21 Electrochemistry Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies. . Electricity. Ancient people noticed electricity
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Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies.
Ancient people noticed electricity
1746 B. Franklin demonstrated lightening as electric effect and performed the kite experiment in 1751. Two people tried to repeat his kite flying experiment were killed by thunder.
1767 L. Galvani inserted two different metals in frog fluid and constructed a electric cell
1800 A. Volta substituted frog fluid; made batteries, consisted of several cells.
1802 G. Romagnosi noticed magnetism related to electricity
Michael Faraday 1791-1867 discovered many theories of electricity and magnetism
Luigi Galvani (1737-1798): erroneously concluded that the frog's nervous system generated an electrical charge, his work stimulated much research into the electrochemistry.
The depiction of his laboratory
A. Volta (1745-1827) experimented with different materials, and made voltaic piles (batteries)
William Nicholson (1753-1815) observed bubbles forming on the surfaces of metals submerged in water when they are connected to a voltaic pile
Humphry Davy (1778-1829) observe electrolysis of water and metal salts. Following that, … Michael Faraday (1791-1860) studied electrolysis, and discovered the relationship between charges and chemical stoichiometry
qe = –1.60217733e-19 C F = 96485 C me = 0.00054856 amu = 9.1093897e-31 kgspin = ½ (two state)magnetic moment = 9.284770e–24 J/tesla
Voltaic piles (batteries) made the following study possible
W. Crookes (1832-1919) observed cathode rays in low-pressure tubes.
1897:J.J. Thomson determined the charge to mass ratio (e– / me) of cathode rays (electrons).
1916R. Millikan (1868-1953) measured the amount of charge of e–.
Energy drives chemical reactions.
Redox reactions involve the transfer of electrons.
Loss of electron (increase oxidation state) is oxidation, (Leo).Gain of electron (decrease oxidation state) is reduction, (ger).Zn Zn2+ + 2 e–leo Cu2+ + 2 e – Cu gernet: Zn + Cu2+ Cu + Zn2+redox
Chemical energy in redox reactions may be convert to electric energy by applying electrochemistry.
In the meantime, we should learn to balance the redox reaction equations.
A galvanic cell consists of two different metals inserted into a solution of an electrolyte (salt, acid or base), simulated
Representation:Zn | Zn2+ || Cu2+ | Cu
0 for any element
1 for H in compounds, but –1 for LiH, NaH, etc
– 2 for O in compounds, but –1 for H2O2, Na2O2
+1 for alkali metals, +2 for alkaline earth metals
The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion.
Oxidation and reduction can be written as half-reaction equations such as
Zn Zn2+ + 2 e–leo
Cu2+ + 2 e – Cuger
net: Zn + Cu2+ Cu + Zn2+redox
Demonstrate how to balance these
Fe2+ Fe3+ + __ e–
C2O42- 2 CO2 + __e –
MnO4 – + __e– Mn2+
Cr2O72– + __e– + __ H+ 2 Cr3+ + __ H2O
Steps to balance half reaction
Assign oxidation number
Figure out what is oxidized or reduced.
Add electrons according to oxidation number change
Balance charge with H+ (acid) or OH – (base)
Balance atoms with H2O
2 I– I2 + __e–
ClO2 + __ OH– ClO3- + __e– + H2O (in basic solution)
2 S2O32– S4O62- + __e–
HS(=S)O3–S + __e- + HSO4–
__ H3O+ + __e – H2(g) + __ H2O
H2O2 + __e – 2 H2O
ClO2 + __e – ClO2–
NO3– + __e – NH4+
An electrochemical series is a list of metals in order of decreasing strength as reductant, or increasing strength as oxidant.An example of an activity series of metals based on the Standard Potentials given would be:
K > Ba > Ca > Na > Mg > Al > Mn > Zn > Fe > Ni > Sn > Pb > Cu > Ag
In this series the most active metal is potassium (K) and the least active metal is silver (Ag)
Reactions and cells are illustrated in 21-1 of Text (PHH).
A half cell consists of an oxidizing and its oxidized species
Zn | Zn2+Cu | Cu2+Pt | H2 | H+(Pt as conductor)Pt |Fe2+ , Fe3+Cl– | Cl2 | Pt
Student cell set from School Master Science $30
Explain the cell convention and reactions of cells.
