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On the basis of standard electrode potential values, suggest which of the following reactions would take place ? (Consult the book for E^(Ө) value) (a) Cu+Zn^(+2)toCu^(+2)+Zn (b) Mg+Fe^(+2)toMg^(+2)+Fe ( c) Br_(2)+2Cl^(-)toCl_(2)+2Br^(-) (d) Fe+Cd^(+2)toCd+Fe^(+2) |
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Answer» Solution :`E_(Cu^(+2)//Cu)^(@)=0.34V,E_(Zn^(+2)//Zn)^(@)=-0.76V` `E_(Mg^(+2)//Mg)^(@)=-2.37V,E_(Fe^(+2)//Fe)^(@)=-0.74V` `E_(Br_(2)//Br^(-))^(@)=+1.08V,E_(Cl_(2)//Cl^(-))^(@)=+1.36V` `E_(CD^(+2)//Cd)^(@)=-0.44V` (a) `E_(Cu^(+2)//Cu)^(@)=-0.34V,E_(Zn^(+2)//Zn)^(@)=-0.76V` `Cu+Zn^(+2)TOCU^(+2)+Zn` `E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)` = `E_(Zn^(+2)//Zn)^(@)-E_(Cu^(+2)//Cu)^(@)` = `-0.76-(+0.34)=-1.10V` Value of `E_("cell")^(@)` is negative so reaction is not possible. (B) `Mg+Fe^(+2)toMg^(+2)+Fe` `E_("cell")^(@)=E_(Fe^(+2)//Fe)^(@)-E_(Mg^(+2)//Mg)^(@)` = `-0.74-(-2.37)=+1.63V` Value of `E_("cell")^(@)` is positive so reaction is possible. ( c) `Br_(2)+2Cl^(-)toCl_(2)+2Br^(-)` `E_("cell")^(@)=E_(Br^(-)//Br_(2))^(@)-E_(Cl^(-)//Cl_(2))^(@)` = `+1.08-(+1.36)=-0.28V` Value of `E_("cell")^(@)` is negative so reaction is not possible. (d) `Fe+Cd^(+2)toCd+Fe^(+2)` `E_("cell")^(@)=E_(Cd^(+2)//Cd)^(@)-E_(Fe^(+2)//Fe)^(@)` = `-0.44-(-0.74)=+0.30V` Value of `E_("cell")^(@)` is positive so reaction is not possible. |
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