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40.05 mL of 1.0 M Ce^(4+) are required to titrate 20.0 mL of 1.0 M Sn^(2+) " to "Sn^(4+) . What is the oxidatin state of cerium in the reduction product ?

Answer» <html><body><p></p>Solution :The reactions occurring are : <br/> `Ce^(4+)+ "ne"^(-) rarr Ce^((4-n)+)` <br/> `Sn^(2+) rarr Sn^(4+)+2e^(-)` <br/> To balance the equation , (the no . Of electrons <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> = no of electrons <a href="https://interviewquestions.tuteehub.com/tag/gained-2665852" style="font-weight:bold;" target="_blank" title="Click to know more about GAINED">GAINED</a> ) multiply eq. (i) by 2 eq (ii) by n and add <br/> `2Ce^(4+)+ nSn^(2+) rarr Ce^((4-n)+)+Sn^(4+)` <br/> Moles of `Ce^(4+)` in 40.05mL of 1.0 M solution , <br/> `=(1.0)/(1000)xx40.05=40.05 xx10^(-3)mol` <br/> Now 2 mol of `Ce^(4+)` will oxidise n mole of `Sn^(2+)` <br/> `40.05 xx10^(-3) " mol of " Ce^(4+) ` will oxidise `Sn^(2+)` <br/> `=n/2 xx40.05xx10^(-3)` mol `=20.02n xx 10^(-3)` mol = `20.02 n xx10^(-3)` mol<br/> But moles of `Sn^(2+)` in 20.0 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of 1.0 M solution <br/>`=(1.0)/(1000)xx20.0=20.0xx10^(-3)` mol <br/> `:. 20.02n xx10^(-3) mol = 20.0 xx10^(-3)` mol <br/> `:. n = 1` <br/> Hence 1 mol of electrons are required in the <a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a> of each mol of `Ce^(4+)` ion. <br/> `Ce^(4+)+e^(-) rarr Ce^(3+)` <br/> `Ce^(3+)` is the reduction <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a>.</body></html>


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