Home
Class 12
CHEMISTRY
Calculate the equilibrium constant for t...

Calculate the equilibrium constant for the reaction : `Fe^(2+)+Ce^(4+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(Fe^(3+)//Fe^(2+))^(@)=0.68V`

Text Solution

Verified by Experts

The correct Answer is:
`7.236xx10^(12)`
For the cell reaction `:`

Half cell reaction are `:`
`a. Fe^(2+)(aq) rarr Fe^(3+)(aq) +e^(-)" "E^(c-)=-0.68V`
`b. Ce^(4+)(aq)+e^(-) rarr Ce^(3+)(aq) " "E^(c-)=+1.44V`
`overlineunderline(Fe^(2+)(aq)+Ce^(4+)hArrFe^(3+(aq)+Ce^(3+)E_("Cell")^(Θ)=+0.76V)`
At equilibrium,
`E^(c-)._(cell)=(0.0591)/(n)log K_(eq)(` because at equilibrium, `E_(cell)=0)`.
`+0.76=(0.0591)/(1)log K_(eq)`
`logK_(eq)=(0.76)/(0.0591)=12.895impliesK_(eq)=7.236xx10^(12)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate Delta_rG^@ for the following reactions: Fe^(+2) (aq)+Ag^(+) (aq) to Fe^(+3) (aq)+Ag (s)

Which one has the larger size : Fe^(2+) " or " Fe^(3+) ?

Calculate the standard e.m.f. of the reaction Fe^(3+)+3e^(-)rarrFe_((s)) . Given the e.m.f. values of Fe^(3+)+e rarr Fe^(2+) and Fe^(2+)+2e rarr Fe_((s))" as "+0.771 V and =0.44V " respectively."

The equilibrium constant for the reaction : Fe^(3+)(aq)+SCN^(-)(aq)hArr Fe SCN^(2+)(aq) is 140 at 298 K. The equilibrium constant for the reaction : 2Fe^(3+)(aq)+2SCN^(-)(aq)hArr 2Fe SCN^(2+)(aq) is :

Identify the reducing agent in the following reaction : Fe_(2)O_(3)+3COrarr2Fe+3CO_(2)