1.

A current of 0.2 ampere is passed for 600 sec. through 100 mL of 0.1M NaCl. If hydrogen and chlroine gas is produced at the cathode and anode respectively, calculate

Answer»

Solution :Anode : `2CL^(-) rarr Cl^(2)+2e`
Cathode : `2H^(-)+2e rarr H_(2)`
`[H^(+)]` of `H_(2)O` are used up at cathode
`= [OH^(-)]` of `H_(2)O` left free
`:.` EQ. of `Cl^(-)` used = Eq. of `[H^(+)]` used = Eq, of `[OH^(-)]` left
`= (i.t)/(96500) = (0.2 xx 600)/(96500) = 1.24 xx 10^(3)`
`:.` Mole of `{OH^(-)]` left `= 1.24 xx 10^(-3)`
`:.` MOLARITY of `[OH^(-)] = (1.24 xx 10^(-3))/(100//1000) = 1.24 xx 10^(-2)`
Also Eq. of`NaCl` left = Initial eq.-Eq. of `Cl^(-)` lost
`= 0.1 xx 0.1 - 1.24 xx 10^(-3)`
`= 8.76 xx 10^(-3) =`Mole of `Cl^(-)`
`M_(NaCl) = (8.76 xx 10^(-3))/(100) xx 1000`
`= 8.76 xx 10^(-20) M`


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