| Slide1 : 21 Electrochemistry
Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies. |
| Electricity : Electricity 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 |
| Galvani : Galvani 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 |
| Electrochemistry : Electrochemistry 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 |
| Electrons : Electrons 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).
1916 R. Millikan (1868-1953) measured the amount of charge of e–. qe = –1.60217733e-19 C F = 96485 C me = 0.00054856 amu = 9.1093897e-31 kg spin = ½ (two state) magnetic moment = 9.284770e–24 J/tesla |
| Redox reactions and electrons : Redox reactions and electrons 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 ger net: 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. |
| Galvanic Cell : Galvanic Cell 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 |
| Assign oxidation states : Assign oxidation states –3 NH3
–2 N2H4
–1 NH2OH
0 N2
+1 N2O
+2 NO
+3 NO2–
+4 NO2
+5 NO3– 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. –1 Cl–
0 Cl2
+1 ClO–
+3 ClO2–
+4 ClO2
+5 ClO3–
+7 ClO4– |
| Half reaction equations : Half reaction equations Oxidation and reduction can be written as half-reaction equations such as
Zn Zn2+ + 2 e– leo
Cu2+ + 2 e – Cu ger
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 |
| More half-reaction equations : More half-reaction equations 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+ |
| Electrochemical Series : Electrochemical Series 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). |
| Constructing half cells : Constructing half cells 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. |
| Galvanic cells : Galvanic 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. |
| Cell convention : Cell convention Oxidation takes place always at the anode Zn (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 cathode Cu2+ (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 and work : Electric energy and work 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, G G = – n F E (n F = q is the charge) electromotive force (emf, or V) number of electrons in the reaction equation n F is the charge q See slides in 16 Equilibria |
| Standard cell emf’s and electrode potentials : Standard cell emf’s and electrode potentials 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 |
| Gibb’s Free Energy in a Cell : Gibb’s Free Energy in a Cell How much energy is available for the cell Zn | Zn2+ || Ag+ | Ag operating at standard condition when one mole of Zn is consumed ?
Solution Eo Zn = Zn2+ + 2 e 0.762 V 2 Ag+ + 2e = 2 Ag 0.799 V (from table) Zn + 2 Ag+ = Zn2+ + 2 Ag Eo = 1.561 V
Go = – n F E = – 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? |
| Table of standard reduction potential : Table of standard reduction potential Reaction E o (V)
Li+ + e– = Li (s) – 3.04 Na+ + e– = Na (s) – 2.71 Mg2+ + 2 e– = Mg (s) – 2.38 Zn2+ + 2 e– = Zn (s) – 0.76 (reference) 2 H+ + 2 e– = H2 (g) 0.000 (reference) H2 (g) = 2 H+ + 2 e– 0.000 Cu2+ + 2 e– = Cu (s) 0.34 Cu+ + e– = Cu (s) 0.52 I2 (s) + 2e– = 2 I– (aq) 0.54 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 F2 (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? See 21-2 |
| Table of standard reduction potential : Table of standard reduction potential Reaction E o (V)
F2 (g) + 2 e– = 2 F– (aq) 2.87 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 I2 (s) + 2e– = 2 I– (aq) 0.54 Cu+ + e– = Cu (s) 0.52 Cu2+ + 2 e– = Cu (s) 0.34 (reference) H2 (g) = 2 H+ + 2 e– 0.000 (reference) 2 H+ + 2 e– = H2 (g) 0.000 Zn2+ + 2 e– = Zn (s) – 0.76 Mg2+ + 2 e– = Mg (s) – 2.38 Na+ + e– = Na (s) – 2.71 Li+ + 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. |
| Strength of oxidation : Strength of oxidation 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. _____ |
| Reaction direction and emf : Reaction direction and emf What is the emf for the reaction, Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)?
Solution: Know what data to look for Zn2+ + 2 e Zn Eo = – 0.76 V Fe3+ + e Fe2+ Eo = + 0.77 V + 2 Fe2+ 2 Fe3+ + 2 e Eo = – 0.77 V Zn2+ (aq) + 2 Fe2+ (aq) Zn (s) + 2Fe3+ (aq) Eo = – 1.53 V non-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical
The reverse reaction is spontaneous, Zn (s) + 2Fe3+ (aq) Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 V Zn | Zn2+ || Fe3+ | Fe2+ | Pt |
| Free energy and emf : Free energy and emf 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+ + 2e Eo = 0.76 Ag+ + e Ag Eo = 0.80 2 Ag+ + 2e 2Ag Eo = 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. |
| General cell emf : General cell emf G o is the standard energy change. G is for non-standard conditions.
