Unit 7: Notes One Chapter 17 Oxidation and Reduction Redox

Unit 7: Notes One Chapter 17 Oxidation and Reduction 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) Cl2 + 2 e > 2 Cl-1

Redox: Reduction occurs when an atom gains one or more electrons.
Ex: O + 2e-1  O2- Oxidation Foccurs when an atom or ion loses one or more electrons. Ex: Fe  Fe e-1 LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.

Redox reactions involve electron transfer:
Lose e - =Oxidation Cu (s) + 2 Ag+1 (aq) Cu +2 (aq) + 2 Ag(s) Gain e - =Reduction

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: C6H12O6 + 6O2 6CO2 + 6H2O+ Heat Rusting Iron:
4Fe + 3O22Fe2O3 + Heat Oxidation - Loss of e-1. Na(s)Na+1 +1e-1 Reduction - Gain of e-1. Cl2+ 2e-1 2Cl-1 Number line (Oxidation…Left or right?)

Oxidation #’s: Key Elements
(99%) H H-1 (Hydrides, only with Metals) (99%)O 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 (NO3-1) ion is always +5 (SO4-2) ion is always +6

Finding Oxidation Numbers
The sum of the oxidation numbers must be equal to _____ for a compound. zero Find Ox#’s for H2O? +1 -2 H2O 2 (H)+ 1 (O) = Zero 2 (+1)+ 1 (-2) = Zero Find Ox#’s for H3PO4? +1 +5 -2 H3PO4 3 (H)+ 1 (P)+ 4 (O) = Zero 3 (+1)+ 1 (+5)+ 4 (-2) = Zero

Identifying OX, RD, SI Species
Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0  Ca+2, so Ca0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1  H0, 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 H2 from a solution by the reactions: Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq) Zn(s) + 2 H+(aq) → H2(g) + Zn2+(aq) Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction: Mg(s) + Zn2+ → Zn(s) + Mg2+

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
3Cu+2 (aq) + 2Fe (s)  3 Cu (s) + 2Fe+3 (aq) 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. 6 e-

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 Seven Activity Series of Metals How to use ½ RXN sheet
Electrolysis Lab A-Electrolysis

Activity Series of Metals ELA701
Three chemical RXNS to make metals: Heating only: 2PbO(s)  2Pb(s) O2(g) Heating with Coke (Carbon) 2MnO3(s) + 3C(s)  4Mn(s) + 3CO2(g) Electricity through molten oxide 2Li2O(s)  4Li(s) + 1O2(g)

How to use the ½ Reaction Resource:
Notice they are all written as Reductions (gaining of electrons) 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 can cancel electrons and write the WHOLE CELL RXN. Let’s Practice????? X

Weaker Oxidizing Agent
Stronger 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Weaker Br2(l)+2e-2Br- +1.06 Oxidizing Agent Reducing Agent NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e- Ag(l) +0.80 Hg2++2e- Hg(l) +0.78 NO3-+2H++e-NO2(g)+H2O +0.78 Fe3++e-Fe2+ +0.77 Gains e- I2(s)+2e-2I- +0.53 Loses e- Cu++e-Cu(s) +0.52 Cu2++2e-Cu(s) +0.34 SO42-+4H++2e-SO2(g) +0.17 Sn4++2e-Sn(s) +0.15 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 Sn2++2e-Sn(s) -0.14 Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 2H+(pH=7)+2e-H2(g) -0.41 Fe2++2e-Fe(s) -0.44 Weaker Oxidizing Agent Stronger Reducing Agent Cr3++3e-Cr(s) -0.74 Zn2++2e- Zn(s) -0.76 2H2O+2e-2OH-+H2(g) -0.84

Electrolysis Electrolysis -electric current produced chemical reaction
Cathode -reduction Anode -oxidation Which is the…  anode Cathode=? Anode=? Electron flow? Mass Gain=? Cathode(spoon) Mass Loss=? Cathode Copper

Electrolysis Lab- Demo KI (aq)
What is available to react? K+1 I-1 H2O Cathode Reaction highest reaction on left. Anode Reaction lowest reaction on right. 

KI(aq) K+1 I-1 H2O Cath: highest on left. An: lowest on right.
1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 K+1 I-1 H2O 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e-Ag(l) +0.80 Cath: highest on left. Hg2++2e-Hg(l) +0.78  NO3-+2H++e-NO2(g)+H2O +0.78 Fe3++e-Fe2+ +0.77 I2(s)+2e-2I- +0.53 We see pink. We saw bubbles An: lowest on right. Cu++e-Cu(s) +0.52 Cu2++2e-Cu(s) +0.34  2I-I2(s)+2e- SO42-+4H++2e-SO2(g) +0.17 Sn4++2e-Sn(s) +0.15 We see brown:I2(s) 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 Sn2++2e-Sn(s) -0.14 Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 2H+(pH=7)+2e-H2(g) -0.41 Fe2++2e-Fe(s) -0.44 Cr3++3e-Cr(s) -0.74 Zn2++2e-Zn(s) -0.76 2H2O+2e-2OH-+H2(g) -0.84 K++e-K(s) -2.92

