Question

For the complete redox reactions given here, write the half-reactions and identify the oxidizing and re- ducing agents: (a) 4Fe 3O2 ¡ 2Fe2O3
(b) Cl2 2NaBr ¡ 2NaCl Br2 (c)Si2F2¡SiF4 (d) H2 Cl2 ¡ 2HC

Answer 1

Answer:

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

O₂           O                0          Atoms bonded to the same type of atoms

                                              have oxidation number 0

Fe₂O₃     O                -2          In compounds, except peroxides, oxygen

                                              has oxidation number -2                

               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:

• $O^0+2e^-->O^{-2}$

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:

• $Fe^0-3e^-->Fe^{+3}$

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

           

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:

• $Cl^0+1e^-->Cl^{-1}$

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:

• $Br^{-1}-1e^-->Br^{0}$

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

F₂            F               0           Atoms bonded to the same type of atoms

                                             have oxidation number 0

SiF₄         F               -1            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:

• $F^0+1e^-->F^{-1}$

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:

• $Si^{0}-4e^-->Si^{+4}$

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

H₂            H               0           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:

• $Cl^0+1e^-->Cl^{-1}$

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:

• $H^{0}-1e^-->H^{+1}$

4) Oxidizing agent: chlorine

    Reducing agent: hydrogen