Unit 6 Notes One Redox Oxidation Numbers Identifying
Unit 6: Notes One • Redox • Oxidation Numbers • Identifying what is oxidized and what is reduced • Activity Series of Metals Chapter 17 Oxidation and Reduction
Oxidation Reduction Chemisty: Redox Chemistry Oxidation and Reduction reactions always take place simultaneously. Loss of electrons – oxidation (Increase in Oxidation Number) Ex: Na ------> Na+1 + e-1 Gain of electrons - reduction ( Decrease in Oxidation Number) Cl 2 + 2 e-1 ------> 2 Cl-1
Redox: Reduction occurs when an atom gains one or more electrons. Ex: O + 2 e-1 O 2 Oxidation Foccurs when an atom or ion loses one or more electrons. Ex: Fe Fe+3 + 3 e-1 LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO 3) 2 Ag + Cu(NO 3)2.
n. Redox reactions involve electron transfer: Lose e - =Oxidation Cu (s) + 2 Ag+1 (aq) Cu +2 Gain e - =Reduction (aq) + 2 Ag(s)
Oxidation occurs when a molecule does any of the following: Loses electrons Loses hydrogen Gains oxygen If a molecule undergoes oxidation, it has been oxidized and it is the reducing agent (aka reductant).
Reduction occurs when a molecule does any of the following: Gains electrons Gains hydrogen Loses oxygen If a molecule undergoes reduction, it has been reduced and it is the oxidizing agent (aka oxidant).
zinc is being oxidized while the copper is being reduced. Why?
Redox Burning: C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O+ Heat Rusting Iron: 4 Fe + 3 O 2 2 Fe 2 O 3 + Heat Oxidation - Loss of e-1. Na(s) Na+1 +1 e-1 Reduction - Gain of e-1. Cl 2+ 2 e-1 2 Cl-1 Number line (Oxidation…Left or right? )
Oxidation #’s: Key Elements • • (99%) H+1 H-1 (Hydrides, only with Metals) (99%)O-2 O-1 (Peroxides) (Always) Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1 (Always) Be+2, Mg+2, Ca+2, Ba+2, Sr+2, Ra+2 (Always) Al+3 (with only a metal) F-1, Cl-1, Br-1, I-1 (NO 3 -1) ion is always +5 (SO 4 -2) ion is always +6
Finding Oxidation Numbers The sum of the oxidation numbers must zero for a compound. be equal to _____ Find Ox#’s for H 2 O? +1 -2 H 2 O 2 (H)+ 1(O) = Zero 2 (+1)+ 1(-2) = Zero Find Ox#’s for H 3 PO 4? +1 +5 -2 H 3 PO 4 3 (H)+ 1 (P)+ 4(O) = Zero 3(+1)+1(+5)+ 4(-2) = Zero
Identifying OX, RD, SI Species • Ca 0 + 2 H+1 Cl-1 Ca+2 Cl-12 + H 20 • Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca 0 Ca+2, so Ca 0 is the species that is oxidized. • Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1 H 0, so the H+1 is the species that is reduced. • Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl 1 Cl-1, so the Cl-1 is the spectator ion.
Oxidizing Agents and Reducing Agents: Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why? Spectator Ions are ions that do NOT change their oxidation number from the reactant side of a RXN to the product side of a RXN. They are just “hanging out”.
Activity Series of Metals (in your packet) The activity series of metals is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction. It can be used to predict the products in similar reactions involving a different metal. The activity series is a chart of metals listed in order of declining relative reactivity. The top metals are more reactive than the metals on the bottom. For example, both magnesium and zinc can react with hydrogen ions to displace H 2 from a solution by the reactions: Mg(s) + 2 H+(aq) → H 2(g) + Mg 2+(aq) Zn(s) + 2 H+(aq) → H 2(g) + Zn 2+(aq) Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction: Mg(s) + Zn 2+ → Zn(s) + Mg 2+
If Metal is higher On chart than Ion RXN will happen
Reduction: Cu+2(aq) + 2 e- Cu(s) The Cation becomes a solid metal (the + charge GAINS ELECTRONS to become a zero charge. Oxidation: Cu(s) Cu+2(aq) + 2 e. The metal becomes a cation (the zero charge metal LOSES ELECTRONS To become a + charge.
