Calculate the equilibrium constant for the reaction below at 25 ^$\circ$C,
Co(s) + 2 Cr³⁺(aq) $\to$ Co²⁺(aq) + 2 Cr²⁺(aq)
Given the following thermodynamic information.
Co²⁺(aq) + 2 e^- $\to$ Co(s) $~~~~~~~~$ E^$\circ$= -0.28 V
Cr³⁺(aq) + e^- $\to$ Cr²⁺(aq) $~~~~~~~~$ E^$\circ$= -0.41 V

Question

Calculate the equilibrium constant for the reaction below at 25 ^$\circ$C,
Co(s) + 2 Cr³⁺(aq)

\to

Co²⁺(aq) + 2 Cr²⁺(aq)
Given the following thermodynamic information.
Co²⁺(aq) + 2 e^-

\to

Co(s)

~~~~~~~~

E^$\circ$= -0.28 V
Cr³⁺(aq) + e^-

\to

Cr²⁺(aq)

~~~~~~~~

E^$\circ$= -0.41 V

Quizplus · Accepted Answer

The equilibrium constant for a reaction is given by the exponential of the negative of the standard free energy change divided by the product of the gas constant and temperature. The standard free energy change is given by the sum of the standard reduction potentials of the half-reactions multiplied by the number of electrons transferred. In this case, the standard reduction potential of the Co²⁺/Co half-reaction is -0.28 V and the standard reduction potential of the Cr³⁺/Cr²⁺ half-reaction is -0.41 V. The number of electrons transferred is 2. Therefore, the standard free energy change is -2 * (0.28 V + 0.41 V) = -1.38 V. The equilibrium constant is then given by:K = exp(-ΔG°/RT) = exp(-(-1.38 V) * (96485 C/mol) / (8.314 J/mol*K) * (298 K)) = 4.0 * 10^-5

Calculate the Equilibrium Constant for the Reaction Below at 25 ∘\circ∘

Calculate the Equilibrium Constant for the Reaction Below at 25 $\circ$