Answer
a) $N_{2}$ (g) is oxidized. $O_{2}$ (g) is reduced.
Oxidizing agent: $O_{2}$
Reducing agent: $N_{2}$
b) $CO$ (g) is oxidized. $O_{2}$ (g) is reduced.
Oxidizing agent: $O_{2}$
Reducing agent: $CO$
c) $SbCl_{3}$ (g) is oxidized. $Cl_{2}$ (g) is reduced.
Oxidizing agent: $Cl_{2}$
Reducing agent: $SbCl_{3}$
d) $K$ (s) is oxidized. $Pb^{2+}$ (aq) is reduced.
Oxidizing agent : $Pb^{2+}$
Reducing agent: $K$
Work Step by Step
First, find oxidation state of each atom in every redox reaction. After that, look at the increase or decrease in the oxidation state for any atom in the product side. Increase in oxidation state means oxidation. Decrease in oxidation state means Reduction. Substance which is oxidized is the reducing agent. Substance which is reduced is the oxidizing agent for that particular reaction.
a) In the left side we have $N_{2}$ and $o_{2}$. So, both N and O atom have oxidation state of '0'. (oxidation state of atom in a free element = 0).
In the product side we have $NO$. O have an oxidation state of -2 in a compound generally. So, N will have oxidation state of +2. So, Oxidation state of N is increased. So, $N_{2}$ is oxidized. Oxidation state of O is decreased. So, $O_{2}$ is reduced.
Similarly, we can find oxidation states of atoms for other equations by applying this method.
See below.
