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