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
