
Starting Solution
 In Order to balance the electrons we must





double the first equation, and reverse the
second equation, so we end up with:
2VO2+ + 4H+ + 2e-  2VO22+ + 2H2O
Zn  Zn2+ + 2e2VO2+ (aq) + 4H+ (aq) + Zn (s) 
2VO22+ (aq) + 2H2O (l) + Zn2+ (aq)
Ecell = 1.76 V
Note: Because we are using the Nernst
Equation we use base values for E
Final Solution
 At 25 C R = 0.0521
 E = Ecell –R/n log (Q)
 E = 1.76 –(0.0521/2) log ([Zn2+][VO22+]2)
[VO2+]2[H+]4
 E = 1.76 –(0.0521/2) log ([0.10][0.010]2)
[2.0]2[0.50]4
 E = 1.76 –(0.0521/2) log 4.0 x 10-5
 E = 1.76 + 0.13 = 1.89V
Homework
 17.61 and 17.63
Batteries are Galvanic Cells
 Car batteries are lead storage batteries.
 Pb +PbO2 +H2SO4 PbSO4(s) +H2O
 Dry Cell
Zn + NH4+ +MnO2  Zn+2 + NH3 + H2O
 Alkaline
Zn +MnO2  ZnO+ Mn2O3 (in base)
 NiCad
 NiO2 + Cd + 2H2O  Cd(OH)2 +Ni(OH)2
Corrosion
 Rusting - spontaneous oxidation.
 Most structural metals have reduction
potentials that are less positive than O2 .
 Fe  Fe+2 +2eEº= 0.44 V
 O2 + 2H2O + 4e- 4OH- Eº= 0.40 V
 Fe+2 + O2 + H2O Fe2 O3 + H+
 Reaction happens in two places.
Salt speeds up process by increasing
conductivity
Water
Rust
e-
Iron Dissolves- Fe  Fe+2
Preventing Corrosion
 Coating to keep out air and water.
 Galvanizing - Putting on a zinc coat
 Has a lower reduction potential, so it is
more. easily oxidized.
 Alloying with metals that form oxide
coats.
 Cathodic Protection - Attaching large
pieces of an active metal like
magnesium that get oxidized instead.
Electrolysis
 Running a galvanic cell backwards.
 Put a voltage bigger than the potential
and reverse the direction of the redox
reaction.
 Used for electroplating.
1.10
e-
e-
Zn
Cu
1.0 M
Zn+2
Anode
1.0 M
Cu+2
Cathode
e-
A battery
>1.10V
Zn
e-
Cu
1.0 M
Zn+2
Cathode
1.0 M
Cu+2
Anode
Calculating plating
 Have to count charge.
 Measure current I (in amperes)
 1 amp = 1 coulomb of charge per second
 q=Ixt
 q/nF = moles of metal
 Mass of plated metal
 How long must 5.00 amp current be applied to
produce 15.5 g of Ag from Ag+
Other uses
 Electroysis of water.
 Seperating mixtures of ions.
 More positive reduction potential means
the reaction proceeds forward.
 We want the reverse.
 Most negative reduction potential is
easiest to plate out of solution.