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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"`. Among the following, identify the correct statement :A. `Cl^(-)` ion is oxidised by `O_(2)`B. `Fe^(2+)`ion is oxidised by iodine.C. `I^(-)` ion is oxidised by chlorine.D. `Mn^(2+)` ion is oxidised by chlorine. |
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Answer» Correct Answer - C (c ) `2I^(-)+Cl_(2) to 2Cl^(-)+l_(2)` `I^(-)` ions is oxidised by `Cl_(2)` because e.m.f. of the cell comes out to be positive. `E_(cell)^(@)=E_(Cl_(2)//2Cl^(-))^(@)-E_(2I^(-)//I_(2))^(@)` `=1.36-0.54=0.82" V"` |
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