1.

The passage of current liberates H_2 at cathode and Cl_2 at anode. The solution is

Answer»

copper chloride in water
NACL in water
`H_2SO_4`
Water

SOLUTION :Since DISCHARGE potential of watier is greater than that of sodium so water is reduced at cathode instead of `Na^(+)`.
Cathode: `H_(2)O+e^(-)to(1)/(2)H_(2)+OH^(-)`
Anode: `CL^(-)to(1)/(2)Cl_(2)+e^(-)`.


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