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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» <html><body><p>25 mL, <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> mL<br/>25 mL, 15 mL<br/>20.8 mL, 33.3 mL<br/>10.4 mL, 16.7 mL</p>Solution :m.eqts of `Fe^(+2)` = mEq.ts of `KMnO_(4)` <br/> implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` <br/> In <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of Zn, <br/> m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` <br/> In pressure of Zn, <br/> me.eqtsof `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> m.eqts of `Fe^(+3)=40xx0.02xx5=4` <br/> Now in the second case `K_(2)Cr_(2)O_(7)` <br/> used instead of `KMnO_(4)`, <br/> m eq. `K_(2)Cr_(2)O_(7)` =m eq of `Fe^(+2)` <br/> `0.02xx6xxv=2.5` <br/> `implies V=(2.5)/(0.12)=20.8ml` <br/> In pressure of Zn <br/> m. eqts of `K_(2)Cr_(2)O_(7)` = m eq of `Fe^(+3)` <br/> `0.02xx6xxv=2` <br/> `impliesV=(4)/(0.12)=33.3ml`</body></html>


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