1.

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

Answer»

Molten NaCl
Dilute solution of NaCl
Concentration solution of NaCl
Solid NaCl.

Solution :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))^(-))`


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