21. A Daniell cell initially has and (). Sufficient is added to the cathode to make the free at equi — Electrochemistry Chemistry Question

💡 Solution & Explanation

The $Cu^{2+}$ is heavily complexed. $K_f = 10^{12} = \frac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}][NH_3]^4}$. Assume all Cu forms complex, so $[Cu(NH_3)_4^{2+}] \approx 1.0 M$. $[Cu^{2+}] = 1 / (10^{12} \times 2^4) = 6.25 \times 10^{-14} M$. $E = 1.10 - (0.0591/2)\log(1.0 / 6.25 \times 10^{-14}) = 1.10 - 0.02955 \times 13.204 \approx 0.71 V$.