Using corns for a galvanic cells is illustrated by an Internet site: (schoolnet.ca/general/electric-club/e/page9.html)
This picture illustrates a way to make a pact of battery using coins of different metals.
Apply the principles you have learned regarding electrochemical series and electrolytes to make such a pile will be an interesting exercise.
Oxidation takes place always at the anodeZn (s) Zn2+ (aq) + __ e– Zn(s) | Zn2+(aq)
Fe2+ (aq) Fe3+ (aq) + e– Pt | Fe2+ , Fe3+
H2 (g) 2 H+ (aq) + __ e– Pt | H2(g) | H+(aq)
Reduction takes place at the cathodeCu2+ (aq) + __ e– Cu (s) Cu2+ | Cu(s)
Cl2 (g) + __ e– Cl– (aq) Cl2(g) | Cl –(aq) | Pt
Fe3+ (aq) + __ e – Fe2+ (aq) Fe3+(aq), Fe2+(aq)| Pt
2 H+ (aq) + __ e– H2 (g) H+(aq) | H2(g)| Pt
Concentration (1.1 M) and pressure (0.9 atm) are also included in cell notations.
Electric energy or electric work = charge * potential difference
W = q * V (1 J = 1 Coulomb Volt, C V)compare W = m g h
The Faraday constant F is the charge for one mole of electrons,F = 96485 C; F / NA = 96485 C / 6.022e23 = 1.602177e-19 C, charge per e–
The maximum chemical energy of the cell that can be converted to electric work is the Gibbs free energy change, GG = – n F E (n F= q is the charge)electromotive force (emf, or V)number of electrons in the reaction equationn F is the charge q
See slides in 16 Equilibria
The standard cell emf is the emf of a voltaic cell operating under standard conditions (1 M, 1 atm, 25oC etc). E.g.
Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 1.10 V
Mg (s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 2.90 V
The absolute potential of the electrode cannot be determined. Only relative potential can be measured. The standard reduction potential is measured against a SHE, for which Eo = 0.0000 V
Zn (s) | Zn2+(aq, 1 M) || H+(aq, 1 M) | H2(1 atm)(g) Eo = 0.76 V
Zn = Zn2+ + 2 e–(oxidation, displace H+ possible)Eo = 0.76 V
Zn2+ + 2 e– Zn (reduction)Eo = – 0.76 V
How much energy is available for the cell Zn | Zn2+ || Ag+ | Agoperating at standard condition when one mole of Zn is consumed ?
SolutionEo Zn = Zn2+ + 2 e0.762 V 2 Ag+ + 2e = 2 Ag 0.799 V (from table) Zn + 2 Ag+ = Zn2+ + 2 Ag Eo = 1.561 V
Go = – nFE = – 2 * 96485 C * 1.561 V = 301226 J (1J = 1 C V) = 301.2 kJ
How much silver is consumed?How much energy is available if 6.5 g of Zn is consumed?
Reaction E o (V)
Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52I2 (s) + 2e– = 2 I– (aq) 0.54Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36F2 (g) + 2 e– = 2 F– (aq) 2.87
Standard cell potentials
Cell E o
Li | Li+ || Cu2+ | Cu____
Mg | Mg2+ || I2 | I– | Pt ____
Zn | Zn2+ || Br2 | Br– | Pt ____
Cu | Cu2+ || Zn2+ | Zn____
Which is not spontaneous?
Reaction E o (V)
F2 (g) + 2 e– = 2 F– (aq) 2.87Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36Br2 (l) + 2 e– = 2 Br– (aq) 1.07I2 (s) + 2e– = 2 I– (aq) 0.54Cu+ + e– = Cu (s) 0.52Cu2+ + 2 e– = Cu (s) 0.34(reference)H2 (g) =2 H+ + 2 e– 0.000(reference)2 H+ + 2 e– = H2 (g) 0.000Zn2+ + 2 e– = Zn (s) – 0.76Mg2+ + 2 e– = Mg (s) – 2.38Na+ + e– = Na (s) – 2.71Li+ + e– = Li (s) – 3.04
The listing order in the table may be different in different text books. However, the principles and methods of application remain the same.