Redox: Oxidation Reduction Reaction 3 Cu+2 (aq) + 2 Fe (s) 3 Cu (s) + 2 Fe+3 (aq) 6 e- The paired reduction and oxidation Electrons transfer from the metal to the cation if the metal Is above (ie higher) on the Activity Series Chart in your packet. This reaction will occur because Fe (metal) is higher than Cu so the reaction will occur.
Cu+2 (aq) + Mg (s) Cu (s) + Mg+2 (aq) Mg is higher than Cu on the activity series Zn+2 (aq) + Ag (s) No RXN Because Ag is lower than Zn on The activity series
Notes Two Unit Six • • Activity Series of Metals Balancing by Redox Electrolysis Lab A-Electrolysis
Activity Series of Metals ELA 601 • Three chemical RXNS to make metals: • Heating only: 2 Pb. O(s) 2 Pb(s) + 1 O 2(g) • Heating with Coke (Carbon) 2 Mn. O 3(s) + 3 C(s) 4 Mn(s) + 3 CO 2(g) • Electricity through molten oxide 2 Li 2 O(s) 4 Li(s) + 1 O 2(g)
Finding Oxidation #’s for Compounds +1 -2 H 2 O +1+5 -2 H 3 PO 4 +1+5 -2 HNO 3 +1+6 -2 H 2 SO 4 +1 +6 -2 Hg 2 SO 4 +1 +6 -2 Na 2 Cr 2 O 7 +1+4 -2 H 2 CO 3 -3 +1 +4 -2 (NH 4)2 CO 3 +2 +5 -2 Ca 3(As. O 4)2 +3 +6 -2 Fe 2(SO 4)3 +2 +7 -2 Ba(Cl. O 4)2 +3 +4 -2 Al 2(CO 3)3
Balancing By Redox Example One #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1 -2 0 +1 +6 -2 +1 +5 -2 5 e-1 lost X 8 +1 -2 12 H 2 O + 2 P 4+ 5 H 2 SO 4 8 H 3 PO 4+ 5 H 2 S 8 e-1 Gained X 5 Multiply by 1. 12 H 2 O + 2 P 4+ 5 H 2 SO 4 8 H 3 PO 4+ 5 H 2 S
Balancing By Redox Example Two #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1 +5 -2 0 0 5 e-1 Gained X 3 +1 -2 +1 +3 -2 3 K 3 PO 4 +5/2 Cl 2 3/4 P 4+ 2 K 2 O+ 5 KCl. O 2 3 e-1 Lost X 5 Multiply by 4. 12 K 3 PO 4 +10 Cl 2 3 P 4+ 8 K 2 O+ 20 KCl. O 2
Applications of Oxidation-Reduction Reactions
Batteries
Alkaline Batteries
Hydrogen Fuel Cells
Corrosion and…
…Corrosion Prevention
How to use the ½ Reaction Resource: Notice they are all written as Reductions (gaining of electrons) 1. Find the highest one on the left hand side and write it in the forward direction. 2. Find the lowest one on the right and write it backwards as an oxidation , along with changing the sign of the voltage. 3. Make sure to balance electrons lost and gained (you DO NOT MULTIPLY the voltages !!!) 4. So you will have two ½ cell reactions and you cancel electrons and write the WHOLE CELL RXN. Let’s Practice? ? ? X
