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 ?

40.05 mL of 1.0 M Ce^(4+) are required to titrate 20.0 mL of 1.0 M Sn^

1.	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» Solution :The reactions occurring are : `Ce^(4+)+ "ne"^(-) rarr Ce^((4-n)+)` `Sn^(2+) rarr Sn^(4+)+2e^(-)` To balance the equation , (the no . Of electrons LOST = no of electrons GAINED ) multiply eq. (i) by 2 eq (ii) by n and add `2Ce^(4+)+ nSn^(2+) rarr Ce^((4-n)+)+Sn^(4+)` Moles of `Ce^(4+)` in 40.05mL of 1.0 M solution , `=(1.0)/(1000)xx40.05=40.05 xx10^(-3)mol` Now 2 mol of `Ce^(4+)` will oxidise n mole of `Sn^(2+)` `40.05 xx10^(-3) " mol of " Ce^(4+) ` will oxidise `Sn^(2+)` `=n/2 xx40.05xx10^(-3)` mol `=20.02n xx 10^(-3)` mol = `20.02 n xx10^(-3)` mol But moles of `Sn^(2+)` in 20.0 ML of 1.0 M solution `=(1.0)/(1000)xx20.0=20.0xx10^(-3)` mol `:. 20.02n xx10^(-3) mol = 20.0 xx10^(-3)` mol `:. n = 1` Hence 1 mol of electrons are required in the REDUCTION of each mol of `Ce^(4+)` ion. `Ce^(4+)+e^(-) rarr Ce^(3+)` `Ce^(3+)` is the reduction PRODUCT.