This is the order given on the Exam Data Sheet, that is different from the text.
The ability of a chemical to oxidize is its ability to take electrons from other species,Oxidizing agent + n e Reduced species
Strength of oxidation of an oxidizing agent is measured by its reduction potential.
Similarly, strength of reduction of a reducing agent is measured by its oxidation potential.
Oxidized species Reducing agent + n e
Be able to order the species according to oxidizing strength. _____
What is the emf for the reaction, Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)?
Solution: Know what data to look forZn2+ + 2 e ZnEo = – 0.76 VFe3+ + e Fe2+Eo = + 0.77 V +2 Fe2+2 Fe3+ + 2 eEo = – 0.77 VZn2+ (aq) + 2 Fe2+ (aq) Zn (s) + 2Fe3+ (aq) Eo = – 1.53 Vnon-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical
The reverse reaction is spontaneous, Zn (s) + 2Fe3+ (aq) Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 VZn | Zn2+ || Fe3+ | Fe2+ | Pt
What is the free-energy change for the cell, Zn | Zn2+ (aq) 1 M || Ag+ (aq) 1 M | Ag?
Solution: Reduction potential required,Zn Zn2+ + 2eEo = 0.76 Ag+ + e AgEo = 0.80 2 Ag+ + 2e 2AgEo = 0.80
Cell reaction 2 Ag+ + Zn 2Ag + Zn2+ Eo = 1.56 V
Go = –n F Eo = – 2 * 96485 * 1.56 = – 3.01e5 J or – 302 kJ
Negative indicate energy is released.
The free energy for the cell is –301 kJ per mole of Zn, what is the emf?
Condition for spontaneous reaction is – G or + E.
G o is the standard energy change. G is for non-standard conditions.
G = – n F E
Similarly, Eo is the standard emf whereas E is general emf.
G = Go + R T ln Q E = Eo – R T / n Fln Qreaction quotient
When a system is at equilibrium (Q = K), G = 0. Therefore,
G = Go + R T lnK= 0 E = Eo – R T / n Fln K = 0equilibrium constant
Go = – R T lnKE o = R T / n Fln K
orGo = – ln(10) R T logKE o = 2.303 R T / n FlogK
Text uses Ecell instead of E
At 298 K 0.0592E o = ———— log K n
For a general reaction,a A + b B = c C + d D
R T[C]c [D]dE = E o – ——— ln ————n F[A]a [B]b
This Nernst equation correlates cell emf with [ ] or reactivities of reactants and products as well as T
Units for [ ]: mol L-1 for aqueous solution, atm (for gas), and constant for solid and liquid.
R T[C]c [D]dE = E o – ——— ln ————n F[A]a [B]b
At 300 K, evaluate the cell emf for Zn | Zn2+ (0.100 M) || H+ (0.200 M) | H2 (1.111 atm) | Pt
Solution:Look up:Zn Zn2+ (aq) + 2e, E o = 0.76 V2 H+ (aq) + 2 e H2 (g), E o = 0.00 V (R = 8.314 J mol-1 K-1, F = 96485 C mol-1)E o = 0.76 VThe reaction is Zn(s) + 2 H+ (aq) Zn2+ (aq) + H2(g)
8.314 J mol-1 K-1 * 300 K(0.100) (1.111) E = 0.76 – —————————— ln —————— 2 * 96485 C mol-1 (0.200)2
= 0.76 – 0.0129 * (1.02) = 0.76 – 0.013 = 0.75 V
See example 19.12
Problem: At 298 K, evaluate the emf of the cellCu | Cu2+ (0.10 M) | | Cu2+ (1.0 M) | Cu Cu(s) Cu2+(0.1 M) + 2e; Cu2+ (1.0 M) + 2e Cu(s)
Solution:The standard emf (Eo = 0.00)for Cu | Cu2+ || Cu2+ | Cu The reaction is actually Cu (s) + Cu2+ (1.0 M) = Cu2+ (0.1 M) + Cu (s)
R T [Cu2+] R T 0.10E = 0.00 – ——- ln ——— = – –––– ln ––––– 2 F [Cu2+] 2 F 1.00
8.3145 * 298 1.0 = +–––––––––– ln –––– = 0.0295 V 2*96485 0.1The voltage is purely due to concentration difference. Solutions in the two compartment try to become equal.
