Redox Reactions

Redox (portmanteau of the words: reduction-oxidation) reactions include all chemical reactions in which atoms have their oxidation state changed.

Simply put, redox reactions are a family of reactions that are concerned with the transfer of electrons between species.

• Oxidation: the loss of electrons and/or increase in oxidation state

• Reduction: the gain of electrons or a decrease in oxidation state 

• Oxidizing agent : Is an “electron acceptor” i.e removes electrons from a substance and is reduced in the process.

•  Reducing agent : Donates electrons and gets oxidized in the process.

Quick Tip: Think 69..go it? Good, now stop thinking and concentrate (thank you Benny sir! and a shout out to Z(J)ubin)

• If an element is in its highest oxidation state then it cannot be oxidized any further therefore it cannot act as a reducing agent and will always behave like an oxidizing agent (i.e will get reduced)

eg: KMnO₄  Mn is in +7 oxidation state

• Similarly, if an element is in its lowest oxidation state cannot be reduced any further and thus will always act as a reducing agent and be oxidized.

eg: H₂S S is in -2 oxidation state

• Element which is present in an intermediate oxidation state will act as a reducing as well as an oxidizing agent.

eg: Peroxide

Balancing Redox Reactions

• In acidic medium

Consider the following reaction:

Fe²⁺ + MnO₄^-  è Mn²⁺ + Fe₂³⁺     

Step 1: Make note of atoms that have undergone change (in oxidation state)

Fe²⁺  è  Fe₂³⁺          

Mn⁺⁷ è Mn²⁺

# Fe has been oxidised ( reducing agent) and Mn has been reduced (oxidizing agent) 

^{Step 2: Balance the atoms}

2 x Fe²⁺  è  Fe₂³⁺          

Mn⁺⁷ è Mn²⁺

Step 3: Make note of the the electron involved and multiply by suitable whole nos. to balance the electrons.

 2 x Fe²⁺  è  Fe₂^{3+     | 2 e-  multiplying by 5}

Mn⁺⁷ è Mn^{2+              |  5 e-  multiplying by 2}

Step 4: Add the two equations and compare with original equation

10Fe²⁺ + 2MnO₄^-  è 2Mn²⁺ + 5Fe₂³⁺

STOP: Before proceeding ensure that all atoms except oxygen and hydrogen are balanced.

Step 5: Balance Oxygen by adding water to the oxygen deficient side. (LHS in this case) (Since, the RHS contains 8 oxygen atoms we add 8 molecules of water to the LHS)

10Fe²⁺ + 2MnO₄^-  è 2Mn²⁺ + 5Fe₂³⁺  + 8H₂O

Step 6: Balance Hydrogen by adding H+ to the hydrogen deficient side.

Final Balanced Equation (acid medium):  16H⁺ + 10Fe²⁺ + 2MnO₄^-  è 2Mn²⁺ + 5Fe₂³⁺  + 8H₂O

Checking Your Work: Ensure that the no. of atoms and the net charge on both the sides are equal

• In basic medium

Consider the following reaction:

Al + NO₃^--  è Al(OH)₄^-  +NH₃

Follow steps 1 to 4: We obtain

8Al + 3NO₃^--  è 8Al(OH)₄^-  + 3NH₃ 

Step 5: Balance Oxygen by adding water to the oxygen deficient side. 

23H₂O +  8Al + 3NO₃^--  è 8Al(OH)₄^-  + 3NH₃

Step 6: Balance Hydrogen also by adding water to the the hydrogen deficient side and then add an equal amount of of OH- to the opposite side. ( In this case we add 5H₂O to the LHS and 5 OH- to the RHS)

5OH^- + 23H₂O +  8Al + 3NO₃^--  è 8Al(OH)₄^-  + 3NH₃  + 5H₂O

Upon simplification,

 Final Balanced Equation (basic medium):  5OH^- + 18H₂O +  8Al + 3NO₃^--  è 8Al(OH)₄^-  + 3NH₃  

Checking Your Work: Ensure that the no. of atoms and the net charge on both the sides are equal

Disproportionation Reactions: Oxidation and reduction in the same element. An element must exist in atleast 3 oxidation states for disproportionation to take place.

Consider the following disproportionation reaction (follow the above described steps to balance this equation)

HNO₂  è HNO₃ + NO

N⁺³ è N⁺⁵ (oxidized to HNO₃)       | 2 e- multiplying by 1

N⁺³  è N⁺² (reduced to NO)           | 1 e- multiplying by 2  

________________________________adding the two eqns.

3 N⁺³ è N⁺⁵ + 2 N⁺²

Comparing with original equation we get,

3HNO₂  è HNO₃ + 2NO

Balancing Oxygen (Final Equation)

3HNO₂  è HNO₃ + 2NO +H₂O

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That's all folks!

The Passive Observer Out!