Saved Bookmarks
| 1. |
Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned. I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V " Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V" Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V" Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V" O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V". While Fe^(2+) ion is stable, Mn^(2+) ion is not stable in acid solution because : |
|
Answer» `O_(2)` OXIDISES `MN^(2+)` to `Mn^(3+)` `Mn^(3+)+E^(-) to Mn^(2+)]xx4 "" cathode` `2H_(2)Oto 4H^(+)+O_(2)+4e^(-) "" ANODE` `4Mn^(3+)+2H_(2)O to 4Mn^(2+)+O_(2)+4H^(+)` `E_(cell)^(@)=E_(Mn^(3+)//Mn^(2+))^(@)-E_(H_(2)"O"//O_(2))^(@)` `=1.50+(-1.23)=0.27" V"` |
|