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Redox Reaction
- 2. Introduction
• Redox reactions include reactions which involve the
loss or gain of electrons.
• The reactant giving away (donating) electrons is called
the reducing agent (which is oxidized)
• The reactant taking (accepting) electrons is called the
oxidizing agent (which is reduced)
• Both oxidation and reduction happen simultaneously,
however each is considered separately using ion-
electron equations.
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- 3. Oxidation-Reduction
Oxidation: Loss of electrons.
Reduction: Gain of electrons.
Reductant: Species that loses
electrons.
Oxidant: Species that gains
electrons.
Oxidation Reduction
Valence: the electrical charge an atom would acquire if it
formed ions in aqueous solution.
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- 4. Points to Remember
Oxidizing agent Reducing agent
Is itself reduced Is itself oxidized
Gains electrons Loses electrons
Causes oxidation Causes reduction
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- 5. VARIABLE VALENCE ELEMENTS
Sulfur: SO4
2-(+6), SO3
2-(+4), S(0), FeS2(-1), H2S(-2)
Carbon: CO2(+4), C(0), CH4(-4)
Nitrogen: NO3
-(+5), NO2
-(+3), NO(+2), N2O(+1), N2(0), NH3(-3)
Iron: Fe2O3(+3), FeO(+2), Fe(0)
Manganese: MnO4
-(+7), MnO2(+4), Mn2O3(+3), MnO(+2), Mn(0)
Copper: CuO(+2), Cu2O(+1), Cu(0)
Tin: SnO2(+4), Sn2+(+2), Sn(0)
Uranium: UO2
2+(+6), UO2(+4), U(0)
Arsenic: H3AsO4
0(+5), H3AsO3
0(+3), As(0), AsH3(-1)
Chromium: CrO4
2-(+6), Cr2O3(+3), Cr(0)
Gold: AuCl4
-(+3), Au(CN)2
-(+1), Au(0)
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- 6. Balancing Redox equations (pH acid)
Identify the atoms that are oxidized and reduced, using ox. no.
Write the half-reactions for oxidation and reduction processes.
Balance mass, so that the number of atoms of each element
oxidized/reduced is the same on both sides.
If there is any O atoms present, balance them adding H2O.
If there is H atoms present, balance them adding H+.
Balance charges with e- (electron).
Multiply the half-equations so that the number of e- lost in one
equation is equal to the e- gained in the other.
Add the two half-equation together and cancel out any
substances that appear on both side.
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- 7. Cl- + MnO4
- Mn2+ + Cl2 (Acid Medium)
Identify atoms oxidised and reduced, using ox. no.
Cl : -1 Mn : +7 O: -2 Mn2+: +2 Cl: 0
Cl oxidises (goes from -1 to 0) Mn reduces (from +7 to +2)
Write the half-reactions. Balance mass, if necessary using
H2O (to add O) and/or H+ (to add H)
2Cl- Cl2 Oxidation (reducing agent)
8H+ + MnO4
- Mn2+ + 4H2O Reduction (oxidising agent)
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- 8. Balance the charges using electrons (e-). Each e- adds a -1
2Cl- Cl2 + 2e- -2 = 0 -2
8H+ + MnO4
- + 5e- Mn2+ + 4H2O +8 -1 -5 = +2 +0
Multiply the half-equations so that e- lost in one = e- gained
in the other
5x(2Cl- Cl2 + 2e-)
2x(8H+ + MnO4
- + 5e- Mn2
+ + 4H2O)
Cl- + MnO4
- Mn2+ + Cl2 (Acid Medium)
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- 9. Add the two half-equation together and cancel out any
substances that appear on both sides
10Cl- 5Cl2 + 10e-
16H+ + 2MnO4
- + 10e- 2Mn2+ + 8H2O)
10Cl- + 16H+ + 2MnO4
- + 10e- 5Cl2 + 10e- + 2Mn2+ + 8H2O
Answer:
10Cl- + 16H+ + 2MnO4
- 5Cl2 + 2Mn2+ + 8H2O
+
Cl- + MnO4
- Mn2+ + Cl2 (Acid Medium)
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- 10. Balancing redox equations (pH basic)
Identify the atoms that are oxidised and reduced, using ox. no.
Write the half reactions for oxidation and reduction processes
Balance mass, so that the number of atoms of each element
oxidised/reduced is the same on both sides
If there is any O atoms present, balance them adding OH-
If there is H atoms present, balance them adding H2O
Balance charges with e-
Multiply the half-equations so that the number of e- lost in one equation
is equal to the e- gained in the other
Add the two half-equation together,cancel out any substances that
appear on both side.
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- 11. CrO4
2- + Fe2+ Cr3+ + Fe3+ (basic pH)
Identify the atoms oxidised and reduced, using ox. no.
Fe : +2 Cr : +6 O: -2 Cr3+: +3 Fe3+: +3
Fe oxidises (goes from +2 to +3) Cr reduces (goes from +6 to +3)
Write the half-reactions. Balance mass, if necessary using OH-
(to add O, double the number needed) and/or H2O (to add H)
Fe2+ Fe3+ Oxidation (reducing agent)
4H2O + CrO4
-2 Cr3+ + 8OH- Reduction (oxidising agent)
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- 12. CrO4
2- + Fe2+ Cr3+ + Fe3+ (basic pH)
Balance the charges using electrons (e-). Each e- adds a -1
Fe2+ Fe3+ + 1 e- (+2 = +3 – 1)
4H2O + CrO4
-2 + 3 e- Cr3+ + 8OH-(0 -2 -3 = +3 -8)
Multiply the half-equations so that e- lost in one = e-
gained in the other
3x(Fe2+ Fe3+ + 1e-)
1x(4H2O + CrO4
-2 + 3e- Cr3+ + 8OH-)
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- 13. CrO4
2- + Fe2+ Cr3+ + Fe3+ (basic pH)
Add the two half-equation together and cancel out
Any substances that appear on both side
3Fe2+ 3Fe3+ + 3e-
4H2O + CrO4
-2 + 3e- Cr3+ + 8OH-
3Fe2+ + 4H2O + CrO4
-2 + 3e- 3Fe3+ + 3e- +Cr3+ + 8OH-
Answer:
3Fe2+ + 4H2O + CrO4
-2 3Fe3+ +Cr3+ + 8OH-
+
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- 14. The End
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