For the standard Daniell cell , the standard cell potential is at . Initially, and are present. Suff — Electrochemistry Chemistry Question
Question
For the standard Daniell cell $Zn(s) \| Zn^{2+}(aq) \|\| Cu^{2+}(aq) \| Cu(s)$, the standard cell potential $E^\circ$ is $1.10\text{ V}$ at $298\text{ K}$. Initially, $1.0\text{ M } Zn^{2+}$ and $1.0\text{ M } Cu^{2+}$ are present. Sufficient concentrated aqueous $NH_3$ is added to the cathode compartment to complex the copper ions as $[Cu(NH_3)_4]^{2+}$, leaving a final equilibrium concentration of uncomplexed $NH_3$ of $2.0\text{ M}$. What is the new operating cell potential in Volts? (Given: Formation constant $K_f \text{ for } [Cu(NH_3)_4]^{2+} = 10^{12}$, and $\frac{2.303RT}{F} = 0.0591\text{ V}$)
💡 Solution & Explanation
Initially $[Cu^{2+}] = 1.0\text{ M}$. $NH_3$ complexes almost all $Cu^{2+}$ via $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$. Thus, $[[Cu(NH_3)_4]^{2+}] \approx 1.0\text{ M}$. Free $[NH_3] = 2.0\text{ M}$. Using the formation constant: $K_f = \frac{[[Cu(NH_3)_4]^{2+}]}{[Cu^{2+}][NH_3]^4} \implies 10^{12} = \frac{1.0}{[Cu^{2+}](2.0)^4} \implies [Cu^{2+}] = \frac{1}{16 \times 10^{12}} = 6.25 \times 10^{-14}\text{ M}$. Applying the Nernst equation for the full cell: $E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 1.10 - 0.02955 \log \frac{1.0}{6.25 \times 10^{-14}} = 1.10 - 0.02955 \times (\log 10^{14} - \log 6.25) = 1.10 - 0.02955 \times (14 - 0.796) = 1.10 - 0.39 = 0.71\text{ V}$.