Oxid agents/Red Agents Solutions

1.a. reducing agent: CH4

oxidizing agent: O2, because each atom gets reduced from 0 to -2. By stealing electrons, it causes the oxidation of CH4.

b. reducing agent: C2O4-2, because each C gets oxidized from 3 to 4. By losing electrons it makes it possible for the Mn to get reduced.

oxidizing agent: MnO4-1, because each Mn atom gets reduced from +7 to +2

c. reducing agent: Li

oxidizing agent: CO2, because each C gets reduced from +4 to 0

d. reducing agent: Na

oxidizing agent: Cl2, because each Cl atom gets reduced from 0 to -1

e. reducing agent: Cl-1, because the Cl goes from -1 up to 0

oxidizing agent: MnO4-1, because each Mn atom gets reduced from +7 to +2

 

2. a. good reducing agents: Li, Na, and K

b. excellent oxidizing agents: F, O and Cl

 

3.a. CH4 + 2H2O CO2 + 8e + 8H+

2[O2 + 4e + 4H+ 2H2O]

Overall: CH4 + 2O2 CO2 + 2H2O

b. C2O4-2 + MnO4-1 Mn+2 + CO2

2[8H+ +5e + MnO4-1 Mn+2 + 4 H2O]

5[C2O4-2 2 CO2 + 2e]

Overall: 16 H+ + 2 MnO4-1 + 5 C2O4-2 10 CO2 + 2 Mn+2 + 8 H2O

c. MnO4-1 + H+1 + Cl-1-->Cl2 +Mn+2 + H2O

2 [8H+ +5e + MnO4-1 Mn+2 + 4 H2O]

5[2Cl-1Cl2 + 2e]

Overall: 16 H+ + 2 MnO4-1 + 10 Cl-12 Mn+2 + 8 H2O + 5 Cl2

 

4.               2 [4 H2O + 5e + MnO4-1 Mn+2 + 8 OH-]

5 [2Cl-1Cl2 + 2e]

Overall: 8 H2O + 2 MnO4-1 + 10 Cl-12 Mn+2 + 5 Cl2 + 16 OH-

5. MnO4-1 is a remarkably good oxidizing agent because Mn is in an unusually high oxidation state of +7. To get to something more stable(like +2,3,or 4 those found in nature), it has to gain electrons.

6. In the water molecule, both hydrogen and oxygen already exist in their common and stable oxidation states of +1 and -2 respectively. This is why water is normally a stable molecule.

 

But in the presence of a strong reducing agent like Na, water will act as an oxidizing agent and get reduced to hydrogen gas:

 

2 H2O + 2e- H2 + 2OH-

 

And if sunlight is available for chlorophyll and its molecular companions, water will act as a reducing agent:

 

2 H2O O2 + 4e- + 4 H+