Cu+2 SO4-2 H2O CuSO4(aq) Cath: highest(left) An: lowest(right)
1/2O2(g)+2H+(pH=7)+2e-H2O CuSO4(aq) +1.23 Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Cu+2 SO4-2 H2O Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Cath: highest(left) NO3-+2H++e-NO2(g)+H2O +0.78  Fe3++e-Fe2+ +0.77 I2(s)+2e-2I- +0.53 We saw copper on the pencil tip! Cu++e-Cu(s) +0.52 Cu2++2e-Cu(s) +0.34 An: lowest(right) SO42-+4H++2e-SO2(g) +0.17  Sn4++2e-Sn(s) +0.15 H2O1/2O2(g)+2H+(pH=7)+2e- 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 We saw bubbles! Sn2++2e-Sn(s) -0.14 Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 2H+(pH=7)+2e-H2(g) -0.41 Fe2++2e-Fe(s) -0.44 Cr3++3e-Cr(s) -0.74 Zn2++2e-Zn(s) -0.76 2H2O+2e-2OH-+H2(g) -0.84 Na++e-Na(s) -2.71

Na+1 SO4-2 H2O Na2SO4(aq) Cath: highest(left) An: lowest(right)
1/2O2(g)+2H+(pH=7)+2e-H2O Na2SO4(aq) +1.23 Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Na+1 SO4-2 H2O Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Cath: highest(left) NO3-+2H++e-NO2(g)+H2O +0.78  Fe3++e-Fe2+ +0.77 I2(s)+2e-2I- +0.53 Cu++e-Cu(s) +0.52 We saw pink. We saw bubbles Cu2++2e-Cu(s) +0.34 An: lowest(right) SO42-+4H++2e-SO2(g) +0.17  Sn4++2e-Sn(s) +0.15 H2O1/2O2(g)+2H+(pH=7)+2e- 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 We saw bubbles. Sn2++2e-Sn(s) -0.14 Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 2H+(pH=7)+2e-H2(g) -0.41 Fe2++2e-Fe(s) -0.44 Cr3++3e-Cr(s) -0.74 Zn2++2e-Zn(s) -0.76 2H2O+2e-2OH-+H2(g) -0.84 Na++e-Na(s) -2.71

Electrolysis lab A

Notes #3 Balancing by Oxidation-Reduction

Finding Oxidation #’s for Compounds
+1 -2 +1 +4 -2 H2O H2CO3 +1 +5 -2 -3 +1 +4 -2 H3PO4 (NH4)2CO3 -2 +1 +5 -2 +2 +5 Ca3(AsO4)2 HNO3 +1 +6 -2 +3 +6 -2 H2SO4 Fe2(SO4)3 +1 +6 -2 +2 +7 -2 Hg2SO4 Ba(ClO4)2 +3 +4 -2 +1 +6 -2 Na2Cr2O7 Al2(CO3)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 +1 +6 -2 +1 +5 -2 +1 -2 5e-1 lost X8 H2O + P4+ H2SO4 H3PO4+ H2S 12 2 5 8 5 8e-1 Gained X5 Multiply by 1. 12H2O + 2P4+ 5H2SO4 8H3PO4+ 5H2S

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 +1 -2 +1 +3 -2 5e-1 Gained X3 K3PO Cl2 P4+ K2O+ KClO2 3 5/2 3/4 2 5 3e-1 Lost X5 Multiply by 4. 12K3PO4 + 10Cl2  3P4+ 8K2O+ 20KClO2

Applications of Oxidation-Reduction Reactions

Batteries

Alkaline Batteries

Hydrogen Fuel Cells

Corrosion and…

…Corrosion Prevention

Notes Three Unit Seven Electrolysis Lab Results
Concept Check Assignment Quiz-Balancing/Electrolysis

CuSO4(aq) Cu+2 SO4-2 H2O Na2SO4(aq) Na+1 SO4-2 H2O HCl(aq) H+1 Cl-1
Cl2(g)2e-2Cl-1 +1.36 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Br2(l)+2e-2Br- +1.06 Cu+2 SO4-2 H2O NO3-+4H++3e-NO(g)+2H2O +0.96 Na2SO4(aq) 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Na+1 SO4-2 H2O NO3-+2H++e-NO2(g)+H2O +0.78 I2(s)+2e-2I- +0.53 HCl(aq) Cu++e-Cu(s) +0.52 Cu2++2e-Cu(s) +0.34 SO42-+4H++2e-SO2(g) +0.17 H+1 Cl-1 H2O Sn4++2e-Sn(s) +0.15 2H++2e-H2(g) 0.00 Mg(s) Cu(s) Pb2++2e-Pb(s) -0.13 Cath: highest(left) Sn2++2e-Sn(s) -0.14 Ni2++2e-Ni(s) -0.25  Co2++2e-Co(s) -0.28 2H+(pH=7)+2e-H2(g) -0.41 An: lowest(right) Fe2++2e-Fe(s) -0.44  Cr3++3e-Cr(s) -0.74 Mg(s)2e-1+Mg+2 2H2O+2e-2OH-+H2(g) +2.37 -0.84 Mg+2+e-Mg(s) -2.37 Cu+2+ Mg(s) Cu(s)+ Mg+2 +2.71 Na+1+e-Na(s) -2.71

