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 Zinc aded in the second titration wil

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 Zinc aded in the second titration wil
Answer» <html><body><p>oxidize `Fe^(2+)` to `Fe^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>+)` <br/><a href="https://interviewquestions.tuteehub.com/tag/reduce-1181332" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCE">REDUCE</a> `Fe^(3+)` to `Fe^(2+)`<br/>reduce `Fe^(3+)` to Fe<br/>reduce `Fe^(2+)` to Fe</p>Solution :In presence of <a href="https://interviewquestions.tuteehub.com/tag/zn-751409" style="font-weight:bold;" target="_blank" title="Click to know more about ZN">ZN</a> `Fe^(+3)` reduced to `Fe^(+2)`</body></html>