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I2 and Br2 are added to a solution conta...

`I_2` and `Br_2` are added to a solution containing `Br^–` and `I^–` ions. What reaction will occur if,
`I_(2) + 2e^(-) rarr 2I^(-)`, `E^(0) = + 0.54V` and `Br_(2) + 2e^(-) rarr 2Br^(-)`, `E^(0) = +1.09 V?`

Text Solution

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Since the reduction potential of `Br_2` is more than that of `I_2` , it means that bromine can readily be reduced . Therefore , `I^(-)` will be oxidised to `I_2` and this reaction should be written as oxidation . Therefore , the following reactions will occur.
`Br_2+2e^(-)rarr 2Br^(-)`
`2I^(-) rarr I_2+2e^(-)`
`Br_2+2I^(-) rarr 2Br^(-)+I_2`
Alternatively , the problem can also be solved by considering the half cell reactions.
(i) `I_2+2e^(-)rarr2I^(-) " "E^@=+0.54V`
(ii) `Br_2+2e^(-)rarr2Br^(-) " "E^@=+1.02V`
Since for the feasibility of the reaction , the e.m.f . should be+ve , therefore to get +ve value for the overall cell reaction , subtract the equation representing lower value of `E^@` from the equation representing the higher value of `E^@` Thus , subtract Eq. (i) from (ii).
`Br_2+2I^(-) rarr 2Br^(-) +I_2`
`E^@=1.02-0.54=0.48V`
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