(a) Explain why electrolysis of aqueous solution of NaCl gives H_2 at cathode and Cl_2 at anode. Write overall reaction. (b) The resistancee of a conductivity cell containing 0.000M KCl solution at 298K is 1500 Omega. Calculate the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 times 10^-3 S cm^-1.

(a) Explain why electrolysis of aqueous solution of NaCl gives H_2 at

1.	(a) Explain why electrolysis of aqueous solution of NaCl gives H_2 at cathode and Cl_2 at anode. Write overall reaction. (b) The resistancee of a conductivity cell containing 0.000M KCl solution at 298K is 1500 Omega. Calculate the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 times 10^-3 S cm^-1.
Answer» Solution :(a) Criteria: The Criteria is as FOLLOWS: The substance which possess higher standard reduction potential is reduced first at the cathode. Whereas the substance which possesses LOWER reduction potential than water will get oxidised at the anode. `NaClleftrightarrowNa^(+) (AQ)+Cl^(-) (aq)` `2H_2O (l) to O_2(g)+4H^++4e^-` AT cathode: `2H_2O(l)+2e^(-) to H_2+2OH^-) :E^@=-0.83V` `Na^(+) (aq) +e^(-) toNa(s), E^@=-2.71V` Atanode: `1/2 O_2 (g)+H_(2) (g) to H_2O, E^@=1.23V` `2CL^(-) (aq)+2e^(-) to Cl_2(g),E^@=-1.36V` (b) `R=1500 OmegaMolarity =0.001 KCl` Cell const.=? K (conductivity)=`0.146 times 10^-3 S cm^-1` Cell const.= Conductivity `times` OBS. resistance `=0.146 times 10^-3 times 1500` `=219 times 10^-3 cm^-1`