A galvanic cell is built from a zinc electrode in 1.0 M Zn^2+ and a copper electrode dipping in a solution prepared from 0.010 M Cu^2+ with…

The cell reaction is $\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}$ with $E^\circ_{cell}=0.34-(-0.76)=1.10\ \mathrm{V}$ and $n=2$. Because $\mathrm{NH_3}$ binds copper strongly, free $\mathrm{Cu^{2+}}$ must be found from the complex equilibrium: $[\mathrm{Cu^{2+}}]=\dfrac{[\mathrm{Cu(NH_3)_4^{2+}}]}{K_f[\mathrm{NH_3}]^4}=\dfrac{0.010}{10^{13}\times1^4}=1.0\times10^{-15}\ \mathrm{M}$. The Nernst equation gives $E=E^\circ-\dfrac{0.0592}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=1.10-0.0296\log\dfrac{1.0}{10^{-15}}=1.10-0.0296\times15=1.10-0.444=0.656\ \mathrm{V}$. Hence $E\approx0.66\ \mathrm{V}$. Complexation of the oxidant lowers its electrode potential by 0.444 V, which is why the cell voltage drops from 1.10 V. The calculation assumes that essentially all copper is present as the tetraammine complex, so its concentration is taken as 0.010 M.