1.

A current of 0.20A is passed for 482 s through 50.0mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis?

Answer»

0.0159M
0.0199M
0.10M
0.030M

Solution :The QUANTITY of CHARGED passed `=(0.20 xx 482)/(96500)F=9.99 xx 10^(-4)F`
The reaction at cathode is `2H_(2)O+2e to H_(2)+2OH^(-)`
1F `=1" of mol "OH^(-)`
`9.99 xx 10^(-4)F=9.99 xx 10^(-4)" mol of OH"`
Concentration of `OH^(-)` in solution
After electrolysis `=(9.99 xx 10^(-4)"mol")/((50)/(1000)L)=0.0199M`


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