Stronger 1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O +1. 23 - 2 Br. Br (l)+2 e +1. 06 2 Oxidizing NO 3 -+4 H++3 e- NO(g)+2 H 2 O +0. 96 Agent 1/2 O (g)+2 H+(p. H=7)+2 e- H O +0. 82 2 2 Ag++e- Ag(l) +0. 80 Hg 2++2 e- Hg(l) +0. 78 NO 3 -+2 H++e- NO 2(g)+H 2 O +0. 78 3++e- Fe 2+ Fe +0. 77 Gains e I 2(s)+2 e 2 I +0. 53 Loses e +0. 52 Cu++e- Cu(s) Cu 2++2 e- Cu(s) +0. 34 SO 42 -+4 H++2 e- SO 2(g) +0. 17 Sn 4++2 e- Sn(s) +0. 15 2 H++2 e- H 2(g) 0. 00 Pb 2++2 e- Pb(s) -0. 13 -0. 14 Sn 2++2 e- Sn(s) Ni 2++2 e- Ni(s) -0. 25 -0. 28 Co 2++2 e- Co(s) -0. 41 2 H+(p. H=7)+2 e- H 2(g) -0. 44 Fe 2++2 e- Fe(s) Weaker Cr 3++3 e- Cr(s) -0. 74 Oxidizing 2+ -0. 76 Zn +2 e - Zn(s)2 H 2 O+2 e 2 OH +H 2(g) -0. 84 Agent Weaker Reducing Agent Stronger Reducing Agent
Electrolysis -electric current produced chemical reaction Cathode -reduction Anode -oxidation Which is the… Cathode=? Anode=? Electron flow? Mass Gain=? Cathode(spoon) Mass Loss=? Copper e-1 anode Cathode
Electrolysis Lab- Demo KI (aq) What is available to react? +1 K -1 I Cathode Reaction highest reaction on left. Anode Reaction lowest reaction on right. H 2 O
KI(aq) 1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O Br 2(l)+2 e- 2 Br. NO 3 -+4 H++3 e- NO(g)+2 H 2 O - H O K+1 I-1 H 2 O 1/2 O 2(g)+2 H+(p. H=7)+2 e 2 + +e Ag(l) Cath: highest on left. Hg. Ag 2++2 e- Hg(l) NO 3 -+2 H++e- NO 2(g)+H 2 O Fe 3++e- Fe 2+ - 2 II (s)+2 e We see pink. 2 We saw. Cu bubbles ++e- Cu(s) An: lowest on right. 2++2 e- Cu(s) Cu 2 I- I 2(s)+2 e. SO 42 -+4 H++2 e- SO 2(g) Sn 4++2 e- Sn(s) 2 H++2 e- H 2(g) We see brown: I 2(s) Pb 2++2 e- Pb(s) Sn 2++2 e- Sn(s) Ni 2++2 e- Ni(s) Co 2++2 e- Co(s) 2 H+(p. H=7)+2 e- H 2(g) Fe 2++2 e- Fe(s) Cr 3++3 e- Cr(s) Zn 2++2 e- Zn(s) - 2 OH-+H (g) 2 H 2 O+2 e 2 K++e- K(s) +1. 23 +1. 06 +0. 96 +0. 82 +0. 80 +0. 78 +0. 77 +0. 53 +0. 52 +0. 34 +0. 17 +0. 15 0. 00 -0. 13 -0. 14 -0. 25 -0. 28 -0. 41 -0. 44 -0. 76 -0. 84 -2. 92
1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O Cu. SO 4(aq) Br 2(l)+2 e- 2 Br. NO 3 -+4 H++3 e- NO(g)+2 H 2 O 1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O ++e- Ag(l) +2 -2 Ag Cu SO 4 H 2 O Hg 2++2 e- Hg(l) Cath: highest(left) NO -+2 H++e- NO (g)+H O 3 2 2 3+ 2+ Fe +e Fe I 2(s)+2 e- 2 IWe saw copper on the pencil Cu tip!++e- Cu(s) Cu 2++2 e- Cu(s) An: lowest(right) SO 42 -+4 H++2 e- SO 2(g) 4++2 e- Sn(s) Sn + H 2 O 1/2 O 2(g)+2 H (p. H=7)+2 e+- 2 H +2 e H 2(g) Pb 2++2 e- Pb(s) We saw bubbles! Sn 2++2 e- Sn(s) Ni 2++2 e- Ni(s) Co 2++2 e- Co(s) 2 H+(p. H=7)+2 e- H 2(g) Fe 2++2 e- Fe(s) Cr 3++3 e- Cr(s) Zn 2++2 e- Zn(s) - 2 OH-+H (g) 2 H 2 O+2 e 2 Na++e- Na(s) +1. 23 +1. 06 +0. 96 +0. 82 +0. 80 +0. 78 +0. 77 +0. 53 +0. 52 +0. 34 +0. 17 +0. 15 0. 00 -0. 13 -0. 14 -0. 25 -0. 28 -0. 41 -0. 44 -0. 76 -0. 84 -2. 71
