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+)).
