(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :
Sn(s) `|Sn^(2+) (0.004M)||H^(+) (0.020M) |H_(2)(g)` (1 bar )| Pt(s)
(Given `E_(Sn^(2+)//Sn)^(@) = -0.14V`).
(b) Give reasons :
(i) On the basis of `E^(@)` values , `O_(2)` gas should be liberated at anode but `Cl_(2)` gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of `CH_(3)COOH` decreases on dilution .

Explanation

(a) `Sn(s) | Sn^(2+) (0.004M) ||H^(+) (0.020M) | H_(2)(g) ` (1 bar ) |Pt(s)
Cell reaction : `Sn + 2H^(+) to Sn^(2+) + H_(2)`
`E_("cell") = overset(n=2)(E_("cell")^(@)) - (0.0591)/(chi)"log" ([Sn^(2+)])/([H^(+)]^(2)) = 0-(-0.14) - (0.0591)/(2) "log" (0.004)/(0.020xx 0.020)`
`= 0.14 - (0.059)/(2) "log" 10 = 0.14 - (cancel0.0591)/(cancel2) = 0.14 - 0.02955 = 0.111V.`
(b) (i) `E_("OX")^(@)` value for `O_(2)` is more than that of `E^(@)` for `Cl_(2)` but `Cl_(2)` is produced not `O_(2)` . This unexpected result is explained on the basis of the concept of ''over voltage'' , i.e., water needs greater voltage for oxidation to `O_(2)` as it is kinetically slow process than that needed for oxidation of `Cl^(-)` ions to `Cl_(2)` .This extra voltage required is called over voltage .
`{:(2Cl^(-)(aq) to Cl_(2) (g) + 2e^(-) " " E_(OX)^(@) = -1.36V) , (H_(2)O(l) to (1)/(2)O_(2)(g) + 2H^(+) (aq)+ 2e^(-) E_(OX)^(@) = -1.23V):}`
(ii) On dilution , although `CH_(3)COOH` is completely dissociated but concentration of ions per unit volume is so low that the conductivity decreases with dilution .