Download Oxidation number and Electrolysis(電解)

Oxidation(氧化) and Reduction(還原
Redox reaction(氧化還原
反應), Oxidation number and Electrolysis(電解) Oxidation: Reaction of an element or a compound with O 2 to give an oxide e.g., 4 Na(s) + O 2 (g) → 2 Na 2 O(s) (sodium oxide) 2 H 2 (g) + O 2 (g) → 2 H 2 O(l) (“hydrogen oxide”?) International Junior Science Olympiad (IJSO) Reduction: Reverse of oxidation e.g., 2 CuO(s) → 2 Cu(s) + O 2 (g) Dr. Yu­San Cheung [email protected] Department of Chemistry The Chinese University of Hong Kong 2 1 Oxidation(氧化) and Reduction(還原) Oxidation (氧化) and Electron Transfer(電子轉移) Reduction: Reverse of oxidation e.g., 2 CuO(s) → 2 Cu(s) + O 2 (g) In the following reaction, copper is oxidized and loses electrons to have a positive charge: e.g., 2 Cu(s) + O 2 (g) → 2 CuO(s) CuO(s) can also be converted to Cu(s) with hydrogen: e.g., CuO(s) + H 2 (g) → Cu(s) + H 2 O(l) Therefore, we may also say that: In an oxidation, an element or a compound loses electron(s) to have a positive charge. Therefore, we may also say that: Reduction is a reaction of an element or a compound with H 2 . Similarly: in an reduction, an element or a compound gains electron(s) to have a positive charge. e.g., Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag(s) e.g., H 2 (g) + Cl 2 (g) → 2 HCl(g) [Cl 2 (g) is reduced] Which ion is oxidized? __________ Which ion is reduced? __________ Here, we consider that H 2 and O 2 have opposite properties. 3 4 Rules of Assigning Oxidation Number (O.N.)
Oxidation Number(氧化數) Consider: H + (aq) + OH – (aq) → H 2 O(l) From high to low priority: Is OH – (aq) oxidized or reduced? (Gaining “hydrogen” but with charge increased) • O.N. for atoms in elements: 0 Overall O.N. for neutral molecules and compounds: 0 Oxidation Number/Oxidation State: Numbers assigned to elements, ions, and compounds to help us to tell whether they are oxidized or reduced in a reaction. e.g., H 2 (g), Na(s), NaCl(s), CO 2 (g) • Overall O.N. for ions: equal to the charges It is also assigned to individual atoms in ions and molecules. e.g., Na + & NH 4 + (O.N.= +1), SO 4 2– (O.N.= –2) 5 6 Rules of Assigning Oxidation Number (con’t) Examples: For individual atoms in neutral molecules and compounds: • F: –1 • Group 1A metals (Li, Na, K, …): +1 • Group 2A metals (Be, Mg, Ca, …): +2 • H: +1 (except in metal hydride, see below) • O: –2 • Cl: –1 • Br, I: –1; SO 4 2– : overall: –2 SO 2 : overall: 0 SF 2 : overall: 0 H 2 S: overall: 0 O.N. for O: –2
O.N. for O: –2
O.N. for F: –1
O.N. for H: +1
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O.N. for S = +6 O.N. for S = +4 O.N. for S = +2 O.N. for S = –2 PCl 5 : overall: 0 PO 4 3– : overall: –3 PO 3 3– : overall: –3 PH 3 : overall: 0 O.N. for Cl: –1
O.N. for O: –2
O.N. for O: –2
O.N. for H: +1
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O.N. for P = +5 O.N. for P = +5 O.N. for P = +3 O.N. for S = –3 H 2 O: overall: 0 O.N. for O: –2 ⇒ O.N. for H = +1 CaH 2 : containing Ca 2+ and H –
⇒ O.N. for H = –1 N, P: –3; 7 S: –2 8 Oxidation Number(氧化數)for Oxygen Examples: O.N. for O is usually –2, except (including but not limited to): ⇒
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ClO 4 – : overall: –1 ClO 3 – : overall: –1 ClO 2 – : overall: –1 ClO – : overall: –1 O.N. for O: –2
