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

Calculate the equilibrium constant for the reaction
`Cu(s)+2Ag+(aq)rarrCu^(+2)(aq)+2Ag(s),E_("cell")^(@)=0.46V`.

Text Solution

Verified by Experts

`E^(C-)._(CELL)=(0.059V)/(2)logK_(c)`
`=0.46V=(0.059V)/(2)log K_(c)`
`:. log K_(c)=(0.46Vxx2)/(0.059V)=15.6`
`K_(c)=`Antilog`(15.6)=3.98xx10^(15)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The equilibrium constant of the reaction : A_((s)) + 2 B_((aq))^(+) hArr A_((aq))^(2+) + 2 B_((s)) , E_("cell")^(@) = 0.0295V is [(2.303 RT)/(F) = 0.059]

The equilibrium constant of the reaction : A_((s))+2B_((aq))^(+)hArr A_((aq))^(2+)+2B_((s)): E_(cell)^(@)=0.0295V is :

The standard electrode potantials of the half cells Ag^(+)//Ag and Fe^(3+),Fe^(2+)//Pt are 0.7991V and 0.771V respectively. Calculate the equilibrium constant of the reaction : Ag_((s))+Fe^(3+)hArr Ag^(+)+Fe^(2+)

Calculate standard free energy change for reaction Zn(S)+2Ag^(+)(aq) Leftrightarrow Zn_(aq)^(2+)+2Ag(s) E_("cell")^(0)=1.56V:" Given "1F=96500C mol^(-1) .