The cell reaction when it is working is:
#sf(Zn+Cu^(2+)rightleftharpoonsZn^(2+)+Cu)#
At #sf(25^@C)# a working form of the Nernst Equation is:
#sf(E_(cell)=E^(@)-0.0592/nlogQ)#
#sf(Q)# is the reaction quotient and is given by:
#sf(Q=([Zn^(2+)])/([Cu^(2+)]))#
#sf(n)# is the number of moles of electrons transferred which, in this case, is 2.
As current is drawn from the cell as it is working, the potential difference between the two 1/2 cells gradually falls and eventually becomes zero.
The cell is now flat and #sf(E_(cell)=0)#.
Putting this in to the Nernst Equation we get:
#sf(0=1.1-(0.0592)/(2)logQ)#
#:.##sf(logQ=(1.1xx2)/(0.0592)=37.16)#
This gives #sf(A)# to be the correct response.
The system has now reached equilibrium at which point #sf(K=Q)#.
This means that #sf(K~=10^(37))# which is a fantastically high number and effectively tells us that the reaction has gone to completion as you would expect from our knowledge of displacement reactions.
