Calculate the equilibrium constant for the reaction:
`3Sn(s) +2Cr_(2)O_(7)^(2-) +28 H^(+) rarr 3Sn^(+4) +4Cr^(3+) +14H_(2)O`
`E^(@)` for `Sn//Sn^(+2) = 0.136 V E^(@)` for4 `Sn^(2+)//Sn^(4+) =- 0.154V`
`E^(@)` for `Cr_(2)O_(7)^(2-)//Cr^(3+) = 1.33V`

Answer

`K = 10^(268)`

Explanation

`Sn rarr Sn^(2+) +2e^(-) :. DeltaG_(1) =- 2 xx E_(1) xx F`
`Sn^(2+) rarr Sn^(4+) +2e^(-) :. DeltaG_(2) =- 2 xx E_(2) xx F`
`Sn rarr Sn^(4-) +4e^(-) :. DeltaG_(3) =- 4 xx E_(3) xx F`
`DeltaG_(3) = DeltaG_(1) +DeltaG_(2)`
`E_(3) = (2 xx 0 xx 136 +2 +(-0.154))/(4)`
`E_(3) rArr 0.009`
`E = E^(@)`
at equilibrium `E = 0`
`E^(@) = (0.0591)/(12) log K_(eq)`
`-0.009 +1.33 = (0.0591)/(12) xx log K_(eq)`
`1.321 =(0.0591)/(12) log K_(eq)`
`K_(eq) = 10^(268)`