O.N. for O: –2
O.N. for O: –2
O.N. for O: –2
MnO 4 – : overall: –1 O.N. for O: –2 ⇒ O.N. for Mn = +7 O.N. for Cl = +7 O.N. for Cl = +5 O.N. for Cl = +3 O.N. for Cl = +1 • Note: Mn has bonding with oxygen atoms. It does not exist as Mn 7+ ion. How about FIO 3 ? O.N. = –1 for F, O.N. = –2 for O, finally we have +7 for I • 9 Redox reaction(氧化還原反應) (Oxidation­Reduction) O.N. for O atom: –1 O.N. for O atom: –1/2 10 e.g., Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) Ag + : gaining electron, being reduced Fe 3+ : losing electron, being oxidized In a reaction, O.N. increase of an atom must be accompanied with O.N. decrease of another atom In general: ­ gaining electron, being reduced ­ losing electron, being oxidized
e.g., Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) Fe: +2 → +3 In some ions: Peroxide: O 2 2– Superoxide: O 2 – O.N. and Electron Transfer(電子轉移) Oxidation of an atom: O.N. increases Reduction of an atom: O.N. decreases O.N. change: In fluorine­oxygen compounds e.g.,: OF 2 Neutral molecule: overall O.N. = 0 Then assign O.N. = –1 for F So, O.N. = +2 for O Ag: +1 → 0 We say that: Fe 2+ (aq) is oxidized to Fe 3+ (aq) Ag + (aq) is reduced to Ag(s) 11 12 Oxidizing Agent(氧化劑) and Reducing Agent(還原劑) Common Oxidizing Agents (氧化劑) Ions of metal at low position in the reactivity series (e.g., Ag + → Ag) Oxidizing agent (oxidant): ­ oxidizing another species ­ being reduced (O.N. decreased) MnO 4 – & Cr 2 O 7 – in acidic medium: MnO 4 – → Mn 2+ ; Cr 2 O 7 – → Cr 3+ Reducing agent (reductant) : reducing another species ­ being oxidized (O.N. increased) Conc. HNO 3 (aq): NO 3 – → NO 2 (g) Conc. H 2 SO 4 (aq): SO 4 2– → SO 2 (g) ­ e.g., Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) Oxidizing Agent: Ag + (aq) Reducing Agent: Fe 2+ (aq) Cl 2 (g) → 2 Cl – ; Br 2 (l) → 2 Br – O 2 (g) → O 2– ; H 2 O 2 (aq) → H 2 O(l) 13 Common Reducing Agents(還原劑) 14 Remarks: Metal at high position in the reactivity series (e.g., Na → Na + ) Some chemical can act as both oxidizing agent and reducing agent. C(s) → CO(g) or CO 2 (g) e.g., CO(g) → CO 2 (g) SO 2 (g) → S(s) SO 2 (g) → SO 4 2– (aq) oxidation / reduction oxidation / reduction SO 3 2– → SO 4 2– Species with high O.N. atom has higher chance to be an oxidizing agent. Fe 2+ → Fe 3+ Similarly, species with low O.N. atom has higher chance to be a reducing agent. 2I – → I 2 H 2 (g) → 2 H + 15 Concentration(濃度)Effect on Oxidizing/Reducing Power Acid­Base Reaction & Redox Reaction O.N.: H O 16 H + (aq) + OH – (aq) → H 2 O(l) Example +1 Dilute nitric acid (HNO 3 ) reacts with magnesium but not copper. Concentrated nitric acid reacts with copper. +1 –2 +1 –2 No atom has its O.N. change. • Dil. HNO 3 (aq) with Mg: 2 H + (aq) + Mg(s) → H 2 (g) + Mg 2+ (aq) In general, acid­base reaction is NOT a redox reaction. • Dil. HNO 3 (aq) with Cu: no reaction • Conc. HNO 3 (aq): (unbalanced equation) NO 3 – (aq) + Cu(s) → NO 2 (g) + Cu 2+ (aq) Exercise Verify the conclusion for the following: HCl(aq) + NaHCO 3 (aq) → NaCl(aq) + CO 2 (g) + H 2 O(l) • Dil. HNO 3 (aq) acts as an acid, conc. HNO 3 (aq) can act as an oxidizing agent. Similarly for H 2 SO 4 (aq).
