Question

Balance the following equation by the oxidation number method.
$C{r_{\left( s \right)}} + Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} \to Cr{\left( {N{O_3}} \right)_{3\left( {aq} \right)}} + P{b_{\left( s \right)}}$

Answer 1

Oxidation number method is used to balance the redox reactions, where they track the number of electrons to balance the equation. The idea behind this is that electrons are transferred between charged atoms.

Complete step by step answer:
We can balance the redox reaction using two methods. One is the oxidation number and the other one is the ion-electron or half equation method. In the oxidation number method we said we track the electrons involved in them.
Let us look into the question
$C{r_{\left( s \right)}} + Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} \to Cr{\left( {N{O_3}} \right)_{3\left( {aq} \right)}} + P{b_{\left( s \right)}}$
First of all we can write the skeletal equation representing the chemical change
$Cr + P{b^{2 + }} \to C{r^{3 + }} + Pb$
Now we can indicate the oxidation number involved in the reaction, so that we can understand the change in oxidation number.
Here, $Cr$and $Pb$ will have an oxidation number of zero as both are in its solid state. And the $P{b^{2 + }}$and $C{r^{3 + }}$will have oxidation numbers of 2 and 3 respectively. Therefore, we can say that the oxidation number of $Cr$increases and oxidation number of $Pb$will increase
i.e.,$C{r_{\left( s \right)}} \to {\text{C}}{{\text{r}}^{3 + }} + 3{{\text{e}}^ - }$ --(1)
$P{b^{2 + }}_{\left( {aq} \right)} + 2{e^ - } \to {\text{P}}{{\text{b}}_{\left( {\text{s}} \right)}}$ --(2)
Here, in order to remove and balance this equation we multiply the equation (1) by 2 and equation (2) by 3. Then we will have
$2C{r_{\left( s \right)}} \to 2{\text{C}}{{\text{r}}^{3 + }} + 6{{\text{e}}^ - }$ --(3)
$3P{b^{2 + }}_{\left( {aq} \right)} + 6{e^ - } \to 3{\text{P}}{{\text{b}}_{\left( {\text{s}} \right)}}$ --(4)
Now, we add (3) and (4), then the $6{{\text{e}}^ - }$ on both sides cancel out to give
$2C{r_{\left( s \right)}} + 3P{b^{2 + }}_{\left( {aq} \right)} \to 2C{r^{3 + }} + 3P{b_{\left( s \right)}}$
By removing the skeleton part we get the whole equation as
$2C{r_{\left( s \right)}} + 3Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} \to 2Cr{\left( {N{O_3}} \right)_{3\left( {aq} \right)}} + 3P{b_{\left( s \right)}}$
Now, the equation is balanced.

Hence, the correct answer is option (A)

Note: Apart from the oxidation number method we can also use the Ion-electron method to balance the reaction. This is based on balancing the half reaction separately. Here, we only balance the H and O atoms on both sides.