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? The following standard reduction potentials apply:
Fe²⁺(aq) + 2e^- $\to$ Fe(s)
$\varepsilon$ ° = -0.44 V
Cu²⁺(aq) + 2e^- $\to$ Cu(s)
$\varepsilon$ ° = +0.34 V

Question

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? The following standard reduction potentials apply:
Fe²⁺(aq) + 2e^-

\to

Fe(s)

\varepsilon

° = -0.44 V
Cu²⁺(aq) + 2e^-

\to

Cu(s)

\varepsilon

° = +0.34 V

Quizplus · Accepted Answer

The reaction between Fe and Cu²⁺ will reach equilibrium when the reaction quotient Q equals the equilibrium constant K. Using the Nernst equation and the given standard reduction potentials, the equilibrium constant K can be calculated, which corresponds to the ratio [Fe²⁺]/[Cu²⁺].

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