Метод полуреакций

видео

1)https://study.com/academy/lesson/redox-oxidation-reduction-reactions-definitions-and-examples.html

2) объяснение 1 примера

http://myweb.astate.edu/mdraganj/Redox.html

Рабочий лист

Balance the following redox reactions:

1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution) 

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution) 

3.   MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution) 

Рабочий лист

Balance the following redox reactions:

1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution) 

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution) 

3.   MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution) 

Рабочий лист

Balance the following redox reactions:

1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution) 

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution) 

3.   MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution) 

Рабочий лист

Balance the following redox reactions:

1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution) 

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution) 

3.   MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution) 

  
  

1.   1.   Bi(OH)₃   +     SnO₂^2-    --------->     SnO₃^2-   +  Bi    (basic solution)

Step 1. Break into half-reactions:

     Bi(OH)₃    ------->    Bi 
      SnO₂^2-     ------->    SnO₃^2-

Step 2. Balance atoms other than H and O

     Bi(OH)₃    ------->    Bi 
      SnO₂^2-     ------->    SnO₃^2-

Step 3. Balance O by adding H₂O

                  Bi(OH)₃    ------->    Bi  +  3 H₂O 
    H₂O  +   SnO₂^2-     ------->    SnO₃^2-

Step 4. Balance H by adding H⁺

       3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3 H₂O 
        H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +    2H⁺

Step 5. Balance charge by adding electron(s)

3 e^-   +  3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3 H₂O 
            H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +   2 H⁺  +  2 e^-

Step 6. Electrons lost = electrons gained

 (3 e^- + 3 H⁺   +   Bi(OH)₃    ------->    Bi  +  3  H₂O) x 2 
 (           H₂O  +   SnO₂^2-     ------->    SnO₃^2-  +   2 H⁺  +  2 e^- ) x 3

Gives: 
 6 e^- +  6 H⁺ + 2 Bi(OH)₃     ------->   2 Bi     +   6 H₂O 
        3 H₂O  +   3 SnO₂^2-     ------->  3 SnO₃^2-  +    6H⁺  +   6 e^-

Step 7. Cancel like terms and add half reactions

 2 Bi(OH)₃    +     3 SnO₂^2-       ------->   2Bi   +  3 SnO₃^2-  +  3 H₂O

Step 8. Since it is a basic solution, add OH^-  for each H⁺ (add OH^- to both sides of eqn)

No H⁺, finished 

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution)

Step 1.  Break into half-reactions:

      S₂O₃^2-    ------->    S₄O₆^2- 
              I₂    ------->     I^-

Step 2.  Balance atoms other than H and O

   2 S₂O₃^2-    ------->     S₄O₆^2- 
              I₂    ------->     2 I^-

Step 3. Balance O by adding H₂O (already balanced)

      2  S₂O₃^2-   ------->     S₄O₆^2- 
                  I₂    ------->      2 I^-

Step 4. Balance H by adding H⁺  (No H's)

   2 S₂O₃^2-    ------->     S₄O₆^2- 
               I₂    ------->      2 I^-

Step 5. Balance charge by adding electron(s)

   2  S₂O₃^2-    ------->     S₄O₆^2-     + 2 e^- 
  2 e^-   +  I₂    ------->      2 I^-

Step 6. Electrons lost = electrons gained  (2 e^- lost, 2 e^- gained)

   2  S₂O₃^2-    ------->    S₄O₆^2-     + 2 e^- 
  2 e^- + I₂    ------->      2 I^-

Step 7. Cancel like terms and add half reactions

      2  S₂O₃^2-      +  I₂           ------->    S₄O₆^2-     +   2 I^-

Answer - Problem 3

3.   MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution)

Step 1. Break into half-reactions:

  MnO₄^-    ------->    MnO₂ 
          I^-     ------->     I₂

Step 2. Balance atoms other than H and O

  MnO₄^-    ------->    MnO₂ 
      2  I^-     ------->     I₂

Step 3. Balance O by adding H₂O

  MnO₄^-    ------->   MnO₂   +   2 H₂O 
      2  I^-     ------->     I₂

Step 4. Balance H by adding H⁺

 4 H⁺ +  MnO₄^-    ------->    MnO₂    +   2 H₂O 
                  2  I^-     ------->     I₂

Step 5. Balance charge by adding electron(s)

3 e^- +  4 H⁺ +  MnO₄^-    ------->   MnO₂       + 2 H₂O 
                               2  I^-     ------->     I₂   +    2 e^-

Step 6. Electrons lost = electrons gained

 (3 e^- +  4 H⁺ +  MnO₄^-    ------->    MnO₂       + 2 H₂O) x 2 
 (                             2  I^-     ------->     I₂   +    2 e^- ) x 3

Gives: 
 6 e^- + 8 H⁺ + 2 MnO₄^-    ------->   2 MnO₂       + 4 H₂O 
          6 I^-      ------->     3  I₂   +   6 e^-

Step 7. Cancel like terms and add half reactions

  8 H⁺ + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂       + 4 H₂O +  3 I₂

Step 8. Since it is a basic solution, add OH^-  for each H⁺ (add OH^- to both sides of eqn)

8 OH^-   +    8 H⁺  + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂    + 4 H₂O +  3  I₂   +   8 OH^-

9. Combine H⁺ with OH^- to form H₂O

8 OH^-   +    8 H⁺  + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂   +  4 H₂O +   3 I₂   +   8 OH^- 
Gives: 
      8 H₂O   +    2 MnO₄^-   +   6 I^-       ------->   2 MnO₂   +   4 H₂O +  3 I₂   +   8 OH^-

10. Cancel like terms:

   4 H₂O   +    2 MnO₄^-   +   6 I^-       ------->   2 MnO₂    +  3  I₂   +   8 OH^- 
 

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