Answer
a)$ 3K (s) + Cr^{3+} (aq)-> 3K^{+} (aq) + Cr(s) $
b)$ Mg (s) + 2Ag^{+} (aq) -> Mg^{2+} (aq) + 2Ag (s)$
c) $2Al (s) + 3Fe^{2+} (aq) -> 2Al^{3+} (aq) + 3Fe (s) $
Work Step by Step
a) First, split the equation into two half reaction i.e. oxidation and reduction reaction.
Oxidation reaction: $K (s) -> K^{+} (aq)$
After balancing charge, we get $K(s) -> K^{+} (aq) + e$ .........(1)
Reduction reaction: $Cr^{3+} -> Cr(s) $
After balancing charge, we get $Cr^{3+} + 3e -> Cr(s) $ ..........(2)
Now, to eliminate e from reaction we have to multiply equation (1) with 3 & equation (2) with 1 and then add both equation we get
$ 3K (s) + Cr^{3+} (aq)-> 3K^{+} (aq) + Cr(s) $
b)Oxidation reaction: $Mg(s) -> Mg^{2+} (aq)$
After balancing charge, we get $Mg(s) -> Mg^{2+} (aq) + 2e$ .........(1)
Reduction reaction: $Ag^{+}(aq) -> Ag(s) $
After balancing charge, we get $Ag^{+} (aq)+ e -> Ag(s) $ ..........(2)
Now, to eliminate e from the equations we have to multiply equation (1) with 1 & equation (2) with 2 and then add both equation we get
$ Mg (s) + 2Ag^{+} (aq) -> Mg^{2+} (aq) + 2Ag (s)$
c) Oxidation reaction: $Al(s) -> Al^{3+} (aq)$
After balancing charge, we get $Al(s) -> Al^{3+} (aq) + 3e$ .........(1)
Reduction reaction: $Fe^{2+}(aq) -> Fe(s) $
After balancing charge, we get $Fe^{2+} (aq)+ 2e -> Fe(s) $ ..........(2)
Now, to eliminate e from the equations we have to multiply equation (1) with 2 & equation (2) with 3 and then add both equation we get
$2Al (s) + 3Fe^{2+} (aq) -> 2Al^{3+} (aq) + 3Fe (s) $
