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Redox reactions play a pivotal role I ch...

Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential `(E^(@))` of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their `E^(@)` values ( with respect to normal hydrogen electrode ) Using this data :
`I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) " "E=1.36V`
`Mn^(3+) + e^(-) to Mn^(2+) E^(@) = 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V`
`O_2 +4H^(+) + 4e^(-) to 2H_2O" " E^(@) = 1.23 ,`
Among the following , identify the correct statement :

A

Chloride ion is oxidised by `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
` I_2 // 2I^(-)(0.54V) , Fe^(+3)//Fe^(+2)(0.77V)`
`O_2 , 4H^(oplus) // 2H_2O (1.23V) , Cl_2 // 2Cl^(Ɵ)(1.36V) , Mn^(+3)//Mn^(+2)(1.50V)`
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Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respect to normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) " "E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@) = 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O" " E^(@) = 1.23 , While Fe^(3+) is stable , Mn^(3+) is not stable in acid solution because :

Name the two half-cell reactions that are taking place in the Daniell cell.

Which is a correct cell reaction ?

Write the electrode reactions in a Daniel cell.

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