Q4.74P - Chemistry: Molecular Nature Of Matter And Change

Question

Identify the oxidizing and reducing agents in the following:

$\left(a\right)8H^{+}\left(aq\right)+6C{l}^{-}\left(aq\right)+Sn\left(s\right)+4N{O}_3^{-}\left(aq\right)\to SnC{l}_6^{2-}\left(aq\right)+4N{O}_2\left(g\right)+4H_{2}O\left(l\right)$

$\left(b\right)2Mn{O}_4^{-}\left(aq\right)+10C{l}^{-}\left(aq\right)+16H^{+}\left(aq\right)\to 5C{l}_2^{}\left(g\right)+2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)$

Step-by-Step Solution

Answer

You need to identify the oxidizing agent and reducing agent of the given reaction.

Step 1: (a) Oxidation state of tin in Sn

The oxidation state of element tin is 0.

Step 2: Oxidation state of tin in SnCl 6 2-

Oxidation state of Chlorine is -1.

Let, oxidation state of tin in SnCl₆^2- is x.

Now you can write,

 ^{$$\begin{gathered} x+6\left(-1\right)=-2 \\ x=+4 \end{gathered}$$}

Step 3: Oxidation state of nitrogen in NO 3 -

Oxidation state of Oxygen is -2.

Let, oxidation state of nitrogen in NO₃^- is y.

Now you can write,

 ^{$$\begin{gathered} y+3\left(-2\right)=-1 \\ y=+5 \end{gathered}$$}

Step 4: Oxidation state of nitrogen in NO 2

Oxidation state of Oxygen is -2.

Let, oxidation state of nitrogen in NO₂ is z.

Now you can write,

 ^{$$\begin{gathered} z+2\left(-2\right)=0 \\ z=+4 \end{gathered}$$}

Step 5: Conclusion

In the given reaction

$8H^{+}\left(aq\right)+6C{l}^{-}\left(aq\right)+Sn\left(s\right)+4N{O}_3^{-}\left(aq\right)\to SnC{l}_6^{2-}\left(aq\right)+4N{O}_2\left(g\right)+4H_{2}O\left(l\right)$

The oxidation state of tin in Sn is 0 which increases to +4 in SnCl₆^2-. Therefore, Sn undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.

The oxidation state of nitrogen in NO₃^- is +5 which reduces to +4 in NO₂. Therefore, NO₃^- undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

Step 6: (b) Oxidation state of manganese in MnO 4 -

Oxidation state of Oxygen is -2.

Let, oxidation state of manganese in MnO₄^- is x.

Now you can write,

 ^{$$\begin{gathered} x+4\left(-2\right)=-1 \\ x=+7 \end{gathered}$$}

Step 7: Oxidation state of manganese in Mn 2+

Oxidation state of manganese in Mn²⁺ is +2.

Step 8: Oxidation state of chlorine in Cl -

Oxidation state of chlorine in Cl^- is -1. 

Step 9: Oxidation state of chlorine in Cl 2

Oxidation state of chlorine in Cl₂ is 0. 

Step 10: Conclusion

In the given reaction

$2Mn{O}_4^{-}\left(aq\right)+10C{l}^{-}\left(aq\right)+16H^{+}\left(aq\right)\to 5C{l}_2^{}\left(g\right)+2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)$

 

The oxidation state of manganese in MnO₄^- is +7 which decreases to +2 in Mn²⁺. Therefore, MnO₄^- undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

The oxidation state of chlorine in Cl^- is -1 which increases to 0 in Cl₂. Therefore, Cl^- undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.