On the basis of standard electrode potential values, suggest which of the following reactions would take place?

1.  Cu+Zn2+Cu2++Zn2.  Mg+Fe2+Mg2++Fe3.  Br2+2Cl-Cl2+2Br4.  Fe+Cd2+Cd+Fe2+


As we know that,
E°Cu2+/Cu=0.34 V, E°Zn2+/Zn=-0.76 V,E°Mg2+/Mg=-2.37 V, E°Fe2+/Fe=-0.74 V,E°Br2/Br-=+1.08 V, E°Cl2/Cl-=+1.36VE°Fe2+/Fe=-0.74 V, E°Cd2+/Cd=-0.44 V
(a) E°Cu2+/Cu=0.34 V and E°Zn2+/Zn=-0.76 V
Cu+Zn2+Cu2++Zn
In the given cell reaction, Cu is oxidised to Cu2+, therefore, Cu2+/Cu couple acts as anode and Zn2+ is reduced to Zn, therefore, Zn2+/Zn couple acts as cathode.
E°cell=E°cathode-E°anodeE°cell=-0.74-(+0.34)=-1.10 V
Positive value of E°cell indicates that the reaction will not occur.
(a) Mg+Fe2+Mg2++Fe
E°Mg2+/Mg=-2.37 V and E°Fe2+/Fe=-0.74 V
In the given cell reaction, Mg is oxidised to Mg2+ hence, Mg2+/Mg couple acts as anode and Fe2+ is reduced to Fe hence, Fe2+/Fe couple acts as cathode.
E°cell=E°cathode-E°anodeE°cell=-0.74-(-2.37)=+1.63 V
Positive value of E°cell 5 indicates that the reaction will occur.
(c) Br2+2Cl-Cl2+2Br-
E°Br2/Br-=+1.08 V and E°Cl2/Cl-=+1.36V
In the given cell reaction, Cl- is oxidised to Cl2 hence, Cl-/Cl2 couple acts as anode and Br2 is reduced to Br- hence; Br-/Br2 couple acts as cathode.
E°cell=E°cathode-E°anodeE°cell=+1.08-(+1.36)=-0.28 V
Negative value of E°cell indicates that the reaction will occur.
(d) Fe+Cd2+Cd+Fe2+
E°Fe2+/Fe=-0.74 V and E°Cd2+/Cd=-0.44 V
In the given cell reaction, Fe is oxidised to Fe2+ hence, Fe2+/Fe couple acts as anode and Cd2+ is reduced to Cd hence, Cd2+/Cd couple acts as cathode.
E°cell=E°cathode-E°anodeE°cell=-0.44-(-0.74)=+0.30 V
Positive value E°cell indicates that the reaction will occur.