Calculate the equilibrium constant for the reaction, `2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_3^-` . The standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for `Fe^(3+)//Fe^(2+)` and `I_3^(-) //I^-` couples.

Answer

`10^(8)`

Explanation

At anode `: 2I^(c-) rarr I_(3)^(c-)+2e^(-)`
At cathode `: 2Fe^(3+)+2e^(-) rarr 2Fe^(2+)`

At equlibrium, `E_(cell)=0`
`E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)`
`=E^(c-)._((Fe^(3+)|Fe^(2+)))-E^(c-)._((I_(3)^(c-)|3I^(c-)))`
`=0.78-0.54=0.24V`
`E_(cell)=E^(c-)._(cell)-(0.06)/(2)logK_(eq)(n_(cell)=2)`
`E^(c-)._(cell)=0.03logK_(eq)`
`logK_(eq)=(E^(c-)._(cell))/(0.03)=(0.24V)/(0.03V)=8.0`
`K_(eq)=10^(8)`