(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 whichis liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of CH_3COOH decreases on dilution.

(a) Write the cell reaction and calculate the emf of the following cel

1.	(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 whichis liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of CH_3COOH decreases on dilution.
Answer» Solution :`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.