Q4.75P - Chemistry: Molecular Nature Of Matter And Change

Question

Identify the oxidizing and reducing agents in the following:

$$\begin{gathered} \left(a\right)8H^{+}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)+3S{O}_3^{2-}\left(aq\right)\to 2C{r}^{3+}\left(aq\right)+3S{O}_4^{2-}\left(aq\right)+4H_{2}O\left(l\right) \\ \left(b\right)N{O}_3^{-}\left(aq\right)+4Zn\left(s\right)+7O{H}^{-}\left(aq\right)+6H_{2}O\left(l\right)\to 4Zn{\left(OH\right)}_4^{2-}\left(aq\right)+N{H}_3\left(aq\right) \end{gathered}$$

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 chromium in Cr 2 O 7 2-

Oxidation state of Oxygen is -2.

Let, oxidation state of chromium in Cr₂O₇^2- is x.

Now you can write,

 ^{$$\begin{gathered} 2x+7\left(-2\right)=-2 \\ 2x=+12 \\ x=+6 \end{gathered}$$}

Step 2: Oxidation state of chromium in Cr 3+

Oxidation state of chromium in Cr³⁺ is +3.

Step 3: Oxidation state of sulfur in SO 3 2-

Oxidation state of Oxygen is -2.

Let, oxidation state of sulfur in SO₃^2- is y.

Now you can write,

 ^{$$\begin{gathered} y+3\left(-2\right)=-2 \\ y=+4 \end{gathered}$$}

Step 4: Oxidation state of sulfur in SO 4 2-

The oxidation state of Oxygen is -2.

Let, oxidation state of sulfur in SO₄^2- is z.

Now you can write,

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

Step 5: Conclusion

In the given reaction

$8H^{+}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)+3S{O}_3^{2-}\left(aq\right)\to 2C{r}^{3+}\left(aq\right)+3S{O}_4^{2-}\left(aq\right)+4H_{2}O\left(l\right)$

 

The oxidation state of chromium in Cr₂O₇^2- is +6 which decreases to +3 in Cr³⁺. Therefore, Cr₂O₇^2- undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

The oxidation state of sulfur in SO₃^2- is +4 which increases to +6 in SO₄^2-. Therefore, SO₃^2- undergoes oxidation in this given reaction. Hence, it acts as a reducing 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.

Now you can write,

 

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

Step 7: Oxidation state of nitrogen in NH 3

Oxidation state of Hydrogen is +1.

Let, oxidation state of nitrogen in NH₃ is y.

Now you can write,

 ^{$$\begin{gathered} y+3\left(+1\right)=0 \\ y=-3 \end{gathered}$$}

Step 8: Oxidation state of zinc in Zn

Oxidation state of element Zn is 0. 

Step 9: Oxidation state of zinc in Zn(OH) 4 2-

Oxidation state of OH^- is -1 

Let, oxidation state of zinc in Zn(OH)₄^2- is z.

Now you can write,

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

Step 10: Conclusion

In the given reaction

$N{O}_3^{-}\left(aq\right)+4Zn\left(s\right)+7O{H}^{-}\left(aq\right)+6H_{2}O\left(l\right)\to 4Zn{\left(OH\right)}_4^{2-}\left(aq\right)+N{H}_3\left(aq\right)$

 The oxidation state of nitrogen in NO₃^- is +5 which decreases to -3 in NH₃. Therefore, NO₃^- undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.

The oxidation state of zinc in Zn is 0 which increases to +2 in Zn (OH)₄^2-. Therefore, Zn undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.