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 + 2H₂Oà CO₂ + 8e + 8H⁺

2[O₂ + 4e + 4H⁺ à 2H₂O]

Overall: CH₄ + 2O₂ à CO₂ + 2H₂O

b. C₂O₄^-2 + MnO₄^-1 à Mn⁺² + CO₂

2[8H⁺ +5e + MnO₄^-1 à Mn⁺² + 4 H₂O]

5[C₂O₄^-2 à 2 CO₂ + 2e]

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

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

2 [8H⁺ +5e + MnO₄^-1 à Mn⁺² + 4 H₂O]

5[2Cl^-1àCl₂ + 2e]

Overall: 16 H⁺ + 2 MnO₄^-1 + 10 Cl^-1à2 Mn⁺² + 8 H₂O + 5 Cl₂

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

5 [2Cl^-1àCl₂ + 2e]

Overall: 8 H₂O + 2 MnO₄^{-1 +} 10 Cl^-1à2 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^- à H₂ + 2OH-

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

2 H₂O à O₂ + 4e^- + 4 H⁺