Determine the Half reactions for oxidation and reduction of carbon

Question

Balance the following half-reaction showing the oxidation of carbon:

CH4​(g)CO2​(g)

Balance the following half-reaction showing the reduction of oxygen: 

O2​(g)H2​O(g)

Answer 1

Step 1:

Solve the half reaction for carbon by writing out the oxidation states of the carbon atom

CH₄ - > CO₂

C of methane (-4), C of carbon dioxide (+4)

The difference of electrons is 8

therefore,

C (4-) -> C (4+) + 8e^-

CH₄ -> CO₂ + 8e _{(adding electrons)}

Adding waters and H⁺s to balance

2H₂O + CH₄ -> CO₂ + 8e^- + 8H⁺

This is the first

Step 2: Solve the half reaction for oxygens by writing out the oxidation states of the oxygen atom

O₂ -> H₂O

O₂ (0) - > 2O^-1

2e^- + O₂ -> 2O^-

Repeat the steps for the previous half reaction

2e^- +O₂ -> 2H₂O

4H⁺ + 4e^- +O₂ -> 2H₂O

Mix the 2 but multiply the 2nd to balance out the electrons between the two

8H⁺ + 8e^- + 2H₂O + CH₄ + 2O₂ -> 4H₂O + CO₂ + 8e^- + 8H⁺

Cancel out products and reactants

CH₄ + 2O₂ -> 2H₂O + CO₂

Hope this helps.