When 2 [ ]’s are equal, E = 0
See p. 841
Calculate the solubility product of AgCl from data of standard cell
Solution: Look up desirable data
Ag+Cl– (s) + e Ag0(s) + Cl–E° = 0.2223 V
Ag+ (aq) + e Ag (s) E° = 0.799 V
Ag (s) Ag+ (aq) + e E° = – 0.799 V
Get the desirable eq’n AgCl (s) Ag+ (aq) + Cl– (aq)DE° = – 0.577 V
Ksp = [Ag+][Cl– ]
log Ksp = – 0.577 / 0.0592 = – 9.75Ksp = 10– 9.75 = 1.8e–10
Show that for this cellAg | Ag+,1 M || Cl–,1 M | AgCl | AgEo = – 0.577 Vbut for this cellAg | AgCl |Cl–, 1 M || Ag+,1 M | AgEo = +0.577 V
At 298 K 0.0592Eo = ———— log K n
See p. 837
Evaluate G o for the reaction
Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s)
Required to look up: Eo VAg+ + e = Ag 0.80
Zn2+ + 2 e = Zn – 0.76
2 Ag+ + 2 e 2Ag 0.80+
Zn Zn2+ + 2 e + 0.76Zn (s) + 2 Ag+ (aq) Zn2+(aq) + 2 Ag (s)1.56 (= E o)
G o = –n F E o= – 2 * 96485 C * 1.56 V = – 301000 J = – 301 kJ
See slides 17 and 22
Write the cell for this rxn
Chemical energyHo, So
Electric energyGo = – n F Eo
Go = Ho – T So
Reaction quotient &equilibrium constant
G = G o + R T ln QG o = – R T ln K
Reaction quotient &equilibrium constant
E = E o – R T/n F ln QE o = R T/n F ln K
Consider the cell,
Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt
From table data, Zn2+ + 2e = Zn Eo = – 0.762 H+ + 2 e = H2Eo = 0.00Zn = Zn2+ + 2e Eo = 0.76Zn + 2H+ = Zn2+ + H2Eo = 0.76
R T [Zn2+] PH2E = Eo – —— ln ————2 F [H+]2
At 298 K (pH meters)
0.76 – EpH = —————0.0592
= Eo + 0.0592 log [H+]
= 0.76 – 0.0592 pH
pH Range: 0-14Temp. Range: 0-100 CInternal Ref: ROSSJunction: CeramicDimensions: 120 mm x 12 mmSlope: 92 - 102%Temp. Accuracy: 0.5 CCatalog Number:8202BN (BNC Connector, 1 meter cable)
More research has gone into pH measurements. Nernst started it.
By use: automobile, flash light, radio, computer, camera, watch, emergency equipment, artificial heart machine, pace makers, hearing aids, calculators, … (portable energy)
By type: alkaline, dry, wet, storage, rechargeable, etc
By chemistry: alkaline, carbon zinc, dry, cell, lithium, lithium ion, lithium polymer, NiCd, etc.
By material: anode material, cathode material, electrode, etc.
Aluminum for battery manufacture
Anode – Negative platePb + SO42- PbSO4 + 2e-
Cathode – Positive plate
A 12-V battery consists of 6 such cells
PbO2 + 4H+ + SO42- + 2e- PbSO4 + 2 H2O
Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O
Fe(s) Fe2+ (aq) + 2e–O2 + H2O (l) + 4e– 4 OH– (aq)
2 Fe(s) + O2 + H2O 2 Fe2+ (aq) + 4 OH– (aq)
What are effective corrosion prevention methods?
CoatingUse sacrifice electrode
Reaction E o (V)
Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52I2 (s) + 2e– = 2 I– (aq) 0.54Ag+ (aq) + e- = Ag (s) 0.80Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36F2 (g) + 2 e– = 2 F– (aq) 2.87
What metal will react with certain ions?