1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O Na 2 SO 4(aq) Br 2(l)+2 e- 2 Br. NO 3 -+4 H++3 e- NO(g)+2 H 2 O 1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O ++e- Ag(l) +1 -2 Ag Na SO 4 H 2 O Hg 2++2 e- Hg(l) Cath: highest(left) NO -+2 H++e- NO (g)+H O 3 2 2 3+ 2+ Fe +e Fe I 2(s)+2 e- 2 I++e- Cu(s) Cu We saw pink. We saw bubbles Cu 2++2 e- Cu(s) An: lowest(right) SO 42 -+4 H++2 e- SO 2(g) 4++2 e- Sn(s) Sn + H 2 O 1/2 O 2(g)+2 H (p. H=7)+2 e+- 2 H +2 e H 2(g) Pb 2++2 e- Pb(s) We saw bubbles. Sn 2++2 e- Sn(s) Ni 2++2 e- Ni(s) Co 2++2 e- Co(s) 2 H+(p. H=7)+2 e- H 2(g) Fe 2++2 e- Fe(s) Cr 3++3 e- Cr(s) Zn 2++2 e- Zn(s) - 2 OH-+H (g) 2 H 2 O+2 e 2 Na++e- Na(s) +1. 23 +1. 06 +0. 96 +0. 82 +0. 80 +0. 78 +0. 77 +0. 53 +0. 52 +0. 34 +0. 17 +0. 15 0. 00 -0. 13 -0. 14 -0. 25 -0. 28 -0. 41 -0. 44 -0. 76 -0. 84 -2. 71
Electrolysis lab A
Notes Three Unit Six • Electrolysis Lab Results • Concept Check Assignment • Quiz-Balancing/Electrolysis
Cu. SO 4(aq) Cl 2(g)2 e- 2 Cl-1 1/2 O 2(g)+2 H+(p. H=7)+2 e- H 2 O - 2 Br. Br (l)+2 e +2 -2 2 Cu SO 4 H 2 O NO -+4 H++3 e- NO(g)+2 H O 3 2 + H 2 O Na 2 SO 4(aq) 1/2 O 2(g)+2 H (p. H=7)+2 e Ag++e- Ag(l) Hg 2++2 e- Hg(l) Na+1 SO 4 -2 H 2 O NO 3 -+2 H++e-- NO 2(g)+H 2 O I 2(s)+2 e 2 I HCl(aq) Cu++e- Cu(s) Cu 2++2 e- Cu(s) 2 -+4 H++2 e- SO (g) SO 4 2 H+1 Cl-1 H 2 O 4+ Sn +2 e Sn(s) 2 H++2 e- H 2(g) Mg(s) Cu(s) Pb 2++2 e- Pb(s) Sn 2++2 e- Sn(s) Cath: highest(left) Ni 2++2 e- Ni(s) Co 2++2 e- Co(s) 2 H+(p. H=7)+2 e- H 2(g) An: lowest(right) Fe 2++2 e- Fe(s) 3++3 e- Cr(s) Cr -1 +2 Mg(s) 2 e +Mg - 2 OH-+H (g) +2. 37 2 H 2 O+2 e 2 +2 Mg +e Mg(s) +2 +2 Cu + Mg(s) Cu(s)+ Mg Na+1+e - Na(s) +2. 71 +1. 36 +1. 23 +1. 06 +0. 96 +0. 82 +0. 80 +0. 78 +0. 53 +0. 52 +0. 34 +0. 17 +0. 15 0. 00 -0. 13 -0. 14 -0. 25 -0. 28 -0. 41 -0. 44 -0. 74 -0. 84 -2. 37 -2. 71
Lab C Voltaic Cell e-1 Mg(s) H+1 Cl-1 SO 4 -2 Na+1 Anode Mg(s) 2 e-1+Mg + - rxn +2 Cathode rxn Cu+2 SO 4 -2 Cu(s) Cu 2++2 e- Cu(s) Spectator Ions SO 4 -2 H+1 Na+1 Cl-1 Current Flow
Notes Four Unit Six • Faraday’s Law Lab B • This as a chemical process that uses electricity to produce industrial quantities of specific chemicals.
