Explain oxidation and reduction , respectively.

Question

Calculate the coefficients of the redox reactions of ①-③, and finish the ionic equations.The equation of oxidation half reaction and the equation of reduction half reaction are multiplied independently in order to equalize the electron count, followed by addition of them to complete the ionic equation
①    MnO₄⁻+ H₂O₂ + H⁺⟶Mn^2₊+O₂+H₂O    (in acidic solution)
oxidation half reaction     MnO₄⁻ + 8H⁺ + 5e⁻ ⟶ Mn²⁺ + 4H₂O

reduction half reaction     H₂O₂ ⟶ O₂ + 2H⁺+2e⁻

 

②    MnO₄⁻ + H₂O₂ ⟶ MnO₂ + O₂ + H₂O + 2OH⁻   (in basic solution)

oxidation half reaction     MnO₄− + 2H₂O+3e⁻ ⟶ MnO₂ + 4OH⁻
reduction half reaction     H₂O₂ + 2OH⁻⟶ O₂+2H₂O+2e⁻

 

 

③   Cr₂O₇²⁻+ C₂H₅OH + H⁺⟶ Cr³⁺ + CH₃COOH + H₂O
oxidation half reaction  Cr₂O₇²⁻ + 14H⁺ +6e⁻⟶2Cr³⁺+7H₂O
reduction half reaction  C₂H₅OH + H₂O ⟶ CH₃COOH + 4H⁺ + 4e⁻

Answer 1

Oxidation is the loss of electrons.

Reduction is the gaining of electrons.

Balance each half reaction separately, and then equalize electrons by multiplying by whatever factor necessary.  In acidic solution, use H⁺ and H₂O to balance.  In basic solution, use OH^- and H₂O to balance.

 

(1)  

MnO₄− + 8H⁺ + 5e⁻ ⟶ Mn²⁺ + 4H₂O [x2] multiply by 2 to get 10 electrons

H₂O₂ ⟶ O₂ + 2H⁺+2e⁻  [x5]  multiply by 5 to get 10 electrons

 

2MnO₄^- + 16H⁺ + 10e^- + 5H₂O₂ ==> 2Mn²⁺ + 8H₂O + 5O₂ + 10H⁺ + 10e^- 

Canceling and combining you get:

2MnO₄^- + 6H⁺ + 5H₂O₂ ==> 2Mn²⁺ + 8H₂O + 5O₂ 

THIS IS THE FINAL BALANCED EQUATION. IT IS BALANCED FOR MASS AND CHARGE.

 

You should be able to do the others by following this example.