We need to use standard electrode potentials to calculate this:
#sf(Zn^(2+)+2erightleftharpoonsZn)# #" "sf(E^@=-0.76color(white)(x)V)#
#sf(Cu^(2+)+2erightleftharpoonsCu" "E^@=+0.34color(white)(x)V)#
#sf(Zn_((s))+Cu_((aq))^(2+)rightleftharpoonsZn_((aq))^(2+)+Cu_((s)))#
#sf(K_c=([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])#
The expression for free energy change is:
#sf(DeltaG=DeltaG^(@)+RTlnQ)#
Where #sf(Q)# is the reaction quotient.
At equilibrium #sf(DeltaG=0)# so this becomes:
#sf(0=DeltaG^(@)+RTlnQ)#
Since we are at equilibrium #sf(Q=K_c)# so we can write:
#sf(DeltaG^(@)=-RTlnK_c" "color(red)((1)))#
If this were an electrochemical cell then the free energy change is the maximum amount of work you can get from the cell.
It is related to the #sf(E_(cell)^@)# value by:
#sf(DeltaG^0=-nFE_(cell)^(@)" "color(red)((2)))#
#sf(F)# is the Faraday Constant and is the charge carried by a mole of electrons which is #sf(9.648xx10^(4)color(white)(x)"C""/"mol)#.
#sf(n)# is the number of moles of electrons transferred.
To find #sf(E_(cell)^@)# for this reaction subtract the least +ve #sf(E^@)# from the most +ve:
#sf(E_(cell)^@=+0.34-(-0.76)=+1.1color(white)(x)V)#
Putting #sf(color(red)((1)))# equal to #sf(color(red)((2)))# we get:
#sf(-nFE_(cell)^@=-RTlnK_c)#
#:.##sf(lnK_c=(nFE^@)/(RT)#
Putting in the numbers:
#sf(lnK_c=(2xx9.648xx10^4xx1.1)/(8.31xx370)=69.03)#
From which #sf(K_c~=10^(30)#
This number is so large as to have little physical significance.
It means that we can regard the reaction as having gone to completion where all the reactants are converted into products.
This is what we see if zinc is added to a solution of #sf(Cu_((aq))^(2+))#.
