InterviewSolution
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InterviewSolution
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`E_(RP)^(@)` and their respective half reactions for some change are given below: i) `Mn^(7+)(aq)+8H^(+)+5erarrMn^(2+)(aq)+4H_(2)O(l),` `E^(@)=+1.52V` ii) `Mn^(7+)(aq)+4H^(+)+3erarrMnO_(2)(s)+2H_(2)O(l),` `E^(@)=+2.26V` iii) `Mn^(7+)(aq)+2erarrMn^(5+)(aq),` `E^(@)=+0.56V("acid med.")` `Mn^(3+)+erarrMn^(2+)`, `E^(@)=+1.51V("acil med.")` v) `Mn^(4+)+erarrMn^(3+),` `E^(@)=+0.95V("acid med.")` The `E^(@)` for, `Mn^(4+)+2erarrMn^(2+),` is equal to:A. `1.23V`B. `2.46V`C. `0.56V`D. `-0.56V` |
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Answer» Correct Answer - A `{:(Mn^(+3)+e^(-)rarrMn^(2+),E^(@)=1.51V),(Mn^(+4)+2e^(-)rarrMn^(3+),E^(@)=0.954V),(Mn^(4+)+2e^(-)rarrMn^(2+),E^(@)=(1.51+0.95)/(2)):}` |
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