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निष्क्रिय इलेक्ट्रोडों का प्रयोग कर सोडि...

निष्क्रिय इलेक्ट्रोडों का प्रयोग कर सोडियम ब्रोमाइड के विद्दुत अपघटन से प्राप्त उत्पादों को प्रकृति की प्रागुक्ति कीजिए । दिया हुआ है -
`E_(Na^(+)//Na)^(@)=-2,71V,E_(Br//Br^-)^(@)=+1.08V,E_(H_2O //(1)/(2)H_(2),OH^(-1))^(@)=-0.83V` तथा ` E_((1)/(2)O_(2),2H^(+)//H_(2)O)^(@)=1.23V`.

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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 ?

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 ?

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 ?

(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.

(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.

Calculate the e.m.f. of the following cell at 298K : Pt(s)|Br_(2)(l)|Br^(-)(0.010M)||H^(+)(0.030M)|H_(2)(g)(1"bar")|Pt(s) Given : E_((1)/(2)Br_(2)//Br^(-))^(@)=+1.08V .