17 18 Oxidation Number (氧化數) in Chemical Naming O.N. of Atoms(原子) in Various Species Exercise: verify the O.N. Roman number is sometimes used for O.N./O.S. Oxidation number e.g., “The Oxidation State of Mn in MnO 4 – is +VII.” Sulphur Nitrogen Carbon Iron Copper Manganese Chromium +7 +6 The Romanic number system is also used in the “Stock system” to distinguish different compounds. KMnO 4 H 2 SO 4 +5 +4 SO 2 +3 e.g., Cu 2 O, containing Cu + ion: copper(I) oxide CuO, containing Cu 2+ ion: copper(II) oxide +2 0 SCl 2 S H 2 S ­3 19 CaCO 3 K 2 Cr 2 O 7 NO MnO 2 FeCl 3 CO FeSO 4 N 2 O ­1 ­2 NO 2 HNO 2 +1 SO 4 2– , sulphate(VI) ion (more common: sulphate) SO 3 2– , sulphate(IV) ion (more common: sulphite) K 2 MnO 4 HNO 3 CrCl 3 MnSO 4 CrCl 2 Mn Cr CuCl N 2 C NH 2 OH C 2 H 2 N 2 H 4 C 2 H 4 NH 3 C 2 H 6 ­4 CuSO 4 Mn 2 O 3 Fe Cu CH 4 20 Electrolysis(電解) of molten PbBr 2 Electrolysis(電解) Charging and discharge of rechargeable battery Anode (“+”, attracting anions): Br ­ àBr + e ­ Cathode (“­”, attracting cations): Pb 2 + + 2e ­ à Pb • Charging: applying voltage à new substances (electrical energy à chemical energy) “An Ox”: anode ­ oxidation • Discharge: giving out electrical energy (chemical energy à electrical energy) “Red Cat”: reduction ­ cathode Electrolysis: chemical reaction by applying voltage 21 Electrolysis(電解) of CuSO 4 solution 22 Preference of ion discharge Ions attracted to anode: OH ­ (aq) & SO 4 2­ (aq) (1) Electrochemical series Oxidation: 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ SO 4 2­ (aq): no reaction Na à Na + + e ­ Ag à Ag + + e ­ Ions attracted to cathode: Cu 2+ (aq) & H + (aq) Reduction: (More easily) (More difficultly) Therefore, we can infer that sodium ions gain electrons to form atoms more difficultly than silver ions.
Cu 2+ (aq) + 2e ­ à Cu(s) H + (aq): no reaction 23 24 Characteristics • The half­reactions are reductions (i.e., gaining electrons on the left­ hand­side). Different species have different abilities to gain electrons to be reduced. • The half­reactions are equilibriums. • The abilities of gaining electrons are quantified by “standard electrode potentials”. The smaller the potentials (near the top) are, the more difficultly the reductions occur. These abilities are summarized in the electrochemical series. Example: http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29 Compare Na(s) à Na + (aq) + e ­ Ag(s) à Ag + (aq) + e ­ E o = +2.71 V E o = ­0.80 V (Note that when the reactions are reversed, the signs of E o are changed.) 25 Characteristics Characteristics • Higher concentration or pressure: more favorable to go to the opposite side. “Standard” values: 1 M for concentration and 1 atm pressure for pressure. e.g., Ca 2+ (aq) + 2e ­ à Ca(s) 26 • A complete reaction consists of a reduction and an oxidation. e.g., anode: 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ cathode: Cu 2+ (aq) + 2e ­ à Cu(s) (1) (2) E o = ­2.87 V They are combined to form a complete reaction, in which no electron shows up. If increasing Ca 2+ concentration, more favorable to the right­hand­ side, larger E (less negative or even positive). (1): 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ (2)x2: 2Cu 2+ (aq) + 4e ­ à 2Cu(s) e.g., NO 3 ­ (aq) + 2H + (aq) + e ­ à NO 2 (g) + H 2 O(l) E o = +0.78 V If increasing NO 2 (g) pressure, more favorable to the left­hand­side, smaller E. Adding: 4OH ­ (aq) + 2Cu 2+ (aq) à O 2 (g) + 2H 2 O(l) + 2Cu(s) (3) 27 Characteristics http://www.yorku.ca/skrylov/
Teaching/Chemistry1001/
electrochemistry.pdf
