(a)
It helps to use #rightleftharpoons# instead:
#Zn_((s))+2Ag_((aq))^+rightleftharpoonsZn_((aq))^(2+)+2Ag_((s))#
Just by using Le Chatelier's Principle you can predict that by increasing #[Zn^(2+)]# the position of equilibrium will shift to the left thus decreasing #DeltaE_("cell")#.
(c)
#n# is the number of moles of electrons transferred which you can see must be 2 since we have :
#ZnrarrZn^(2+)+2e#
I'll attempt a fuller explanation:
The equation in the question refers to the electrode potential of the the 1/2 cell #E_("red")#, not #E_"cell"#. It should read:
#E_("cell")=E_("cell")^@-(RT)/(nF)lnQ#
At 298K this simplifies to:
#E_("cell")=E_("cell")^@-(0.05916)/(2)logQ#
#Q# is the reaction quotient and is given by:
#Q=([Zn^(2+)])/([Ag^+]^2)#
We can work out #E_("cell")^@# from standard electrode potentials:
#Zn^(2+)"/"Zn " "E^@=-0.76V#
#Ag^(+)"/"Ag" "E^@=+0.8V#
To find #E_("cell")^@# subtract the least +ve #E^@# from the most +ve #rArr#
#E_("cell")^@=0.8-(-0.76)=+1.56V#
If we had standard conditions i.e unit concentrations etc then #Q=1# so #"log"Q=0# and #E=E_("cell")-0=+1.56V#.
What happens if we increase #[Zn^(2+)]# to 2M as might happen in the question?
#E_("cell")=E_("cell")^@-(0.05916)/(2)log([2]/[1])#
#:.E_("cell")=+1.56-0.0089=+1.551V#
So you can see that there is a slight decrease as predicted.
I note the equation has now been corrected. The same reasoning applies. If #[Zn^(2+)]# is increased you can see that #E^@# is made less -ve so #E_("cell")# is reduced.
