(a) The electrical resistance of a column of `0.05 M KOH` solution of length 50 cm and area of cross-section `0.625 cm^2` is `5 xx 10^3` ohm. Calculate its resistivity, conductivity and molar conductivity.
(b) Predict the products of electrolysis of an aqueous solution of `CuCI_2` with platinum electrodes.
[Given : `E_(Cu^(2+)//Cu)^(o)=+0.34V, E_(((1)/(2)CI_(2) // CI^(-) ))^(o)=+1.36V, E_(H^(+) //H_(2) (g),Pt)^(o) = 0.00V, E_(((1)/(2) O_(2)//H_(2) O))=+1.23 V`]
OR
(a) Calculate e.m.f. of the following cell :
`Zn(s)//Zn^(2+) (0.1 M )||(0.01 M)Ag^(+)//Ag(s)`
[Given : `E_(Zn^(2+)//Zn)^(o)=-0.76V, E_(Ag^(+)//Ag)^(o) =+0.80 V, log 10=1`]
(b) X and Y are two electrolytes. On dilution molar conductivity of 'X' increases 2.5 times while that Y increases 25 times. Which of the two is a weak electrolyte and why ?

Explanation

(a) `A=0.625 cm^(2) , R=5 xx 10^(3) Omega, I=50 cm`
Given, `R=(rhol)/(A)` or `rho= (RA)/(l) = (5xx10^(5) Omega xx 0.625 cm^(2) )/( 50 cm)`
or `rho= 0.0625 xx 10^(5) Omega cm =6250 Omega cm`
`k=((1)/( 6250))S cm^(-1) =0.00016 S cm^(-1)`
Molar conductivity, `A_(m) =(k xx 1000)/( C ) =(0.00016 S cm^(-1) xx 1000 cm^(3) L^(-1))/(0.05 "mol"L^(-1))`
Molar conductivity, `A_(m) =3.2 S cm^(2) mol^(-1)`
(b) At cathode
There are two cations in race, `Cu^(2+) and H^(+)
Cu^(2+) (aq) + 2e^(-) to Cu (s), E^(n) =+0.34 V`
`H^(+) (aq) +e^(-) to (1)/(2) H_(2) (g), E^(o) =0.0`
Copper is deposited at the cathode because of higher reduction potential.
At Anode
`CI^(-) to (1)/(2) CI_(2) +e^(-) " "E^(o) =+1.36 V`
`2H_(2) O (i) to O_(2) (g) + 4H^(+) (aq) +4e^(-) " "E^(o) =+1.23 V`
The anion with lower reduction potential i.e., `CI^(-)` ion should be deposited at the anode but due to overpotential oxygen is liberated at the anode.
OR
(a) `E^(o) = E_(Ag^(+) //Ag)^(o) -E_(Zn^(2+)//Zn)^(o)`
`=0.80-(-0.76)=1.56 V`
According to Nernst equation, we have
`E=E^(o) -(0.059)/(2) log (0.1)/(0.01^(2))`
or `E=1.56 -0.0295 log (0.01)/(1 xx 10^(-4) )`
or `E=1.56 0-0.0295 xx 3`
or `E=1.56 -0.0885 = 1.4715 V`
(b) Y is a weak electrolyte.
This is because weak electrolytes have lower degree of dissociation at higher concentration and hence for such electrolytes, the change in `A_m` with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of the solution that contains 1 mol of electrolyte. In such cases, `A_m` increases steeply.