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2.68xx10^(-3) moles of solution containi...

`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Text Solution

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Step 1 To write the reduction and oxidation half reaction
Reducton : `Mno_(4)^(-) + 8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
oxidation `A^(n+)+3H_(2)OrarrAO_(3)^(-)+6H^(+)+(5-n)E^(-)`
Step 2 To find out the value of n
since in a redox reaction number of electrons lost = number of electrons gained therefore multiple oxidant of Eq (i) i.e `MnO_(4)^(-)` by (5-n) and reductant of Eq (ii) i.e `A^(n+)` by 5 adn equate we have (5-n) `MnO_(4)^(-)=5 A^(n+) i.e (5-n)` moles of `MnO_(4)^(-)` will oxidise `An^(n+)` = moles
Equationg the values of moles of `A^(n+)` actually oxidised =`2.68xx10^(-3)` moles
Equating the value of Eq (iii) and (iv) we have `(5)/(5-n) xx1.61xx10^(-3)=2.68xx10^(-3) or 5xxx1.61=(5-n)xx2.68 or 2.68 n = 5(2.68-1.61)=5xx1.07=5.35 or n =(5.35)/(2.68)=2`
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