Use the standard potentials of the couples `Au^(+)//Au(+1.69V),Au^(3+)//Au(+1.40V),` and `Fe^(3+)//Fe^(2+)(+0.77V)` to calculate the equilibrium constnat for the reaction `2Fe^(2+)(aq)+Au^(3+)(aq) leftrightarrow2Fe^(3+)(aq)+Au^(+)(aq)`

Options

A. `4xx10^(16)`
B. `8xx10^(8)`
C. `4xx10^(-16)`
D. `1xx10^(14)`

Answer

1

Explanation

We need to obtain `E^(@)` for the couple.
`(3)" "Au^(3+)(aq)+2e^(-)rarr Au^(+)(aq)`
from the value of `E^(@)` for the couples
`(1)" "Au^(+)(aq)+e^(-)rarrAu(s)" "E^(@)=1.69V`
`(2)" "Au^(3+)(aq)+3e^(-)rarr Au(s)" "E_(2)^(@)=1.40V`
We see that `(3)=(2)-(1),` therefore
`-n(3)FE_(3)^(@)=" "-n_(1)FE_(1)^(@)-n_(2)FE_(2)^(@)`
Solving `E_(3)^(@)` we obtain
`E_(3)^(@)=(n_(2)E_(2)^(@)-n_(1)E_(1)^(@))/(n_(3))=((33)xx(1.40V)-(1)xx(1.69V))/(2)=1.26V`
Then `R:Au^(3+)(aq)+2e^(-)rarr Au^(+)(aq)" "E_(R)=1.26V`
`L:2Fe^(3+)(aq)+2e^(-)rarr 2Fe^(2+)(aq)E_(L)^(@)=0.77V`
`R-L : 2Fe^(2+) (aq)+Au^(3+)(aq)+rarr 2Fe^(3+) (aq)+Au^(+)(aq)`
`E^(@)=E_(R)^(@)-E_(L)^(@)=(1.26V)-(0.77V)=+0.49V`
`2.303RTlogK=nFE^(@)`
`loK=(nFE^(@))/(2.303RT)" "rArr" "logK=(2xx0.49)/(0.059)`
`K=4xx10^(14)`