This is a redox reaction, which involves the transfer of electrons. When this transfer occurs through a wire we can measure the current using an amp meter and if there is a current there must be a voltage or Electromagnetic Force (EMF) that is pushing the electrons through the wire.
We can determine the Eo or of the reaction by looking at a table and using the equation:
Eo = Red (Red) - Oxi (Red)
The (Red) refers to the reaction in the table (that is probably in your textbook somewhere). I looked up each reaction and found this:
Br2 (l) + 2e- —> 2 Br- Eo = 1.066 V
Ca2+ + 2e- —> Ca Eo = -2.868 V
Notice that each of those reaction are the reduced half. That is what I mean when I put (Red) in the equation.
Now we have to look at the reaction in the problem to determine which part of oxidized and which part is reduced. Br goes from an oxidation number of 0 to -1 so it is reduced, which means Ca must be oxidized.
Therefore the answer is:
Eo = 1.066 - -2.868 = 3.934 V
