Which of the following gives `H_(2)` on cathode and `O_(2)` on anode on electrolysis by using platinum electrode ?

Options

A. Molten NaCl
B. Dilute solution of NaCl
C. Concentration solution of NaCl
D. Solid NaCl.

Answer

B

Explanation

Electrolysis of dilute solution of NaCl is actually electrolysis of water only. So on reduction of water `H_(2)` gas liberated on cathode and `O_(2)` gas liberated on anode.
On anode: `H_(2)O_((l)) to 2H_((aq))^(+)+(1)/(2)O_(2(g))` . . . (Oxidation)
`underline("On cathode : "2H_(2)O_((l))+2e^(-) to H_(2(g))+2OH_((aq))^(-) . . . ."(Reduction)")`
`3H_(2)O_((l)) to underset("Near cathode")(H_(2(g))+underset("Near anode")((1)/(2))O_(2(g))+(2H_(aq))^(+)+2OH_((aq))^(-))`