Question

A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: Cu²+ (aq) + Zn (s) → Cu (s) +Zn2+ (aq) Suppose the cell is prepared with 1.61 M Cu?* in one half-cell and 1.83 M Zn in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. DO E x a 5 ?

Answer 1

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

E°(cell) = E°(Cu) - E°(Zn)

E° = 0.339 - (-0.762) = 1.101 V

Reaction Quotient (Q) = [Zn2+] / [Cu2+] = 1.83/1.61 = 1.13665

Number of electrons transferred (n) = 2

Using Nernst equation at 25° C or 298K;

E = E° - (0.0592 * log Q)/n

E = 1.101 - (0.0592 * log (1.13665)) / 2

E = 1.099 OR

E = 1.09 V OR 1.10 ...Answer

Let me know if any doubts.