Chapter U5, Problem 11RE

Interpretation Introduction

Interpretation:The balanced equation for the oxidation of lead, $\text{Pb}$ should be determined.

Concept Introduction:Chemical equation is written in such a way that the symbolic representation of reaction represents the reaction taking place in the system. The reactants are written on the left-hand side and the products are written on the right-hand side of the equation and are separated by an arrow, two or more reactants and products are separated by “+”.

The reactions for those the number of atoms of each element in the reactant and the product, the side are equal, such reactions are said to be a balanced chemical equation.

Answer to Problem 11RE

The balanced equation for the oxidation of lead $\text{Pb}$ is

${\text{2Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2PbO(s)}$

Explanation of Solution

The oxidation of lead means the addition of oxygen to lead which results in the formation ofa compound named lead (II) oxide, $\text{PbO}$ . The reaction for the oxidation of lead is

${\text{Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ PbO(s)}$

The above reaction is not balanced as the number of O atom on the reactant side is 2 and that on the product side is 1. So, to balance the reaction, the coefficient 2 is written before Pb and PbO on the reactant and product side. Hence, the balanced reaction is

${\text{2Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2PbO(s)}$

The oxidation of lead to lead (II) oxide is confirmed by oxidation states as.

The oxidation state is defined as the charge(s) that an atom would have when an electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

While determining the oxidation state of the compound, the element with greater electronegativity is assigned with a negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

Since, $\text{Pb}$ and ${\text{O}}_{\text{2}}$ are in its stable elemental state so, the oxidation state of each is zero for $\text{PbO}$ .The oxidation is determined as:

The oxidation state of $\text{O}$ is assigned as -2 and the oxidation state of $\text{Pb}$ is assigned as x.

Since $\text{PbO}$ has no charge, so the sum of the oxidation state must be zero.

so:

$\begin{array}{c}x+\left(-2\right)=0\end{array}$

   $\begin{array}{c}x=+2\end{array}$

So, the oxidation for each element in the reaction is:

  $\begin{array}{l}{\text{ 2Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2PbO(s) }\end{array}$

  $\begin{array}{c}\text{Oxidation states: 0 0 For Pb = +2 }\end{array}$

   $\begin{array}{c}\text{ For O = -2 }\end{array}$

Since, the oxidation state of $\text{Pb}$ increases from 0 to +2 so, it undergoes oxidation and the oxidation state of $\text{O}$ decreases from 0 to -2, so it undergoes reduction.

Conclusion

The balanced equation for the oxidation of lead $\text{Pb}$ is:

${\text{2Pb(s) + O}}_{\text{2}}\text{(g) }\to \text{ 2PbO(s)}$