Characteristics • An electrochemical cell consists of two half­reactions. A single half­reaction does not exist alone and the absolute values of E o for half­reactions cannot be measured. Therefore, the E o of one of the half­reactions, • Voltage of an electrochemical cell: (1): 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ (2)x2: 2Cu 2+ (aq) + 4e ­ à 4Cu(s) 28 ­0.40 V +0.34 V Metal wire
H 2 (g) at 1 atm Glass tube IMPORTANT: E o does not change when the equation is “doubled”. Adding: E o = ­0.40 + 0.34 = ­0.06 V 2H + (aq) + 2e ­ à H 2 (g), is set to zero. The electrode for this half­reaction is shown on the right and is called “standard hydrogen electrode (SHE)”. The E o of all the others can be determined as values relative to this “standard”. Pt electrode H + (aq, 1 M) 29 30
Characteristics e.g., 2H + (aq) + 2e ­ à H 2 (g) Zn 2+ (aq) + 2e ­ à Zn(s) Consider Zn(s) à Zn 2+ (aq) + 2e ­ Characteristics • If the voltage is positive, the reaction occurs spontaneously. If the voltage is negative, the reaction does not occur spontaneously and an external voltage must be applied. The external voltage must be larger than the magnitude of the cell voltage. E 1 o = 0 V E 2 o = ? ­E 2 o Adding: 2H + (aq) + Zn(s) à H 2 (g) + Zn 2+ (aq) e.g., H 2 (g) + Zn 2+ (aq) à 2H + (aq) + Zn(s) E o = –0.76 V E cell o = E 1 o + (­E 2 o ) = ­E 2 o The external voltage applied must be at least 0.76 V. E cell o of the last electrochemical cell is measured as +0.76 V. Therefore, E 2 o = ­0.76 V. 31 Concentrations of species Characteristics Increasing concentration increases the discharge tendency of an ion. For example, • In electrolysis, the preference of ion discharge depends on E o of the relevant half­reaction potential. For example, 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ 2SO 4 2­ (aq) à S 2 O 8 2­ (aq) + 2e ­ 32 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ 2Cl ­ (aq) à Cl 2 (g) + 2e ­ ­0.40 V ­2.01 V ­0.40 V ­1.36 V For dilute NaCl solution, OH ­ is discharged because the E o value of the first half­reaction is preferred. But for concentrated NaCl solution, Cl ­ concentration is high enough for Cl ­ to be discharged. The first half­reaction is preferred because its E o is larger. 33 Electrodes (電極) 34 Electrodes (電極)
Commonly used graphite and platinum electrodes are inert and have no effect on the preference of ion discharge. But some may. • Metal electrode When an anion discharges at anode, it gives out electrons. If a metal electrode is used as the anode, the metal atoms may also give out electrons to form metal ions, i.e., the metal electrode may compete with the anion in giving out electrons. For example, if copper electrode is used as the anode in the electrolysis of copper sulfate solution, the copper electrode becomes thinner and thinner. • Mercury electrode(汞電極) If graphite or platinum electrodes are used in the electrolysis of concentrated NaCl solution, only H + is discharged at the cathode. But if mercury electrode is used for the cathode, Na + is discharged because sodium metal forms an alloy with mercury. (This method is used in industry for the production of sodium.) 35 36 Electrodes(電極) Summary of common cases: Electrodes cf. Cu(s) à Cu 2+ (aq) + 2e ­ 4OH ­ (aq) à O 2 (g) + 2H 2 O(l) + 4e ­ Solution ­0.34 V ­0.40 V Main products at Cathode Anode Cathode NaNO 3 or NaSO 4 Copper metal of the anode completes with OH ­ . The potential of the first half­reaction is larger. Copper metal, rather than OH ­ , gives out electrons. H 2 SO 4 NaOH In principle, if platinum electrode is used, platinum may also give out electrons to form platinum ion. But in practice, it seldom happens due to the very negative value of E o for this process: Pt(s) à Pt 2+ (aq) + 2e ­ Anode O 2 Graphite H 2 (Dil) NaCl (Conc) ­1.20 V (Conc) CuSO 4 37 Salt Bridge Cl 2 Graphite Graphite