G = – n F E
Similarly, E o is the standard emf whereas E is general emf.
G = G o + R T ln Q E = Eo – R T / n F ln Q reaction quotient
When a system is at equilibrium (Q = K), G = 0. Therefore,
G = G o + R T ln K = 0 E = Eo – R T / n F ln K = 0 equilibrium constant
G o = – R T ln K E o = R T / n F ln K
or G o = – ln(10) R T log K E o = 2.303 R T / n F log K At 298 K 0.0592 E o = ———— log K n Text uses Ecell instead of E |
| The Nernst equation : The Nernst equation R T [C]c [D]d E = E o – ——— ln ———— n F [A]a [B]b For a general reaction, a A + b B = c C + d D 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. See 21-4 |
| Evaluating E : Evaluating E 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 V 2 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 V The 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 R T [C]c [D]d E = E o – ——— ln ———— n F [A]a [B]b See example 19.12 |
| Concentration cell : Concentration cell Problem: At 298 K, evaluate the emf of the cell Cu | 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 0.10 E = 0.00 – ——- ln ——— = – –––– ln ––––– 2 F [Cu2+] 2 F 1.00
8.3145 * 298 1.0 = + –––––––––– ln –––– = 0.0295 V 2*96485 0.1 The 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 |
| Equilibrium Constant K and Eocell : Equilibrium Constant K and Eocell 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 = – 9.75 Ksp = 10– 9.75 = 1.8e–10 At 298 K 0.0592 E o = ———— log K n See p. 837 Show that for this cell Ag | Ag+, 1 M || Cl–, 1 M | AgCl | Ag Eo = – 0.577 V but for this cell Ag | AgCl |Cl–, 1 M || Ag+, 1 M | Ag Eo = +0.577 V |
| Evaluate free-energy change : Evaluate free-energy change Evaluate G o for the reaction
Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s)
Solution:
Required to look up: Eo V Ag+ + e = Ag 0.80
Zn2+ + 2 e = Zn – 0.76
2 Ag+ + 2 e 2Ag 0.80 +
Zn Zn2+ + 2 e + 0.76 Zn (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 |
| Summary of thermodynamics : Summary of thermodynamics Chemical energy Ho, So Go = Ho – T So Electric energy Go = – n F Eo Reaction quotient & equilibrium constant
G = G o + R T ln Q G o = – R T ln K Reaction quotient & equilibrium constant
E = E o – R T/ n F ln Q E o = R T/ n F ln K
Stoichiometry
n
|
| insight from cold denaturation and a two-state water structure : insight from cold denaturation and a two-state water structure By Tsai CJ, Maizel JV Jr, Nussinov R.
…The exposure of non-polar surface reduces the entropy and enthalpy of the system, at low and at high temperatures. At low temperatures the favorable reduction in enthalpy overcomes the unfavorable reduction in entropy, leading to cold denaturation. At high temperatures, folding/unfolding is a two-step process: in the first, the entropy gain leads to hydrophobic collapse, in the second, the reduction in enthalpy due to protein-protein interactions leads to the native state. The different entropy and enthalpy contributions to the Gibbs energy change at each step at high, and at low, temperatures can be conveniently explained by a two-state model of the water structure….
Biochem Mol Biol. 2002;37(2):55-69 |
| pH and emf : pH and emf Consider the cell,
Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt
From table data, Zn2+ + 2e = Zn Eo = – 0.76 2 H+ + 2 e = H2 Eo = 0.00 Zn = Zn2+ + 2e Eo = 0.76 Zn + 2H+ = Zn2+ + H2 Eo = 0.76 R T [Zn2+] PH2 E = Eo – —— ln ———— 2 F [H+]2 = Eo + 0.0592 log [H+] = 0.76 – 0.0592 pH At 298 K (pH meters)
0.76 – E pH = ————— 0.0592 |
| pH electrodes : pH electrodes pH Range: 0-14 Temp. Range: 0-100 C Internal Ref: ROSS Junction: Ceramic Dimensions: 120 mm x 12 mm Slope: 92 - 102% Temp. Accuracy: 0.5 C Catalog Number: 8202BN (BNC Connector, 1 meter cable) |
| Ion selective electrode : Ion selective electrode Concentration;
Temperature;
Electrode surface conditions;
Number of charges of ions (8);
Stirring (6);
Suspension (7);
Zwitterionic nature, net charge density;
Anything changing ionic adsorption;
Isoelectric nature of surface material;
The Nernst equation deals only with concentration and temperature More research has gone into pH measurements. Nernst started it. |
| Battery technology : Battery technology 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 See 21-5 |
| Lead Storage Battery for Autos : Lead Storage Battery for Autos Anode – Negative plate Pb + SO42- PbSO4 + 2e- PbO2 + 4H+ + SO42- + 2e- PbSO4 + 2 H2O Separator Cathode – Positive plate H2SO4 A 12-V battery consists of 6 such cells Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O |
| A mercury battery. : A mercury battery. |
| Corrosion: Unwanted Voltaic Cells : Corrosion: Unwanted Voltaic Cells 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?
