The redox reaction is
[tex]Ca^{+2} + 2Li(s) = Ca(s) + 2Li^{+}[/tex]
Here
Calcium undergoes reduction, and acts as cathode
Lithium undergoes oxidation and acts as anode
The reduction potential of calcium is -2.87 V
The reduction potential of lithium is - -3.05 V
We know that
Ecell = Ecathode - Eanode
Ecell = -2.87 - (-3.05) = 0.18 V
