An excess of finely divided iron is stirred up with a solution that contains Cu²⁺ ion, and the system is allowed to come to equilibrium. The solid materials are then filtered off, and electrodes of solid copper and solid iron are inserted into the remaining solution. What is the value of the ratio [Fe²⁺]/[Cu²⁺] at 25°C?

Question

Quizplus · Accepted Answer

The given solution contains Cu²⁺ ion, which means it has copper ions and sulfide ions. When solid iron is added to this solution, it will react with the sulfide ions to form FeS solid and release S2- ions. The reaction can be written as: Fe(s) + S2-(aq) ⇌ FeS(s) Since the iron is in excess, the equilibrium will shift towards the right side according to Le Chatelier's principle. This means more FeS will be formed. As a result, the concentration of S2- ions in the solution will decrease. Now, when copper electrode is inserted in the solution, it will undergo reduction reaction and gain electrons to form Cu(s) solid. The half-reaction is: Cu2+(aq) + 2e- ⇌ Cu(s) Since the Cu2+ ions are being consumed, the equilibrium of the reaction Cu²⁺ ⇌ Cu2+(aq) + S2-(aq) will shift towards the right side to produce more Cu2+ ions, according to the Le Chatelier's principle. As a result, the concentration of S2- ions will decrease further. So, we can conclude that the [Fe²⁺]/[Cu²⁺] ratio will be greater than 1. To calculate the exact value of the ratio, we need to know the equilibrium constant (K) of the reaction: Cu²⁺ ⇌ Cu2+(aq) + S2-(aq) However, this information is not given in the question. But we can say that the ratio will be greater than 2 (since excess iron will ensure that most of the sulfide ions will react with Fe to form FeS) and less than 4 (since all the Cu2+ ions cannot be consumed due to some sulfide ions remaining in the solution). Therefore, the best choice is B: 2.5 F1F1F1S1F1F1F10 10²⁶.

An Excess of Finely Divided Iron Is Stirred Up with a Solution