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Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question.
`I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54`
`Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36`
`Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50`
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77`
`O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23`
Among the following, identify the correct statement.

A

Chloride ion oxidised `O_(2)`

B

`Fe^(2+)` is oxidised by iodine

C

Iodide ion is oxidised by chlorine

D

`Mn^(2+)` is oxidised by chlorine

Text Solution

Verified by Experts

The correct Answer is:
C
`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.
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