To perform an analysis of a mixture of metal ions by electro-deposition, the second metal to be depo — Electrochemistry Chemistry Question
Question
To perform an analysis of a mixture of metal ions by electro-deposition, the second metal to be deposited must not being plating out until the concentration ratio of the second to the first is about 10^6. What must be the minimum difference in standard potential of the two metals which form dipositive ions in order for such an analysis to be feasible?
💡 Solution & Explanation
Let the metals be M1 and M2. For simultaneous deposition, their reduction potentials must be equal: E°1 + (0.0591/2)*log[M1^2+] = E°2 + (0.0591/2)*log[M2^2+]. Rearranging yields E°1 - E°2 = (0.0591/2) * log([M2^2+]/[M1^2+]). Substituting the required ratio [M2^2+]/[M1^2+] = 10^6 gives E°1 - E°2 = (0.0591/2) * log(10^6) = (0.0591/2) * 6 = 0.0591 * 3 = 0.1773 V. Therefore, correct answer is A.