Coating Use sacrifice electrode See 21-6 |
| Cathodic protection of an underground pipe. : Cathodic protection of an underground pipe. |
| Ion displacement reactions (corrosion) : Ion displacement reactions (corrosion) Zn + 2 Ag+ Zn2+ + 2 Ag
Zn + 2 Cu2+ Zn2+ + Cu Reaction E o (V)
Li+ + e– = Li (s) – 3.04 Na+ + e– = Na (s) – 2.71 Mg2+ + 2 e– = Mg (s) – 2.38 Zn2+ + 2 e– = Zn (s) – 0.76 (reference) 2 H+ + 2 e– = H2 (g) 0.000 (reference) H2 (g) = 2 H+ + 2 e– 0.000 Cu2+ + 2 e– = Cu (s) 0.34 Cu+ + e– = Cu (s) 0.52 I2 (s) + 2e– = 2 I– (aq) 0.54 Ag+ (aq) + e- = Ag (s) 0.80 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 F2 (g) + 2 e– = 2 F– (aq) 2.87 Zn2+ + 2 Ag Zn + 2 Ag+
Zn2+ + Cu Zn + 2 Cu2+ What metal will react with certain ions? See 21-1 |
| Electrolysis of molten salts : Electrolysis of molten salts Battery 2 Cl– Cl2 + 2e–
2 Na+ + 2 e– 2 Na
Net 2 NaCl 2 Na + Cl2
805o C
Molten salts consists of Na+ and Cl– ions CATHODE e e Eo=-2.71V; 2Na+ + 2e– 2Na Eo=-1.36 V; 2 Cl– Cl2 + 2e– Charges required to produce 1 mole Cl2 and 2 moles Na = 2F
Energy required = 2 F E; (E > 4.07 V) reduction oxidation ANODE See 21-7 |
| Electrometallurgy of Sodium : Electrometallurgy of Sodium |
| Electrolysis of NaCl solution : Electrolysis of NaCl solution Battery
Salt solution consists of Na+ and Cl– ions CATHODE e e 2Na+ + 2e– = 2Na 2Na + 2H+ = H2 + 2Na+ 2 Cl– = Cl2 + 2e– Cl2 + H2O = HCl + ½ O2 reduction oxidation ANODE |
| Refining Copper by Electrolsis : Refining Copper by Electrolsis 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)
Solution
5.00 2*96485 C 1 s ------ mol ----------- ------ = work out your answer New 65.5 1 mol 2 C _______ |
| Production of aluminum : Production of aluminum 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 Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and obtained Al metal. This is the Hall process. Charge required for each mole Al = 3 F
Energy required = 3 F E 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. |
| Electrometallurgy of Aluminum : Electrometallurgy of Aluminum |
| Electrolysis of acid solution : Electrolysis of acid solution Battery
Solutions containing H+ and SO42– ions CATHODE e e 2H+ + 2e– = H2 H2O = ½ O2 + 2e– + 2 H+ Charges required to produce 1 mole H2 and ½ moles O2 = 2F
Energy required = 2 F E reduction oxidation ANODE |
| Electrolysis of H2SO4 solution : Electrolysis of H2SO4 solution 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 are 2 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) |
| Electrolysis of H2SO4 solution : Electrolysis of H2SO4 solution Battery E o = – 1.23 V 2 H2O (l) = 4 e– + O2 (g) + 4 H+ E o = – 2.01 V 2 SO42– = [SO3O–OSO3]2– + 2 e–
Solution consists of H+ and SO42– ions CATHODE ANODE e e 2H+ + 2e = H2 reduction oxidation |
| Electroplating of metals : Electroplating of metals Galvanizing Zn2+ + 2 e– Zn onto metal surface
Copper purification Cu2+ + 2 e– Cu onto pure Cu electrode
Silver plating Ag+ + 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. Over a half century of extensive and innovative research has made us one of the nation's leading experts in plating on magnesium |
| Summary : Summary 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. |
| Skills for Electrochemistry (review) : Skills for Electrochemistry (review) 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 |