These are four questions. See each answer with its explanation below.
Explanation:
(a) 4Fe + 3O₂ →  2Fe₂O₃
1) Determine the oxidation numbers of each atom in every compound or molecule:
Unit    atom   oxidation   rule
              number
       Fe        0      Isolated atoms have oxidation number 0
Oâ‚‚ Â Â Â Â Â O Â Â Â Â Â Â Â Â 0 Â Â Â Â Â Atoms bonded to the same type of atoms
                       have oxidation number 0
Fe₂O₃   O         -2      In compounds, except peroxides, oxygen
                       has oxidation number -2        Â
        Fe        +3      Charge balance: 2(+3) + 3(-2) = 0
2) Reducing half-reaction:
i) Identify the oxidizing agent: the oxidizing agent gets reduced, reducing its oxidation number, by gaining electrons. Oxygen reduced its oxidation number from 0 to -2, so this is the oxidizing agent.
ii) Write the half-reaction:
3) Oxidizing half-reaction
i) Idenfity the reducing agent: the reducing agent gets oxidized, increasing its oxidation number by releasing electrons. Iron increased its oxidation number from 0 to +3, so it is the reducing agent.
ii) Write the half-reaction:
4) Oxidizing agent: oxygen
  Reducing agent: iron
(b) Cl₂ + 2NaBr → 2NaCl + Br₂
1) Determine the oxidation number of each atom in every compound or molecule:
Unit    atom   oxidation   rule
              number
Cl₂      Cl        0      Atoms bonded to the same type of atoms
                       have oxidation number 0
NaBr     Na       +1      In compounds alkaly metals have oxidation
                       number +1
       Br        -1      Charge balance: +1 + (-1) = 0
     Â
NaCl    Na        +1      Alkaly metal
       Cl         -1      Charge balance: +1 + (-1) = 0
Br₂     Br         0      Atoms bonded to the same type of atoms
                       have oxidation number 0
2) Reducing half-reaction:
i) Identify the oxidizing agent: the oxidizing agent gets reduced, reducing its oxidation number, by gaining electrons. Chlorine reduced its oxidation number from 0 to -1, so this is the reducing agent.
ii) Write the half-reaction:
3) Oxidizing half-reaction
i) Idenfity the reducing agent: the reducing agent gets oxidized, increasing its oxidation number by releasing electrons. Bromine increases its oxidation number from -1 to 0, so it is the reducing agent.
ii) Write the half-reaction:
4) Oxidizing agent: chlorine
  Reducing agent: bromine
(c) Si₂ + F₂ → SiF₄
1) Determine the oxidation number of each atom in every compound or molecule:
Unit    atom   oxidation   rule
              number
Si₂      Si        0      Atoms bonded to the same type of atoms
                       have oxidation number 0
Fâ‚‚ Â Â Â Â Â Â F Â Â Â Â Â Â Â 0 Â Â Â Â Â Atoms bonded to the same type of atoms
                       have oxidation number 0
SiFâ‚„ Â Â Â Â F Â Â Â Â Â Â Â -1 Â Â Â Â Â Â When F react with a metaloid its oxidation
                        number is - 1
        Si        +4      Charge balance: (+4) + 4(-1) = 0
2) Reducing half-reaction:
i) Identify the oxidizing agent: the oxidizing agent gets reduced, reducing its oxidation number, by gaining electrons. Fluorine reduced its oxidation number from 0 to -1, so this is the reducing agent.
ii) Write the half-reaction:
3) Oxidizing half-reaction
i) Idenfity the reducing agent: the reducing agent gets oxidized, increasing its oxidation number by releasing electrons. Silicon increases its oxidation number from 0 to +4, so it is the reducing agent.
ii) Write the half-reaction:
4) Oxidizing agent: fluorine
  Reducing agent: silicon
(d) H₂ + Cl₂ → 2HCl
1) Determine the oxidation number of each atom in every compound or molecule:
Unit    atom   oxidation   rule
              number
Hâ‚‚ Â Â Â Â Â Â H Â Â Â Â Â Â Â 0 Â Â Â Â Â Atoms bonded to the same type of atoms
                        have oxidation number 0
Cl₂      Cl        0      Atoms bonded to the same type of atoms
                        have oxidation number 0
HCl      H        +1       Hydrogen has oxidation number +1 when
                        forms acids.
        Cl        -1       Charge balance: (+1) + (-1) = 0
2) Reducing half-reaction:
i) Identify the oxidizing agent: the oxidizing agent gets reduced, reducing its oxidation number, by gaining electrons. Chlorine reduced its oxidation number from 0 to -1, so this is the reducing agent.
ii) Write the half-reaction:
3) Oxidizing half-reaction
i) Idenfity the reducing agent: the reducing agent gets oxidized, increasing its oxidation number by releasing electrons. Hydrogen increased its oxidation number from 0 to +1, so it is the reducing agent.
ii) Write the half-reaction:
4) Oxidizing agent: chlorine
  Reducing agent: hydrogen
