InterviewSolution
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InterviewSolution
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Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}` While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:A. `O_(2) "oxidises lt" n^(2+) "to" Mn^(3+)`B. `O_(2) "oxidises both" Mn^(2+) "to" Mn^(3+) "and" Fe^(2+) to Fe^(3+)`C. `Fe^(3+) "oxidises" H_(2)O to O_(2)`D. `Mn^(3+) "oxidises" H_(2)O "to" O_(2)` |
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Answer» Correct Answer - D In this case also, only the e.m.f of the cell involving the oxidation of `H_(2)O to O_(2)` by `Mn^(3+)` ion is positive `{:(Mn^(3+)+e^(-)rarrMn^(2+)xx4,E^(@)=+1.50 V),(2H_(2)Orarr4H^(+)+O_(2)+4e^(-),E^(@)=-1.23V),(Mn^(3+)+2h_(2)Orarr4Mn^(2+)+O_(2)+4H^(+),E^(@)=+0.27V):}` |
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