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Calculate the euilibrium constant for th...

Calculate the euilibrium constant for the reaction,
`2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-)`.
The standard reduction potential in acidic conditions are `0.77 V ` and `0.54 V` respectivelu for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.

A

`4.25xx10^7`

B

`7.05xx10^5`

C

`6.25xx10^5`

D

`6.25xx10^7`

Text Solution

Verified by Experts

The correct Answer is:
D
`2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_3^(-)`
For the above change at equilibrium , E=0
Using the relation, `E=E^@-0.059/2"log"K_c`
or `0=E_(Fe^(3+)//Fe^(2+))^@-E_(I_3^- //I^-)^@-0.059/2 "log"K_c`
or 0=0.77-0.54-`0.059/2"log"K_c`
or 0=0.23-`0.059/2"log"K_c`
or `0.059/2"log"K_c=0.23` or log `K_c=(0.23xx2)/0.059`
or log `K_c` =7.796 or `K_c=6.25xx10^7`
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