1.

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. I_(2)+2e^(-)to2I^(-)""E^(0)=0.54 Cl_(2)+2e^(-)to2Cl^(-)""E^(0)=1.36 Mn^(3+)+e^(-)toMn^(2+)""E^(0)=1.50 Fe^(3+)+e^(-)toFe^(2+)""E^(0)=0.77 O_(2)+4H^(+)+4e^(-)to2H_(2)O""E^(0)=1.23 Q. Among the following, identify the correct statement

Answer»

Chloride ion oxidised `O_(2)`
`FE^(2+)` is oxidised by iodine
IODIDE ion is oxidised by chlorine
`Mn^(2+)` is oxidised by chlorine

Solution :`2I^(-)+Cl_(2)toI_(2)+2Cl^(-)`
`E^(o)=E_(I^(-)//I_(2))^(o)+E_(Cl_(2)//CL^(-))^(o)`
`=-0.54+1.36`
`E^(o)=0.82V`
`E^o` is positive HENCE iodide ion is oxidized by chlorine.


Discussion

No Comment Found