(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.

Explanation

(a). Because `H_(2)O` has higher value of `E_(red)^(@)` therefore `H_(2)` is liberated at cathode `Cl_(2)` is liberated at anode because of over voltage.
(b). (i). Cell constant, `G'="conductivity"(K)xx"resistance"(R)`
`G'=1.0xx10^(-4)xx100=10^(-2)cm^(-1)`
(ii). conductivity of solution Y is
`K=(G^(*))/(R)=(10^(-2))/(50)=2x10^(-4)Scm^(-1)`
(iii). `^^_(m)=(Kxx1000)/(M)=(2xx10^(4)xx1000)/(0.01)=20Scm^(2)"mole"^(-1)`