Step 1

$\overline{){\mathbf{K}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\left[\mathrm{Cu}{{\left({\mathrm{NH}}_{3}\right)}_{4}}^{+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]{\left[{\mathrm{NH}}_{3}\right]}^{\mathbf{4}}}}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{Cu}}^{2+}\right]\mathbf{}\mathbf{=}\mathbf{}\frac{\left[\mathrm{Cu}{{\left({\mathrm{NH}}_{3}\right)}_{4}}^{+}\right]}{\mathbf{K}\mathbf{}{\left[{\mathrm{NH}}_{3}\right]}^{\mathbf{4}}}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{Cu}}^{2+}\right]\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{010}}{(1.0\times {10}^{13}){(5.0)}^{\mathbf{4}}}$

**[Cu ^{2+}] = 1.6x10^{-18} M**

Step 2

Cu^{2+}_{ }(aq) + 2 e^{-} → Cu (s) E° = 0.34 V ↓ E → oxidation → anode

Ag^{+} (aq) + e^{-} → Ag (s) E° = 0.80 V ↑ E → reduction → cathode

Step 3

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cathode}}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{anode}}}}\phantom{\rule{0ex}{0ex}}\mathbf{E}{\mathbf{\xb0}}_{\mathbf{cell}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{V}\mathbf{}\mathbf{-}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{}\mathbf{V}$

E°_{cell} = 0.46 V

Step 4

Cu (s) → Cu^{2+}_{ }(aq) + 2 e^{-}

[Ag^{+} (aq) + e^{-} → Ag (s)] x2

2 Ag^{+}_{ }(aq) + Cu (s) → Cu^{2+} (aq) + 2 Ag (s)

n = 2

An electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag ^{+}] = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 M NH_{3} that is also 0.010 M in Cu(NH _{3})_{4} ^{2+}, what is the cell potential at 25°C?

Cu^{2+}(aq) + 4NH_{3} (aq) ⇌ Cu(NH _{3})_{4} ^{2+}(aq) K = 1.0 X 10 ^{13}

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