InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two half-reactions of an electrochemical cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51" V"` `Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-),E^(@)=-0.15" V"` |
|
Answer» Multiply the first equation by 2 and second equation by 5. Add up these equations. `2MnO_(4)^(-)(aq)+16H^(+)(aq)+10e^(-) to 2Mn^(2+)(aq)+8H_(2)O(l),E_(cell)^(@)=+1.51" V"` `(5Sn^(2+)(aq) to 5Sn^(4)(aq)+10e^(-),E^(@)=-0.15" V")/(2MnO_(4)^(-)(aq)+16H^(+)(aq)+5Sn^(2+)(aq) to 2Mn^(2+)(aq)+5Sn^(4+)(aq)+8H_(2)O(l))` `E^(@)=(1.51-0.15)=1.36" V"` Since `E_(cell)^(@)` comes out to be positive, the reaction is reactant favoured. It is likely to proceed in the backward direction. |
|