Chapter 4 Reduction-Oxidation Reactions Redox Reactions.
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Transcript of Chapter 4 Reduction-Oxidation Reactions Redox Reactions.
Chapter 4
Reduction-Oxidation Reactions
Redox Reactions
Sodium chloride
Sodium chloride
Na Na+ + e Cl2 + 2e 2Cl
4
Oxidation–Reduction ReactionsInvolves 2 processes:
Oxidation = Loss of Electrons
Na Na+ + e Oxidation Half-Reaction
Reduction = Gain of electronsCl2 + 2e 2Cl Reduction Half-Reaction
Net reaction:
2Na + Cl2 2Na+ + 2Cl
– Oxidation & reduction always occur together
– Can't have one without the other
Oxidation Reduction ReactionOxidizing Agent - Substance that accepts e's
– Accepts e's from another substance– Substance that is reduced
Cl2 + 2e 2Cl–
Reducing Agent - Substance that donates e's
– Releases e's to another substance– Substance that is oxidized
Na Na+ + e–
Your Turn!
Which species functions as the oxidizing agent in the following oxidation-reduction reaction?
Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)
A. Pt(s)
B. Zn2+(aq)
C. Pt2+(aq)
D. Zn(s)
E. None of these, as this is not a redox reaction.
6
Redox Reactions Very common
– Batteries—car, flashlight, cell phone, computer
– Metabolism of food
– Combustion Chlorine Bleach
– Dilute NaOCl solution
– Cleans through redox
reaction
– Oxidizing agent
– Destroys stains by oxidizing them
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Redox ReactionEx. Fireworks displays
Net: 2Mg + O2 2MgO
Oxidation:
Mg Mg2+ + 2e – Loses electrons = Oxidized– Reducing agent
Reduction:
O2 + 4e 2O2
– Gains electrons = Reduced– Oxidizing agent
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Redox Reaction?
S + O2 SO2
• Combustion: Oxidation in old sense, reaction with oxygen
• But n no ions in SO2
• How can we decide which loses and which gains!!
• Oxidation number (state)!
• If compound were ionic, what would the charges have been.
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Rules for Oxidation States (Numbers)
1. The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion.
2. The oxidation state of elements is zero.
3. Oxidation state for monoatomic ions are the same as their charge.
4. In its compounds fluorine is always –1.
5. Hydrogen is assigned the oxidation state +1.
6. Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide).
7. If two rules conflict, apply higher rule.
Oxidation States Assign the oxidation states to each element
in the following. CO2
NO3-
H2SO4
Fe2O3
Fe3O4
Cr2O72-
O2F2
H2O2
LiH BaO2
Oxidation-ReductionTransfer electrons, so the oxidation states change.
11
2
00
ClNaClNa
Ox
Red
22
2
00
OMgOMg
Ox
Red
Oxidation
Increase in
Oxid. number
Reduction
decrease in
Oxid. number
2
2
4
2
00
OSOS
Ox
Red
2
2
1
2
24
2
0
4
14
OHOCOHC
Ox
RedRedC (CH4) oxidized
→
CH4 reducing agent
O2 reduced →
O2 oxidizing agent
Ox
Red
PbS has been oxidized, PbS is the reducing agent.
O2 has been reduced, O2 is the oxidizing agent.
Ox
Red
PbO has been reduced, PbO is the oxidizing agent.
CO has been oxidized, CO is the reducing agent.
Identify the 1) Oxidizing agent2) Reducing agent3) Substance oxidized4) Substance reduced in the following reactions
Fe (s) + O2(g) ® Fe2O3(s)
Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g)
SO3- + H+ + MnO4
- ® SO4- + H2O + Mn+2
Balancing Redox Reactions
Ion-Electron Method – Acidic Solution
1. Divide equation into 2 half-reactions
2. Balance atoms other than H & O
3. Balance O by adding H2O to side that needs O
4. Balance H by adding H+ to side that needs H
5. Balance net charge by adding e–
6. Make e– gain equal e– loss; then add half-reactions
7. Cancel anything that is the same on both sides
Balance in Acidic Solution
Cr2O72– + Fe2+ Cr3+ + Fe3+
1. Break into half-reactions
Cr2O72 Cr3+
Fe2+ Fe3+
2. Balance atoms other than H & O
Cr2O72 2Cr3+
– Put in 2 coefficient to balance Cr
Fe2+ Fe3+
– Fe already balanced
3. Balance O by adding H2O to the side that needs O.
Cr2O72 2Cr3+
• Right side has 7 O atoms• Left side has none • Add 7 H2O to left side
Fe2+ Fe3+
• No O to balance
+ 7 H2O
4. Balance H by adding H+ to side that needs H
Cr2O72 2Cr3+ + 7H2O
• Left side has 14 H atoms• Right side has none • Add 14 H+ to right side
Fe2+ Fe3+
• No H to balance
14H+ +
5. Balance net charge by adding electrons.
14H+ + Cr2O72 2Cr3+ + 7H2O
– 6 electrons must be added to reactant side
Fe2+ Fe3+
– 1 electron must be added to product side
Now both half-reactions balanced for mass & charge
6e +
+ e
Net Charge = 2(+3)+7(0) = 6
Net Charge = 14(+1) (–2) = 12
6. Make e– gain equal e– loss; then add half-reactions
6e + 14H+ + Cr2O72– 2Cr3+ + 7H2O
Fe2+ Fe3+ + e
7. Cancel anything that's the same on both sides
6[ ]
6e + 6Fe2+ + 14H+ + Cr2O7
2
6Fe3+ + 2Cr3+ + 7H2O + 6e
6Fe2+ + 14H+ + Cr2O7
2
6Fe3+ + 2Cr3+
+ 7H2O
Practice The following reactions occur in acidic
aqueous solution. Balance them:
MnO4- + Fe2+ ® Mn2+ + Fe3+
Cu + NO3- ® Cu2+ + NO(g)
Pb + PbO2 + SO42- ® PbSO4
Mn2+ + NaBiO3 ® Bi3+ + MnO4-
Cr2O72- + C2H5OH ® Cr3+ + CO2
Ion-Electron method in Basic Solution
The simplest way to balance an equation in basic solution
Use steps 1-7 above, then
8. Add the same number of OH– to both sides of the equation as there are H+.
9. Combine H+ & OH– to form H2O
10. Cancel any H2O that you can from both sides
Basic Solution
Ag + CN- +O2 ® Ag(CN)2-
Cr(OH)3 + OCl- + OH- ® CrO42- + Cl- + H2O
CrI3 + Cl2 ® CrO4- + IO4
- + Cl-