Answer
a) Reduction reaction.
$S(s) + 2H^{+} (aq) + 2e^{-} -> H_{2}S(g)$
b) Reduction reaction.
$S_{2}O_{8}^{2-} (aq) + 2e^{-} -> 2SO_{4}^{2-} (aq)$
c) Reduction reaction.
$Cr_{2}O_{7}^{2-} (aq) + 14H^{+} (aq) + 6e^{-} -> 2Cr^{3+} (aq) + 7H_{2}O(l)$
d) Oxidation reaction
$NO (g) + 2H_{2}O(l) -> NO_{3}^{-} (aq) + 4H^{+} (aq) + 3e^{-}$
Work Step by Step
a) first find oxidation state of S in each side. In $S$, oxidation, state of S = 0. In $H_{2}S$, oxidation state of S = -2.
Hence, this is reduction reaction. (oxidation number decreases)
To Balance the half reaction : first balance H. To balance H, add $H^{+}$. Here, add $2H^{+}$ on the left 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 left side.
b) first, find oxidation state of S in each side. In $S_{2}O_{8}^{2-}$, oxidation state of S = +7. (oxidation state of O = -2. Let's say oxidation state of S = $x$, Hence, $2x + (-2)\times8 = -2$ By solving we get $x= +7$)
In $SO_{4}^{2-}$ oxidation state of S = +6.
So, this is reduction reaction. (oxidation number decreases)
To balance, first balance S in both the side. So, we have to put co-efficient 2 before $SO_{4}^{2-}$. Now, all atoms are balanced. After that, balance charge. We have 2 -ve charge on the left side and 4 -ve charges on the right side. so, we have to add $2e$ on the left side.
c) First find oxidation state of Cr in both side. In $Cr_{2}O_{7}^{2-}$, oxidation state of O = -2. Let's say oxidation state of Cr = $x$, Hence, $2x + (-2) \times7 = -2$ by solving we get $x = +6$ So, oxidation state of Cr = +6. On the right side, Cr have oxidation state of +3.
So, this is reduction reaction (oxidation number decreases).
To Balance the half reaction : first balance Cr atom. To balance Cr, add co-efficient 2 before $Cr^{3+}$. Then balance O. To balance O, add same no. of $H_{2}O$ on the opposite side. Here, we add $7H_{2}O$ on the right side because we have 7 O atom present on the left side. After that balance H. To balance H, add $H^{+}$. Here, add $14H^{+}$ on the left side to balance H. After that, balance charge. We have 12+ve charge on the left side and 6 +ve charge on the right side. So, we have to add $6 e$ on the left side.
d) First find oxidation state of N on both side. In $NO$, oxidation state of N = +2. In $NO_{3}^{-}$, oxidation state of N = +5.
So, this is oxidation reaction. (oxidation number increases)
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 no charge on the left side and 3 +ve charge on the right side. So, we have to add $3 e$ on the right side.
