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A current of 1.70 A is passed through 30...

A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of a `ZnSO_(4)` for 230 s with a current efficiency of 90% . Find out the molarity of `Zn^(2+)` after the deposition Zn. Assume the volume of the solution to remain cosntant during the electrolysis.

Text Solution

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Faraday’s passed `=(1.7xx230)/96500=4.052xx10^(-)F`
`implies` Meg. Of `Zn^(2+)` reduced `=3.65`
Initially mrg. Of `Zn^(2+)=300xx0.16xx2=96`
`implies` Meg. Of `Zn^(2+)` remaining `=96-3.65=92.35 implies` Molarity
`Zn^(2+)=(92.35)/2xx1/300=0.154M`
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