1. Oxidation-
Reduction
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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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