Redox reactions involve the transfer of electrons between species. There are two types of agents involved - oxidizing agents that reduce other species by accepting electrons, and reducing agents that oxidize other species by donating electrons. Identification of redox reactions involves looking for a change in oxidation state between reactants and products. Balancing redox reactions uses the ion-electron method of writing and balancing half reactions for oxidation and reduction and combining them. Organic redox reactions use a similar process by writing oxidation and reduction half reactions and balancing mass, charge, and electrons.

1. SUALEHA IQBAL

2. REDOX REACTIONS:
“ Transfer of electrons between two species.”
INVOLVE TWO TYPES OF AGENTS:
OXIDING AGENT:
A compound that reduced is refer to
as oxidizing agent.
EXAMPLES: Br2 ,I2 , H2O2, KMnO4.
REDUCING AGENT:
A compound that is oxidize is refer to
as reducing agent.
EXAMPLES:
I- ,H2S, Zn.
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3. Types Of Species Involve In Redox Reactions:
OXIDIZE
SPECIE
REDUCE
SPECIE
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4. 4

5. IDENTIFICATION OF REDOX REACTIONS:
If the reaction is redox than it involve transfer of electrons from one atom, ion
or molecule to another i.e. change in oxidation state (can observe through
change in color).
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6. OXIDATION STATE
“It is the way to describe the number of electrons that have been transferred or shared
between atoms of different kind.”
Hydrogen =1 (except hydrides).
Oxygen=(-2).
Metal (+).
Compound = 0.
If two elements in a reaction change in oxidation state, one increasing,
the other decreasing,Then the reaction is redox

7. BALANCING REDOX REACTIONS:
From ion electron method:
For example
MnO4- (aq) + I- (aq) Mn2+ (aq) + I2(s)
•Identification of oxidize and reduce species:
Manganese (Mn) goes from a charge of +7 to a charge of +2. Since it gained 5 electrons, it is
being reduced.
Next we see that Iodine (I) goes from a charge of -1 to 0. Thus it lost electrons, and is
being oxidized.
•Divide equation in half reactions:
First half is the reduction half: MnO4- (aq) Mn2+ (aq)
Next we have the oxidation half: I-(aq) I2 (s)
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8. BALANCING PART:
•Four water molecules for 4 oxygen in MnO4.
•Balance hydrogen by adding hydrogen to other side
Reduction half:
MnO4- (aq) + 8H+ + 5e- Mn2+ (aq) +4H2O
•5 electrons on the side having +7 charge are added to balance charge.
Reduction half is balance.
• Now come to oxidation half equation:
•In The oxidation half is iodine, so we can balance it easily by adding another
iodine to the left side.
OXIDATION HALF :
2I- (aq) I 2(S)+2(e-)
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9. • Multiply reduction half cell by 2 and oxidation half equation by 5
for balancing complete equation.
2{MnO4-(aq)+8H +5e- Mn+2(aq) +H2O}
5{2I- (aq) I2(s) + 2e-}
Reduction half becomes:
2MnO4- (aq) + 16H+ (aq) + 10e- 2Mn2+ (aq) + 8H2O (l)
Oxidation half becomes:
10I- (aq) 5I2 (s) + 10e-.
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10. Adding both equations :
2MnO4- (aq) + 16H+ (aq) + 10e- 2Mn2+ (aq) + 8H2O (l)
10I- (aq) 5I2 (s) + 10e-
10I- (aq) + 2MnO4- (aq) + 16H+ (aq) 5I2 (s) + 2Mn2+ (aq) + 8H2O (l)
Final inspection :
Proton :
Oxygen:
Charge:
Balanced
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11. 2nd Example
Through oxidation state method:
EQUATION :
Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O
•Redox or not: (write oxidation states under equation)
Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O
0 +5 +2 +2
• Insert co-efficient so that the total decrease in oxidation state of one
Element equals the total increase in oxidation state of other element.
+2 ×3=+6
Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O
-3×2= -6
3Cu (S)+2HNO3(aq 3Cu+2(aq)+ 2NO (g)
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12. •Balance oxygen: 4H2O on right will balance oxygen.
•Balance hydrogen by adding 6protons on left side.
3Cu (S)+2HNO3(aq)+6H+ 3Cu+2(aq)+ 2NO (g) +4H2O
•Spectators ions: 6 NO3- on each side of equation
3Cu (S)+8HNO3(aq) 3Cu(NO3)2 (aq)+ 2NO (g) +4H2O(l)
BALANCED
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13. Balancing organic redox reactions:
oxidation of n-butyl alcohol with K2Cr2O7
CH3CH2CH2CH2-OH + K2Cr2O7 , H+ CH3CH2CH2COOH
half-reactions for the oxidation and the reduction involved:
oxidation: CH3CH2CH2CH2-OH CH3CH2CH2COOH
reduction: Cr2O72- Cr3+
Mass balance:
oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4 H+
reduction: Cr2O72- + 14H+ 2 Cr3+ + 7 H2O
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14. Charge balance
oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4e-
reduction: 6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O
No. of electron in oxidation half = No. of electron in reduction half
• 3 x{( CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4 e- )}
•2× { (6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O)}
•3 CH3CH2CH2CH2-OH + 3 H2O 3 CH3CH2CH2COOH + 12H+ + 12 e-
•12 e- + 2 Cr2O72- + 28 H+ 4 Cr3+ + 14 H2O
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15. Add the two half-reactions together
3 CH3CH2CH2CH2-OH + 3 H2O + 12e- + 2Cr2O72- + 28 H+
3 CH3CH2CH2COOH + 12 H+ + 12 e- + 4 Cr3+ + 14 H2O
Canceling out the electrons and extra waters, etc.
3 CH3CH2CH2CH2-OH + 2 Cr2O72- + 16 H+
3 CH3CH2CH2COOH + 4 Cr3+ + 11 H2O
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16. Spectator ions
3 CH3CH2CH2CH2-OH + 2K2Cr2O7 + 8H2SO4
3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2 K2SO4
Final equation:
3 CH3CH2CH2CH2-OH + 2 K2Cr2O7 + 8 H2SO4
3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2K2SO4
BALANCED
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17. reduced oxidize
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18. SUALEHA IQBAL..
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