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On the basis of standard electrode potential values suggest which of the following reactionwould take place ? (i) Cu+Zn^(2+)+Zn (ii) Mg+Fe^(2+)+Fe (iii) Br_(2)+2CI^(-)rarrCI_(2)+2Br (iv) Fe +Cd^(2+)rarrCd+Fe^(2+) |
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Answer» Solution :(i) from the table `E_(Cu^(2+)//Cu)=+0.34 V "and" E_(Zn^(2+)//Zn))^(@)=-0.76V` The reaction `Cu+Zn^(2+) rarr Cu^(2+)+Zn` is FEASIBLE if `E_("cell")^(@)` is +v in this cell buy gets oxidised to `Cu^(2+)` therefore `Cu^(2+)//Cu` redox COUPLE acts as the anode further since `Zn^(2+)` getsreduced to Zn therefore `Zn^(2+)//Zn` acts as the cathode `therefore E_(cell)^(@)=E_("anode")^(@)=E_(Cu(Zn^(2+)//Zn)^(@)=-0.76-0.34=-1.10 v` since `E_("cell")^(@)` of the reaction `Cu+zn^(2+)rarrCu^(2+)rarrCu^(2+)Zn` is -ve therefore this reaction does not occur (ii) from `E_(Mg^(2+)//Mg)^(@)=-2.37 v` and `E_(Fe^(2+)//Fe)^(@)=-0.74 V` The reactoin `Mg+Fe^(2+)rarrMg^(2+)+Fe` is possible if `E_(cell)^(@)` is +ve here `Mg^(2+)//Mg`redox couple acts as the anode while `Fe^(2+)//Fe` acts the cathode (iii) From `E_(Br^(-)//Br_(2))^(@)=+1.08 V` and `E_(CI^(-)//CI_(2))^(@)=+1.36 v` The reaction `Br_(2)+21CI^(-)rarrCI_(2)+2Br^(-)` will occur if `E_(cell)^(@)`is +ve in this cel `Br_(2)//Br^(-)` redox couple wil actsas the cahode and `CI_(2)//CI^(-)` will act as the anode `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=E^(2)_(Br^(2+)//Fe)=-0.74 V and E_(Cd^(2+)//Cd)^(@)=-0.44V` The reaction `Fe+Cd^(2+)rarrCd+Fe^(2+)` wil occur if `E_(cell)^(@)` is +ve here in his reaction `Fe^(2+)//Fe` ELECTRODE will act as the anode while`Cd^(2+)//Cd` electrode will act as the cathode `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=E_(Cd^(2+)//Cd)^(@)-e_(Fe^(2+)//Fe)^(@)rarrCd+Fe^(2+)` will occur |
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