Iridium was plated from a solution containing for 2.0 h with a current of 0.075 A. The Iridium depos — Electrochemistry Chemistry Question

💡 Solution & Explanation

Total charge passed $Q = 0.075 \text{ A} \times 2.0 \times 3600 \text{ s} = 540$ C. Number of Faradays = $540 / 96500 \approx 0.005596$ F. Moles of Ir deposited = $0.36 / 192 = 0.001875$ mol. The $n$-factor (oxidation state $x$) = Faradays / Moles = $0.005596 / 0.001875 \approx 2.98 \approx 3$. So, $x = +3$. The complex is $IrCl_6^y$. The sum of oxidation states equals the net charge: $y = x + 6(-1) = +3 - 6 = -3$. Thus, $(x + y) = 3 + (-3) = 0$. Therefore, correct answer is 0.