When silver chloride is dissolved in a large excess of ammonia, practically all silver ion can be as — Electrochemistry Chemistry Question
Question
When silver chloride is dissolved in a large excess of ammonia, practically all silver ion can be assumed to exist in form of a single ionic species [Ag_x(NH3)_y]^+. Compute the values of x and y using the two following cells:<br>Cell I: Ag \| 4.0 × 10^–4 M – AgCl, 1 M – NH3 \|\| 4 × 10^–2 M – AgCl, 1 M – NH3 \| Ag; E_cell = 0.118 V at 298 K.<br>Cell II: Ag \| 3 × 10^–3 M – AgCl, 1 M – NH3 \|\| 3 × 10^–3 M – AgCl, 0.1 M – NH3 \| Ag; E_cell = 0.118 V at 298 K.
💡 Solution & Explanation
From Cell I, keeping [NH3] constant, [Ag^+] is proportional to [complex]^(1/x). E_cell = 0.059 * log([Ag^+]_R / [Ag^+]_L) = 0.059 * log((4e-2/4e-4)^(1/x)) = 0.059 * (2/x). Since E_cell = 0.118 V, 0.118 = 0.118/x, so x = 1. From Cell II, keeping [complex] constant, [Ag^+] is inversely proportional to [NH3]^y. E_cell = 0.059 * log([NH3]_L^y / [NH3]_R^y) = 0.059 * log(1^y / 0.1^y) = 0.059 * y. 0.118 = 0.059 * y, so y = 2. Therefore, correct answer is A.