#1Ag + 1/2Ca^(+2) rarr 1/2Ca + 1Ag^+ #
#2Ag rarr Ag^+ + 2e^- #
We have given the #E_"red" = 0.766V#
Therefore Eox = -Ered
Eox = -0.766V
#Ca^(+2) + 2e^-) rarr Ca #
#E_"red" = -2.76V#
#E^o"cell" = E_"red" + E_"ox"#
#-2.76V + -0.766V = -3.526V#
If the #E^o cell# is negative the reaction is non-spontaneous
As the concentration #Ca^(2+) and Ag^+# is 0.1M and not 1M you will have to solve Nerst Equation .
#Ecell = E^o - "RT"/n" log Q#
-3.526V - 0.0591/2 log Q
#Q = 0.1^(0.5)/0.1^1 = 0.1/0.1^2#
#Ecell = -3.526V - 0.0591/2 log(0.1/0.1^2)#
-3.526V - 0.02955 = -3.55555V
You can also solve by following the ln method but this the easy way
Now as the #E_"cell"# is negative the reaction will not occur as written as #E_"cell"# is negative for non-spontaneous reaction and positive for spontaneous reaction .
The spontaneous reaction opposite of this non spontaneous reaction would be oxidation of Ca and reduction of Ag+
#Ca rarr Ca^(+2) + 2e^-#
#E_"ox" = 2.76V#
#2Ag^+ + 2e^-) rarr Ag#
#E_"red" = 0.766V#
#E^ocell = 3.526V#
Therefore this reaction is a spontaneous one.
The Ecell would remain the same since the concentration are 1M the standard concentration
= #3.526V - 0.0591/2 log (1/1^2)#
= 3.526V
