Match the items of Column-I and Column-II on the basis of data given below: E_(F_(2)|F^(-))^(Theta)=2.87V,E_(Li^(+)|Li)^(Theta)=-3.5V, E_(Au^(3+)|Cu)^(Theta)=1.4V,E_(Br_(2)|Br^(-))^(Theta)=1.09V

Match the items of Column-I and Column-II on the basis of data given b

1.	Match the items of Column-I and Column-II on the basis of data given below: E_(F_(2)\|F^(-))^(Theta)=2.87V,E_(Li^(+)\|Li)^(Theta)=-3.5V, E_(Au^(3+)\|Cu)^(Theta)=1.4V,E_(Br_(2)\|Br^(-))^(Theta)=1.09V
Answer» SOLUTION :`(Ato3),(Bto1),(Cto7),(Dto5),(Eto4),(Fto2),(Gto6)` * The following emf SERIES is obtained by following `E^(@)` value * Any half cell reaction : A+`e^(-)toB` experiencing reduction`to`PRODUCT of reduction. oxidizing agent`to`reducing agent. (A) `F_(2)` : is non-metal and `E^(@)` is maximum. So `F_(2)` is REDUCED easily and is strongest oxidizing agent. (B) Li : is metal, apart from this Au is another metal. Between these two metals Li has high `E^(@)` value so Li is strongest oxidizng agent. (C) `Au^(3+)` is metal ion. it has high reduction potential and so `Au^(3+)` is oxidizing agent. (D) `Br^(-)` is negative ion. Oxidation half reaction: `Br^(-) to (1)/(2)Br` So, oxidation of `Br^(-)` is occur. `3Br^(-) to (3)/(2)Br_(2)+3e^(-)` (Oxidation) `E^(THETA)=1.09V` `Au^(3+)+3e^(-) to Au` (Reduction) `E^(Theta)=1.4V` (E) Au is inert metal because all metals have high `E^(@)` value. (F) `Li^(+)` is metal ions and standard reduction potential is `E^(@)` is low -3.5 V and so it is weak oxidizing agent. (G) `F^(-)` is negative ion and its oxidation `F^(-) to F_(2)` and so `F^(-)` is reducing agent and `E^(@)` is maximum so it is weak reducing agent.