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(a) Explain why electrolysis of aqueous ...

(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`.

Text Solution

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(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`
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