Lab C Voltaic Cell Anode rxn Cathode rxn Spectator Ions Current Flow
Mg(s) + Mg(s)2e-1+Mg+2 Cathode rxn H+1 Cu2++2e-Cu(s) Cu(s) Spectator Ions - Cl-1 Cu+2 SO4-2 SO4-2 H+1 Na+1 Cl-1 SO4-2 Current Flow Na+1

Notes Four Unit Seven 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 = C/mol e-) A x s = C

Faraday’s Law: Lab B F = 96500 C/mole- Amp x second = C Fe(s) Cathode?
Anode? e-1 Fe(s) F = C/mole- Amp x second = C Lose Mass? Fe+3 Gain Mass? 3e-1 Fe+3 NO3-1 NO3-1 NO3-1

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+3e-1Au(s) 2. Calculate Coulombs. 60 min 60 Sec 3.0 amp x 1.5 Hour x x =16000C 1 hour 1 min 3. Calculate moles e-1. 1mole e-1 16000C x = 0.17 mol e-1 96500C 4. Calculate moles of substance. 1 m Au(s) 0.17 mol e-1x =0.056mol Au(s) 3mol e-1 5. Calculate grams. 197.0gAu 0.056mol Au(s)x = 11g Au 1mol 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+1e-1Ag(s) 2. Calculate Coulombs. 60 Sec 2.0 amp x 45 Hour x =5400C 1 min 3. Calculate moles e-1. 1mole e-1 5400C x = mol e-1 96500C 4. Calculate moles of substance. 1 m Ag(s) 0.056 mol e-1x =0.056mol Au(s) 1mol e-1 5. Calculate grams. 107.9gAg mol Au(s)x = 6.0 g Ag 1mol Ag

Salt Bridge Cathode rxn: +0.34 6 e-1 ? 3 Cu+2 + 2e-1 Cu 3 x3 1.08
Anode rxn: +0.74 6 2 Cr Cr+3+3e-1 2 x2 Na+1 SO4-2 Overall rxn: Na+1 2Cr+ 3Cu+2 2Cr+3+ 3Cu emf= 1.08volts Cu(s) Cr(s) Cu+2 SO4-2 Salt Bridge 3e-1 Cr+3 SO4-2 Cu+2 SO4-2 Cr+3 3e-1 Cu+2 SO4-2 Cu+2 Cr+3 SO4-2 SO4-2 Cu+2 SO4-2 Cr+3 Cu+2 SO4-2 SO4-2

Ion-Electron Method for Balancing

#1. Separate Half-rxn. #2. Bal Non-O Elem. #3. + H2O. #4. + H+1 . #5. + e-1 to bal +/-. UO2+2 + I2  U+4 + IO3-1 (Acid) 5X 1 UO2+2 + H+1 4 + e-1 2  U+4 1 + H2O 2 1X 1 I2 + H2O 6  IO3-1 2 + H+1 12 + e-1 10 8 4 5 UO2+2 + H+1 20 + e-1 10  U+4 5 + H2O 10 1 I2 + H2O 6  IO3-1 2 + H+1 12 + e-1 10 5 UO2+2 + H+1 8 + I2 1  IO3-1 2 + U+4 5 + H2O 4

#1. Separate Half-rxn. #2. Bal Non-O Elem. #3. + H2O. #4. + H+1 . #5. + e-1 to bal +/-. IO3-1 + Ti+3 I2 + TiO2+1 (Acid) 1X 2 IO3-1 + H+1 12 + e-1 10  I2 1 + H2O 6 5X 1 Ti+3 + H2O 2  TiO2+1 1 + H+1 4 + e-1 2 2 IO3-1 + H+1 12 + e-1 10  I2 1 + H2O 6 4 8 5 Ti+3 + H2O 10  TiO2+1 5 H+1 20 e-1 10 2 IO3-1 + H2O 4 + Ti+3 5  TiO2+1 5 + I2 1 + H+1 8

Unit 7: Notes One Chapter 17 Oxidation and Reduction Redox

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Unit 7: Notes One Chapter 17 Oxidation and Reduction Redox

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