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In qualitative estimation of any element using an oxidising agent is essential to predict which substance (s) gets oxidised by the oxidising agent. In case more than one substance is getting oxidised then the oxidising agent gets distributed in all the reactions taking place. From this information and the data given below answer the following questions: Fe^(2+) rarr Fe^(3+) +e^(-) E^(@) =- 0.77V MnO_(4)^(-) +5e^(-) rarr Mn^(2+) + 4H_(2)O E^(@) = + 1.51V 2Cl^(-) rarr Cl_(2) +2e^(-) E^(@) =- 1.36V 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-) E^(@) =- 2.0V C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-) Cr_(2)O_(7)^(2-) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_(2)O E^(@) = + 1.33 volt Millimol of FeC_(2)O_(4) in the solution if 50ml of 0.1M KMnO_(4) is used for its oxidation in presence dilute HCl if 3.5 millimoles of Cl_(2) is obtained along with other products. |
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Answer» `(25)/(3)` m. MOL No of meq of `Cl^(-) =` no of meq of `Cl_(2) = 3.5 xx 2 = 7` (n-factor =2) `:.` no of meq of `FeC_(2)O_(4) = 25 - 7 = 18` `:.` no of m mol of `FeC_(2)O_(4) = 18//3 = 6` |
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