(a) Write the cell reaction and calculate the emf of the following cell at 298 K :
`Sn (s) | Sn^(2+) (0.004 M) | | H^(+) (0.020 M) | H^2 (g) (1 bar) | Pt (s)`

[Given : `E_(Sn^(2+)//Sn)^(@) = - 0.14 V`]
(b) Give reasons :
(i) On the basis of E° values, `O_(2)` gas should be liberated at anode but it is `Cl_(2)` gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of `CH_3COOH` decreases on dilution.

Explanation

`Sn(s) to Sn^(2+) (aq) + 2e^(-)`
`2H^(+) (aq) + 2e^(-) to H_(2)(g)`
`Sn(s) + 2H^(+) (aq) to Sn^(2+) (aq) + H_(2)(g)`
emf of the cell `=E^(@) -0.059/2 log 0.004/(0.020)^(2)`
(i) In the electrolysis of NaCl solution, the following oxidation reactions are possible :
`Cl^(-) (aq) to 1/2 Cl_(2) (g) + e^(-), E_("cell"^(@) = 1.36 V`
`2H_(2)O (l) to O_(2) (g) + 4H^(+) (aq) + 4e^(-), E_("cell")^(@) = 1.23 V`
The reaction at anode with lower value of E° is preferred and therefore, water should get oxidised in preference to `Cl^(-)` (aq). However, on account of overpotential of oxygen, reaction (0 is preferred. Therefore `Cl_(2)` gas is liberated.
(ii) Number of ions present in a given volume of the solution decrease on dilution. Therefore the conductivity decreases.