(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

### Step-by-Step Solution

#### Part (a): Electrolysis of Aqueous NaCl

1. **Identify the Electrolytic Reactions**:
- In an aqueous solution of NaCl, the dissociation occurs as follows:
\[
\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-
...