Answer
a) $PbO_{2} (s) + 2I^{-} (aq) + 4H^{+}(aq) -> Pb^{2+} (aq) + I_{2}(s) + 2H_{2}O(l) $
$b) 5SO_{3}^{2-}(aq) + 2MnO_{4}^{-}(aq) + 6H^{+}(aq) -> 5SO_{4}^{2-}(aq) + 2Mn^{2+}(aq) +3H_{2}O(l)$
c) $S_{2}O_{3}^{2-}(aq) + 4Cl_{2}(g) + 5H_{2}O(l) -> 2SO_{4}^{2-}(aq) + 8Cl^{-}(aq) + 10H^{+} (aq) $
Work Step by Step
a) First, find oxidation state of each atom. In $PbO_{2}$, oxidation state of O = -2. Let's say oxidation state of Pb = $x$. hence, $x + (-2) \times2 = 0$ By solving we get, oxidation state of Pb = +7. On the product side we have $Pb^{2+}$. Hence, oxidation state of Pb = +2 on the right side. oxidation state of Pb decreases from reactant to product side. hence, this is a reduction reaction: $PbO_{2} (s) -> Pb^{2+} (aq) $
To Balance the half reaction : first balance O. To balance O, add same no. of $H_{2}O$ on the opposite side. Here, we add 2$H_{2}O$ on the right side because we have 2 O atom present on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $4H^{+}$ on the left side to balance H. After that, balance charge. We have 4 +ve charge on the left side and 2 +ve charge on the right side. So, we have to add $2 e$ on the left side.
So, the reduction reaction : $PbO_{2} (s) + 4H^{+}(aq) + 2e -> Pb^{2+} (aq) + 2H_{2}O(l) $
Now, oxidation state of I in $I^{-}$ is -1 and oxidation state of I in $I_{2}$ is O. So, the oxidation reaction is: $I^{-} (aq) -> I_{2}(s) $
To balance the half reaction: balance atom in both side. Add co-efficient 2 before $I^{-}$. After that balance charge. We have 2 -ve charge on the left side and No charge on the right side. So, we have to add 2e on the right side.
So, the oxidation reaction: $ 2I^{-} (aq) -> I_{2}(s) + 2e^{-} $
Now adding both oxidation and reduction reaction we get
$PbO_{2} (s) + 2I^{-} (aq) + 4H^{+}(aq) -> Pb^{2+} (aq) + I_{2}(s) + 2H_{2}O(l) $
b) 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 reduction reaction: $ MnO_{4}^{-}(aq) -> Mn^{2+}(aq) $
To Balance the half reaction : 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.
$ MnO_{4}^{-}(aq) + 8H^{+}(aq)+5e^{-} -> Mn^{2+}(aq) +4H_{2}O(l)$
Now, Find oxidation state of S on both side. In the left side, oxidation state of S = +4. In the right side, oxidation state of S = +6.
So, this is an oxidation reaction: $SO_{3}^{2-}(aq) -> SO_{4}^{2-}(aq)$
Now, to balance the reaction, first balance O by adding $H_{2}O$ on the opposite side. Here, we add $H_{2}O$ on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $2H^{+}$ on the right side to balance H. After that, balance charge. We have 2 -ve charge on the left side and no charge on the right side. So, we have to add $2 e$ on the right side.
So, the oxidation reaction become: $SO_{3}^{2-}(aq) + H_{2}O -> SO_{4}^{2-}(aq) + 2H^{+} + 2e$
now, we have to multiply oxidation reaction with co-efficient 5 and reduction reaction with co-efficient 2 so that no. of electrons in both the half reactions will be equal.
$ 2MnO_{4}^{-}(aq) + 16H^{+}(aq) + 10e -> 2Mn^{2+}(aq) + 8H_{2}O$
$5SO_{3}^{2-}(aq) +5H_{2}O-> 5SO_{4}^{2-}(aq) + 10H^{+} +10e$
After adding both equations we get
$5SO_{3}^{2-}(aq) + 2MnO_{4}^{-}(aq) + 6H^{+}(aq) -> 5SO_{4}^{2-}(aq) + 2Mn^{2+}(aq) +3H_{2}O(l)$
c) First, find oxidation state of each atom. In $Cl_{2}$, oxidation state of Cl = 0. On the product side we have $Cl^{-}$. Hence, oxidation state of Cl = -1 on the right side. oxidation state of Cl decreases from reactant to product side. hence, this is reduction reaction: $ Cl_{2}-> Cl^{-} $
To Balance the half reaction : first balance atom on both side. Add co-efficient 2 before $Cl^{-}$. 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 left side.
$ Cl_{2} + 2e-> Cl^{-} $
Now, Find oxidation state of S on both side. In the left side, oxidation state of S = +2. In the right side, oxidation state of S = +6.
So, this is an oxidation reaction: $S_{2}O_{3}^{2-}(aq) -> SO_{4}^{2-}(aq)$
Now, to balance the reaction, first balance S atom on both side. Add co-efficient 2 before $SO_{4}^{2-}$. After that, balance O by adding $H_{2}O$ on the opposite side. Here, we add $5H_{2}O$ on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $10H^{+}$ on the right side to balance H. After that, balance charge. We have 2 -ve charge on the left side and 6 +ve charge on the right side. So, we have to add $8 e$ on the right side.
So, the oxidation reaction become: $S_{2}O_{3}^{2-}(aq)+ 5H_{2}O-> 2SO_{4}^{2-}(aq) + 10H^{+} + 8e$
now, we have to multiply oxidation reaction with co-efficient 1 and reduction reaction with co-efficient 4 so that no. of electrons in both the half reactions will be equal.
$ 4Cl_{2} + 8e-> 8Cl^{-} $
$S_{2}O_{3}^{2-}(aq)+ 5H_{2}O-> 2SO_{4}^{2-}(aq) + 10H^{+} + 8e$
After adding both equations we get
$S_{2}O_{3}^{2-}(aq) + 4Cl_{2}(g) + 5H_{2}O(l) -> 2SO_{4}^{2-}(aq) + 8Cl^{-}(aq) + 10H^{+} (aq) $
