Question

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

Answer 1

Explanation:

We know that,

  $E^{o}(Zn^{2+}/Zn(s))$ = -0.7618 V

   = 0.337 V

According to the given reaction/cell notation,  cathode is ($Cu^{2+}/Cu(s))$ and anode is ().

Therefore,  $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

                          = (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 = $E^{o} - (\frac{2.303 \times RT}{nF}) log {\frac{[Zn^{2+}]}^{1}{[Cu^{2+}]^{1}}$

Putting the given values into the above formula as follows.

       E =  $E^{o} - (\frac{2.303 \times RT}{nF}) log {\frac{[Zn^{2+}]}^{1}{[Cu^{2+}]}^{1}$

          = $E^{o} - (\frac{0.0591}{n}) log \frac{[Zn^{2+}]}^{1}{[Cu^{2+}]}^{1}$          

          = $1.099 - \frac{0.0591}{2} log (\frac{7.32}{0.788})$

           = 1.0704 V

          = 1.07 V (approx)

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