A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: Cu2 (aq)+Zn () Cu (s)+Zn2(aq) Suppose the cell is prepared with 0.788 MCu2 in one half-cell and 7.32 M Zn2 in the other Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. 2

Explanation:

We know that,

  = -0.7618 V

   = 0.337 V

According to the given reaction/cell notation,  cathode is ( and anode is ().

Therefore,  

                          = (0.337) - (-0.7618)

                          = 1.0988 V

In the given reaction, number of electrons being transferred in balanced reaction are 2. Hence, n = 2.

Also, we know that

        E =

Putting the given values into the above formula as follows.

       E =  

          =          

          =

           = 1.0704 V

          = 1.07 V (approx)

Thus, we can conclude that the cell voltage under these conditions is 1.07 V.

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