given `E_(S_(2)O_(8)^(2-)//SO_(4)^(2-)^(@)=2.05V` `E_(Br_(2)//Br^(-))^(@)=1.40V` `E_(Au^(3+)//Au)^(@)=1.10V` ,brgt `E_(O_(2)//H_(2)O)^(@)=1.20V` Which of the following is the strongest oxidizing agent ?

Using tha data given below is reducing potenial. E_(Cr_2O_7^(2-)//Cr^(3+))^@ =1.33 V , E_(Cl_2//Cl^-)^@ =1.36 V E_(MnO_4^(-)//Mn^(2+))^@ =1.51 V , E_(Cr^(3+) // Cr)^@ =- 0.74 V find out which of the following is the strongest oxidising agent.

Given E_(Cl_(2)//Cl^(-))^(@)=1.36V,E_(Cr^(3+)//Cr)^(@)=-0.74V E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V Among the following, the strongest reducing agent is

Given that E_(o_(2)//H_(2)O)^(Theta)= +1.23V , E_(S_(2)O_(8)^(2-)//SO_(4)^(2-))^(Theta)=2.05V , E_(Br_(2)//Br^(-))^(Theta)=+1.09V , E_(Au^(3+)//Au^(Theta))=+1.4V The strongest oxidizing agent is :

Given that E_(o_(2)//H_(2)O)^(Theta)= +1.23V , E_(S_(2)O_(8)^(2-)//SO_(4)^(2-))^(Theta)=2.05V , E_(Br_(2)//Br^(-))^(Theta)=+1.09V , E_(Au^(3+)//Au^(Theta))=+1.4V The strongest oxidizing agent is :

If E_(Au^(+)//Au)^(@) is 1.69 V and E_(Au^(3+)//Au)^(@) is 1.40 V, then E_(Au^(+)//Au^(3+))^(@) will be :

Given that E_(O_(2))^(ɵ)//H_(2)O=+1.23V , E_(S_(2)O_(8)^(2-)//SO_(4)^(-2))=2.05V E_(Br_(2)^(ɵ)//Br-=+1.09V , E_(Au^(ɵ)^(3+)//Au=+1.4V The strongest oxidizing agent is :

Given E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V Among the following, the strongest reducing agent is E_(Cr^(3+)//Cr)^(@)=-0.74V^(x),E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(Cl//Cl^(-))^(@)=1.36V Based on the data given above strongest oxidising agent will be

Given E_(cr^(3+)//Cr)^(0)=-0.74 V, E_(MnO_(4)//Mn^(2+))^(0)=1.51 cm E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(0)=1.33 V, E_(Cl//Cl^(-))^(0)=1.36 V Based on the data given above, strongest oxidising agent will be:

(a). Explain why electrolysis of an aqueous solution of NaCl gives H_(2) at cathode and Cl_(2) at anode. Given E_(Na^(+)//Na)^(@)=-2.71V,E_(H_(2)O//H_(2)^(@)=-0.83V E_(Cl_(2)//2Cl^(-))^(@)=+1.36V,E_(2H^(+)//(1)/(2)O_(2)//H_(2)O)^(@)=+1.23V (b). The resistance of a conductivity cell when filled with 0.05 M solution of an electrolyte X is 100Omega at 40^(@)C . the same conductivity cell filled with 0.01 M solution of electrolyte Y has a resistance of 50Omega . The conductivity of 0.05M solution of electrolyte X is 1.0xx10^(-4)scm^(-1) calculate (i). Cell constant (ii). conductivity of 0.01 M Y solution (iii). Molar conductivity of 0.01 M Y solution.

Using the data given below: E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=1.33V E_(Cl_(2)|Cl^(-))^(@)=1.36V E_(MnO_(4)^(-)|Mn^(2+))^(@)=1.51V E_(Cr^(3+)|Cr)=-0.74V In which option the order of reducing power is correct?