Question

Identify the oxidizing and reducing agents in the following:

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

Short answer

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

Step-by-step solution

(a) Oxidation state of chromium in Cr2O72-

Oxidation state of Oxygen is -2.

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

Now you can write,

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

Oxidation state of chromium in Cr3+

Oxidation state of chromium in Cr³⁺is +3.

Oxidation state of sulfur in SO32-

Oxidation state of Oxygen is -2.

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

Now you can write,

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

Oxidation state of sulfur in SO42-

The oxidation state of Oxygen is -2.

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

Now you can write,

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

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.

(b) Oxidation state of nitrogen in NO3-

Oxidation state of Oxygen is -2.

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

Now you can write,

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

Oxidation state of nitrogen in NH3

Oxidation state of Hydrogen is +1.

Let, oxidation state of nitrogen in NH₃is y.

Now you can write,

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

Oxidation state of zinc in Zn

Oxidation state of element Zn is 0.

Oxidation state of zinc in Zn(OH)42-

Oxidation state of OH^- is -1

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

Now you can write,

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

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, Znundergoes oxidation in this given reaction. Hence, it acts as a reducing agent.