Bài 13: Phản ứng oxi hóa – khử

Cân bằng các phản ứng oxi hóa – khử

19/21

Cân bằng các phản ứng oxi hóa – khử sau đây bằng phương pháp thăng bằng electron:

a) NaBr + Cl2 → NaCl + Br2              

b) Fe2O3 + CO →  Fe + CO2

c) CO + I2O5  CO2 + I2                   

d) Cr(OH)3 + Br2 + OH-  CrO42- + Br- + H2O

e) H+ + MnO­4- + HCOOH →  Mn2+ + H2O + CO2

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Giải thích

a) NaBr + Cl2 → NaCl + Br2  

Na+1Br−1    +  Cl02  →  Na+1Cl−1  +  Br02

Br−1  →  Br0    +1eCl0    +1e  →  Cl−1  

1  × 1  × Br−1  →  Br0    +1eCl0    +1e  →  Cl−1 

 

 Br−1  +  Cl0  →  Br0  +  Cl−1

2NaBr + Cl22NaCl + Br2

b) Fe2O3 + CO →  Fe + CO2

Fe+32O−23  +  C+2O−2  →  Fe0  +  C+4O−22

Fe+3  +3e  →  Fe0C+2    →    C+4    +2e

2  × 3  × Fe+3  +3e  →  Fe0C+2    →    C+4    +2e

 

 2Fe+3  +  3C+2  →  2Fe0  +  3C+4

Fe2O3 + 3CO →  2Fe + 3CO2

c) CO + I2O5  CO2 + I2

C+2O−2  +  I+52O−25  →  C+4O−22  +  I02

C+2   →   C+4  +2eI+5   +5e  →  I0

5  × 2  × C+2   →   C+4  +2eI+5   +5e  →  I0

 

 5C+2  +  2I+5  →  5C+4  +  2I0

5CO + I2O5  5CO2 + I2

d) Cr(OH)3 + Br2 + OH-  CrO42- + Br- + H2O

Br0  +1e  →  Br−CrOH3+ 5OH−→   CrO42−+ 4H2O  +3e

3 × 1  ×Br0  +1e  →  Br−CrOH3+ 5OH−→   CrO42−+ 4H2O  +3e

 

 CrOH3+ 3Br+ 5OH−→ CrO42−+ 3Br−+ 4H2O

2Cr(OH)3 + 3Br2 + 10OH-     2CrO42- + 6Br- + 8H2O

e) H+ + MnO­4- + HCOOH →  Mn2+ + H2O + CO2

H++ MnO4−+ H+1C+2O−2O−2H+1  → Mn2++ H2O + C+4O−22

H+1C+2O−2O−2H+1  →  C+4O−22  +2H+ +2e8H++ MnO4− +  5e→   Mn2++ 4H2O

5  × 2 × H+1C+2O−2O−2H+1  →  C+4O−22  +2H+ +2e8H++ MnO4− +  5e→   Mn2++ 4H2O

 

 16H++ 2MnO4− +  5H+1C+2O−2O−2H+1    →   2Mn2++ 8H2O  +  5 C+4O−22  +10H+

6H+ + 2MnO­4- + 5HCOOH →  2Mn2+ + 8H2O + 5CO2