Chapter 7, Problem 41A

Interpretation Introduction

Interpretation:

The given chemical reaction needs to be balanced.

  ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3 }}\text{+ C }\to {\text{Al + CO}}_{\text{2}}$

Concept Introduction:

If the oxidation number is increases, then the species get oxidized.

If the oxidation number is decreased, then the species get reduced.

Answer to Problem 41A

The balanced reaction is shown below:

  ${\text{2Al}}_{\text{2}}{\text{O}}_{\text{3 }}\text{+ 3C }\to {\text{4Al + 3CO}}_{\text{2}}$

Explanation of Solution

The calculation of oxidation number is shown below:

The oxidation number of Al in Al₂O₃ = 2x + 3(-2) = 0

= 2x = 6

x = 3

The oxidation number of C(s) = 0

The oxidation number of Al(s) = 0

The oxidation number of C in CO2 = x + 2(-2) = 0

x = 4

The reduction half reaction is shown below:

  ${\text{Al}}^{\text{3+}}\text{+}{\text{3e}}^{\text{-}}\to \text{Al}$ …………… (1)

The oxidation half reaction is shown below:

  $\text{C}\to {\text{C}}^{\text{4+}}\text{+}{\text{4e}}^{\text{-}}$ ……………… (2)

To balance the number of electrons equation (1) is multiply by 4 and equation (2) is multiply by 3 as shown below:

  ${\text{4Al}}^{\text{3+}}\text{+}12{\text{e}}^{\text{-}}\to 4\text{Al}$ ………….. (3)

  $\text{3C}\to 3{\text{C}}^{\text{4+}}\text{+}12{\text{e}}^{\text{-}}$ ……………… (4)

The overall reaction is written by addition of equation of (3) and (4) as shown below:

  ${\text{4Al}}^{\text{3+}}\text{+}\text{3C}\to \text{4Al}\text{+}{\text{3C}}^{\text{4+}}$

Now transform the atoms in molecules:

  ${\text{2Al}}_{\text{2}}{\text{O}}_{\text{3 }}\text{+ 3C }\to {\text{4Al + 3CO}}_{\text{2}}$

Conclusion

The balanced reaction is shown below:

  ${\text{2Al}}_{\text{2}}{\text{O}}_{\text{3 }}\text{+ 3C }\to {\text{4Al + 3CO}}_{\text{2}}$