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A current of `1.70A` is passed trhough `300.0 mL` of `0.160M` solution of `ZnSO_(4)` for `230s` with a current efficiency of `90%`. Find out the molarity of `Zn^(2+)` after the deposition of `Zn`. Assume the volume of the solution to remain constant during the electrolysis. |
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Answer» Correct Answer - `0.154M` `I=(1.70xx90)/(100)A` `:.` Equivalent of `Zn^(2+)` lost `=(It)/(96500)` `=(1.70xx90xx230)/(100xx96500)` `=3.646xx10^(-13)` Milliequivalent of `Zn^(2+)` lost `=3.646xx10^(-3)` Initially milliequivalent of `Zn^(2+)=300xx0.160xx2=96` `( :. Mxx2=N fo r Zn^(2+)impliesmEq=NxxV_((mL)))` `:.` Milliequivalent of `Zn^(2+)` left in solution `=96-3.646` `=92.354` `:. [ZnSO_(4)]=(92.254)/(2xx300)=0.154M` |
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