A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O IF 0.02 M K_(2)Cr_(2)O_(7) is used instead of 0.02 M KMnO_(4) its volume required in these titrations are respectively

A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+)

1.	A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O IF 0.02 M K_(2)Cr_(2)O_(7) is used instead of 0.02 M KMnO_(4) its volume required in these titrations are respectively
Answer» 25 mL, 40 mL 25 mL, 15 mL 20.8 mL, 33.3 mL 10.4 mL, 16.7 mL Solution :m.eqts of `Fe^(+2)` = mEq.ts of `KMnO_(4)` implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` In PRESSURE of Zn, m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` In pressure of Zn, me.eqtsof `Fe^(+3)` = m eqts of `KMnO_(4)` m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` m.eqts of `Fe^(+3)=40xx0.02xx5=4` Now in the second case `K_(2)Cr_(2)O_(7)` used instead of `KMnO_(4)`, m eq. `K_(2)Cr_(2)O_(7)` =m eq of `Fe^(+2)` `0.02xx6xxv=2.5` `implies V=(2.5)/(0.12)=20.8ml` In pressure of Zn m. eqts of `K_(2)Cr_(2)O_(7)` = m eq of `Fe^(+3)` `0.02xx6xxv=2` `impliesV=(4)/(0.12)=33.3ml`