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Given below are a set of half-cell react...

Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}`
Among the following, identify the correct statement:

A

Chloride ion is oxidised by `O_(2)`

B

`Fe^(2+)` ios oxidised by iodide

C

lodide ion is oxidised by chlorine

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
On calculating the emf of the cell we find that only the cell involving the oxidation of `1^(-) "ion by " CI_(2)` is positive
`(2I^(-)rarrI_(2)+2e^(-),E^(@)=0.54 V)`
`{:(CI_(2)+2e^(-)rarr2CI^(-),E^(@)=1.36V),(CI_(2)+2I^(-)rarr2CI^(-)+l_(2),E^(2)_(cell)=1.36-0.54):}``
"In all other cases the" `E^(@)` is -ve
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