Question

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.

Answer 1

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"`