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
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.
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Amount of charge passed `=1.70xx230` coulomb
Amount of actual charge passed `=90/100xx1.70xx230`
`=351.9` coulomb
No. of moles of Zn deposited by passing 351.9 coulomb of charge
`=1/(2xx96500)xx351.9=0.000182`
Molarity of `Zn^(2+)` ions after deposition of zinc
`=[0.160-(0.000182xx1000)/300]M`
`=0.154 M`
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