Copper Mercury Graphite Cl 2 Na O 2 Copper Copper Cu 2+ 38 Salt Bridge For example, an electrochemical cell can be constructed for the following reaction: The container in which a half­reaction occurs is called a “half­cell”. In the diagram shown, the two half­cells are in the same beaker. Cu(s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag(s) But in some cases they must be separated physically, because the species of the two half­cells react directly without electrons going through the external circuit. We can break it into two half­reactions: Cu(s) → Cu 2+ (aq) + 2e – –0.34 V 2Ag + (aq) + 2e – → 2Ag(s) +0.80 V If we put everything into the same beaker, it does not work as expected. 39 Salt Bridge 40 Salt Bridge Reason: when Ag + is in contact with Cu, they react on the surface of the copper plate. Electrons are given out by Cu to Ag + directly and they do not go through the external circuit. Therefore, the Ag + solution and Cu 2+ solution must be separated in two beakers. Voltmeter Salt­bridge: • “connecting” the two half­ cells. • providing ions to keep the half­cells electrically neutral • simplest version: a strip of filter­paper soaked in KNO 3 or NH 4 NO 3 Voltmeter Cu(s) + 2Ag + (aq) → Cu 2 + (aq) + 2Ag(s) Electron flow Electrical current NO 3 – K + Cu Ag Salt bridge Cu 2+ Cu 2+ 1 M Cu(NO 3 ) 2 (aq) NO 3 – Ag + 1 M AgNO 3 (aq)
Modified from:
http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
41 42 Fuel Cell(燃料電池) H 2 fuel cell Half­reactions: (discharging) • Burning fuel → … → electrical energy: some energy is lost in heating • Fuel cell: chemical energy → electrical energy H 2 (g) → 2H + (aq) + 2e – O 2 (g) + 4H + (aq) + 4e – → 2H 2 O(l) 0.00 V +1.23 V The cell gives an electrical potential of 1.23 V. Example: H 2 fuel cell Fuel Cell – Car & Experiment Kit Lab Manual, Thames & Kosmos (2000) H 2 fuel cell is reversible and it can act as a rechargeable battery H 2 Charging (storing up electrical energy): 2H 2 O(l) → 2H 2 (g) + O 2 (g) O 2 Discharging (releasing electrical energy): 2H 2 (g) + O 2 (g) → 2H 2 O(l) H 2 O Anode PEM Cathode PEM: proton exchange membrane 43 Fuel Cell(燃料電池) 44 Silver­Zinc(鋅)Battery(1.8V) Other than H 2 , some other compounds can also be used for fuel cell. For example: Half reactions of discharge: • Methanol (CH 3 OH) • Ethanol (CH 3 CH 2 OH) Reduction: Ag 2 O(s) + H 2 O(l) + 2e ­ à 2Ag(s) + 2OH ­ (aq) Oxidation: Zn(s) + 2OH ­ (aq) à ZnO(s) + H 2 O(l) + 2e ­ Overall: These kinds of fuel cell may not be reversible, but easier to handle and more energy­rich. Zn(s) + Ag 2 O(s) à ZnO(s) + 2Ag(s) http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
45 The Nickel­Cadmium Rechargeable Battery (1.4 V) 46
E o and K (Equilibrium Constant(平衡常數)) nFE o = RT·ln(K) e.g, Half reactions of discharge: Reduction: Oxidation: 2NiO(OH)(s) + 2H 2 O(l) + 2e ­ à 2Ni(OH) 2 (s) + 2OH ­ (aq) ­ Cd(s) + 2OH ­ (aq) à Cd(OH) 2(s) + 2e Overall: Cd(s) + 2NiO(OH)(s) + 2H 2 O(l) à 2Ni(OH) 2 (s) + Cd(OH) 2 (s) Calculate K for the equilibrium: Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Solution: Cu(s) → Cu 2+ (aq) + 2 e – 2 Ag + (aq) + 2 e –→ 2 Ag(s) –0.34 V +0.80 V E o = –0.34 V + 0.80 V = +0.46 V
http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf
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48 E o and K (Equilibrium Constant(平衡常數))
n = no. of electrons in the half­reactions = 2 F = 96485 C mol –1 (Faraday constant) Put into the equation, nFE o = RT∙ln(K), (2) (96485 C mol –1 ) (0.46 V) = (8.314 J mol –1 K –1 ) (298 K) ln(K) K = 3.6 x 10 15 dm 3 mol –1 Note: the unit of K is determined by the expression: K = [Cu 2+ (aq)]/[Ag + (aq)] 2 (mol dm –3 for concentration, atm for pressure) 49