Zn + 2 Ag+ Zn2+ + 2 Ag
Zn + 2 Cu2+ Zn2+ + Cu
Zn2+ + 2 Ag Zn + 2 Ag+
Zn2+ + Cu Zn + 2 Cu2+
Eo=-2.71V;2Na+ + 2e– 2Na
Eo=-1.36 V; 2 Cl– Cl2 + 2e–
Molten salts consists of Na+ and Cl– ions
2 Cl– Cl2 + 2e–
2 Na+ + 2 e– 2 Na
Net2 NaCl 2 Na + Cl2
Charges required to produce 1 mole Cl2 and 2 moles Na = 2F
Energy required = 2 FE; (E > 4.07 V)
2 Cl– = Cl2 + 2e–Cl2 + H2O = HCl + ½ O2
2Na+ + 2e– = 2Na2Na + 2H+ = H2 + 2Na+
Salt solution consists of Na+ and Cl– ions
Copper can be purified by electrolysis. Raw copper is oxidized
Cu = Cu2+ + 2e
and purer copper deposited on to the cathode from a solution containing CuSO4
Cu2+ + 2e = Cu
If a current of 2 amperes pass through the cell, how long will it take to deposit 5.00 g of copper on the cathode? (1 ampere = 1 C s–1)
5.00 2*96485 C 1 s------ mol ----------- ------ = work out your answer New65.5 1 mol 2 C _______
Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al2O3, remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some Al metal by reducing Al2O3 with potassium vapore.
In 1886, two young men electrolyzed molten cryolite Na3AlF6 (melting point 1000° C), but did not get aluminum.
Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and obtained Al metal. This is the Hall process.
AlF63– + 3 e– Al + 6 F– . . . Cathode 2 Al2OF62– + C(s) + 12 F– + 4 AlF63– + CO2 + 4 e– . . . Anode
2 Al2O3 + 3 C 4 Al + 3 CO2. . . Overall cell reaction
Charge required for each mole Al = 3 F
Energy required = 3 FE
H2O = ½ O2 + 2e– + 2 H+
2H+ + 2e– = H2
Solutions containing H+ and SO42– ions
Charges required to produce 1 mole H2 and ½ moles O2 = 2F
Energy required = 2 FE
Pure water is not a good electric conductor. In the presence of electrolytes, water can be decomposed by electrolysis.
On the other hand, electrolysis of electrolyte solutions may reduce H+ and oxidize O2– in H2O.
In an H2SO4 solution,cathode reductions are2 H2O (l) + 2 e– = H2 (g) + 2 OH–(same as 2H+ + 2e– = H2)
Anode oxidation:2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 1.23 V (observed)2 SO42– = [SO3O–OSO3]2– + 2 e–E o = – 2.01 V (not observed)
E o = – 1.23 V2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 2.01 V2 SO42– = [SO3O–OSO3]2– + 2 e–
2H+ + 2e = H2
Solution consists of H+ and SO42– ions
GalvanizingZn2+ + 2 e– Zn onto metal surface
Copper purification Cu2+ + 2 e– Cu onto pure Cu electrode
Silver platingAg+ + e– Ag onto metal surface
Since 1971, Cal-Aurum has provided electroplating services to the electronic component industry with the highest standards of quality, performance, and competitive pricing.
Miller specializes in plating metals such as magnesium, aluminum, zinc, copper, powdered metals, steel and various other substrates.
The 20th century belongs to electrons. They continue affecting our lives the 21st century.
Chemistry studies the drama played by electrons, and electrochemistry is the finale.
Energy directs and produces the show, but you set the magic stage for a great performance.
Leo and Ger tell electrons to get in and out of your stage, and you must skillfully provide paths to balance the flow.
Cells are the stages for the performance, you must construct, represent, figure out the potentials, and control the show.
Chemical reaction, equilibrium, (acid, base, heterogeneous, and complex formation) and electrochemistry guide us using simple rules.
Apply rules you have learned in Chem1235 to understand what is happening around you and may your live be full of happiness.
Make up a Daniel cell using Pb and Ag as the electrodes. Draw a diagram for it.
Use short notation to represent the cell for the spontaneous reaction
Write half reaction equations for both cathode and anode and explain the reactions
Write balanced redox equations
Calculate emf for a nonstandard cell and its energy
Calculate equilibrium constant K from Eo