Q4.72P - Chemistry: Molecular Nature Of Matter And Change

Question

Identify the oxidizing and reducing agents in the following:

$$\begin{gathered} \left(a\right)5H_2{C}_2{O}_4\left(aq\right)+2Mn{O}_4^{-}\left(aq\right)+6H^{+}\left(aq\right)\to 2M{n}^{2+}\left(aq\right)+10C{O}_2\left(g\right)+8H_{2}O\left(l\right) \\ \left(b\right)3Cu\left(s\right)+8H^{+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)\to 3C{u}^{2+}\left(aq\right)+2NO\left(g\right)+4H_{2}O\left(l\right) \end{gathered}$$

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 carbon in H 2 C 2 O 4

Oxidation state of Hydrogen is +1.

Oxidation state of Oxygen is -2.

Let, oxidation state of carbon in H₂C₂O₄ is x.

Now you can write,

 $$\begin{gathered} 2\left(+1\right)+2x+4\left(-2\right)=0 \\ 2x=+6 \\ x=+3 \end{gathered}$$

Step 2: Oxidation state of carbon in CO 2

Oxidation state of Oxygen is -2.

Let, oxidation state of carbon in CO₂ is y.

Now you can write,

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

Step 3: Oxidation state of manganese in MnO 4 -

Oxidation state of Oxygen is -2.

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

Now you can write,

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

Step 4: Oxidation state of manganese in Mn 2+

The oxidation state of manganese in Mn²⁺ is +2.

Step 5: Conclusion

In the given reaction,

$5H_2{C}_2{O}_4\left(aq\right)+2Mn{O}_4^{-}\left(aq\right)+6H^{+}\left(aq\right)\to 2M{n}^{2+}\left(aq\right)+10C{O}_2\left(g\right)+8H_{2}O\left(l\right)$

The oxidation state of carbon in H₂C₂O₄ is +3 which increases to +4 in CO₂. Therefore, H₂C₂O₄ undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.

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

Step 6: (b) Oxidation state of nitrogen in NO 3 -

Oxidation state of Oxygen is -2.

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

No you can write,

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

Step 7: Oxidation state of nitrogen in NO

Oxidation state of Oxygen is -2.

Let, oxidation state of nitrogen in NO is y.

No you can write,

 

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

Step 8: Oxidation state of copper in Cu

The oxidation state of the element copper in Cu is 0.

Step 9: Oxidation state of copper in Cu 2+

Oxidation state of copper in Cu²⁺ is +2.

Step 10: Conclusion

In the given reaction,

$3Cu\left(s\right)+8H^{+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)\to 3C{u}^{2+}\left(aq\right)+2NO\left(g\right)+4H_{2}O\left(l\right)$

The oxidation state of nitrogen in NO₃^- is +5 which decreases to +2 in NO. Therefore, NO3-undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

The oxidation state of copper in Cu is 0 which increases to +2 in Cu²⁺. Therefore, Cu undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.