Application of Faraday’s law F = 96500 C/mol e-) Axs=C
Faraday’s Law: Lab B Fe(s) Cathode? Anode? e-1 Fe(s) Fe+3 3 e-1 Fe+3 NO 3 -1 F = 96500 C/mole. Amp x second = C Lose Mass? Gain Mass?
Faraday’s Law Calculation One How many grams of Gold will be plated, using a current of 3. 0 amps for 1. 5 hours? 1. Balanced Equation Au+3+3 e-1 Au(s) 2. Calculate Coulombs. 60 min 60 Sec 3. 0 amp x 1. 5 Hour x x =16000 C 1 hour 1 min 3. Calculate moles e-1. -1 1 mole e 16000 C x = 0. 17 mol e-1 96500 C 4. Calculate moles of substance. 1 m Au(s) -1 0. 17 mol e x =0. 056 mol Au(s) -1 3 mol e 5. Calculate grams. 197. 0 g. Au 0. 056 mol Au(s)x = 11 g Au 1 mol Au
Faraday’s Law Calculation Two How many grams of Silver will be plated, using a current of 2. 0 amps for 45 minutes? 1. Balanced Equation Ag+1+1 e-1 Ag(s) 2. Calculate Coulombs. 60 Sec 2. 0 amp x 45 Hour x =5400 C 1 min 3. Calculate moles e-1. -1 1 mole e 5400 C x = 0. 056 mol e-1 96500 C 4. Calculate moles of substance. 1 m Ag(s) -1 0. 056 mol e x =0. 056 mol Au(s) -1 1 mol e 5. Calculate grams. 107. 9 g. Ag 0. 05596 mol Au(s)x 1 mol Ag = 6. 0 g Ag
e-1 ? Salt Bridge 1. 08 Na+1 -2 SO +1 4 Na Cu(s) Cu+2 -2 SO 4 Cu+2 -2 SO +2 4 Cu SO 4 -2 Cu+2 SO 4 -2 Cathode rxn: +0. 34 6 -1 +2 3 Cu + 2 e 3 Cu x 3 Anode rxn: +0. 74 6 2 Cr 2 Cr+3+3 e-1 x 2 Overall rxn: 2 Cr+3 Cu+2 2 Cr+3+3 Cu emf= 1. 08 volts Cr(s) SO 4 -2+3 3 e-1 Salt Bridge Cr Cr+3 3 e-1 Cr+3 SO 4 -2
Ion-Electron Method for Balancing
Ion-Electron Method for Balancing #1. Separate Half-rxn. #2. Bal Non-O Elem. #3. + H 2 O. #4. + H+1. #5. + e-1 to bal +/-. UO 2+2 + I 2 U+4 + IO 3 -1 (Acid) 5 X 1 UO 2+2 + 4 H+1 + 2 e-1 1 U+4 + 2 H 2 O 1 X 1 I 2 + 6 H 2 O 2 IO 3 -1 + 12 H+1 + 10 e-1 8 4 +2 5 UO 2 + 20 H+1 +10 e-1 5 U+4 + 10 H 2 O 1 I 2 + 6 H 2 O 2 IO 3 -1 + 12 H+1 + 10 e-1 5 UO 2+2+ 8 H+1+ 1 I 2 2 IO 3 -1 + 5 U+4+ 4 H 2 O
Ion-Electron Method for Balancing #1. Separate Half-rxn. #2. Bal Non-O Elem. #3. + H 2 O. #4. + H+1. #5. + e-1 to bal +/-. IO 3 -1 + Ti+3 I 2 + Ti. O 2+1 (Acid) 1 X 2 IO 3 -1 +12 H+1+ 10 e-1 1 I 2 + 6 H 2 O 5 X 1 Ti+3+ 2 H 2 O 1 Ti. O 2+1+ 4 H+1 + 2 e-1 2 IO 3 -1 + 12 H+1 +10 e-1 1 I 2 + 6 H 2 O 8 4 5 Ti+3 +10 H 2 O 5 Ti. O 2+1 + 20 H+1+ 10 e-1 2 IO 3 -1 + 4 H 2 O +5 Ti+3 5 Ti. O 2+1 + 1 I 2 + 8 H+1
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