Answer
a) Reduction reaction.
$MnO_{4}^{-} (aq) + 8H^{+} (aq) + 5e^{-} -> Mn^{2+} (aq) + 4H_{2}O (l) $
b) Oxidation reaction.
$Pb^{2+} (aq) + 2H_{2}O (l) -> PbO_{2} (s) + 4H^{+} (aq) + 2e^{-} $
c) Reduction reaction.
$2IO_{3}^{-} (aq) + 12H^{+} (aq) + 10e^{-} -> I_{2} (s) + 6H_{2}O (l) $
d) Oxidation reaction.
$SO_{2} (g) + 2H_{2}O (l) -> SO_{4}^{2-}(aq) + 4H^{+} (aq) + 2e^{-}$
Work Step by Step
a) First, find oxidation state of each atom. In $MnO_{4}^{-}$, oxidation state of O = -2. Let's say oxidation state of Mn = $x$. hence, $x + (-2) \times4 = -1$ By solving we get, oxidation state of Mn = +7. On the product side we have $Mn^{2+}$. Hence, oxidation state of Mn = +2 on the right side. oxidation state of Mn decreases from reactant to product side. hence, this is a reduction reaction.
To Balance the half reaction in acidic aqueous solution : first balance O. To balance O, add same no. of $H_{2}O$ on the opposite side. Here, we add 4$H_{2}O$ on the right side because we have 4 O atom present on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $8H^{+}$ on the left side to balance H. After that, balance charge. We have 7 +ve charge on the left side and 2 +ve charge on the right side. So, we have to add $5 e$ on the left side.
b) Find oxidation state of Pb on both side. In the left side, oxidation state of Pb = +2. In the right side, oxidation state of Pb = +4. (oxidation state of O= -2. so, the equation is $x + (-2)\times2 = 0$ by solving we get, $x= +4$)
So, this is an oxidation reaction.
Now, to balance the reaction, first balance O by adding $H_{2}O$ on the opposite side. Here, we add $2H_{2}O$ on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $4H^{+}$ on the right side to balance H. After that, balance charge. We have 2 +ve charge on the left side and 4 +ve charge on the right side. So, we have to add $2 e$ on the right side.
c) Find oxidation state of I on both side. On the left side, oxidation state of I = +5 and on the right side oxidation state of I = 0.
So, this is reduction reaction.
To Balance the half reaction : first balance I atom on both side. Add co-efficient 2 before $IO_{3}^{-}$. After that balance O. To balance O, add same no. of $H_{2}O$ on the opposite side. Here, we add 6$H_{2}O$ on the right side because we have 6 O atom present on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $12H^{+}$ on the left side to balance H. After that, balance charge. We have 10 +ve charge on the left side and no charge on the right side. So, we have to add $10 e$ on the left side.
d) First find oxidation state of S on the both side. In $SO_{2}$, oxidation state of S = +4 and In $SO_{4}^{2-}$, oxidation state of S= +6.
So, this is oxidation reaction.
Now, to balance the reaction, first balance O by adding $H_{2}O$ on the opposite side. Here, we add $2H_{2}O$ on the left side because we have 2 excess O on the right side. After that balance H. To balance H, add $H^{+}$. Here, add $4H^{+}$ on the right side to balance H. After that, balance charge. We have no charge on the left side and 2 +ve charge on the right side. So, we have to add $2 e$ on the right side.
