We are asked to find the **equilibrium ****constant K** for the spontaneous reaction between Ni^{2+}(aq) and Cd(s).

We will use the **Gibbs Free Energy**** ****Equation** to calculate the equilibrium constant. The Gibbs Free relates the potential of compounds and cell equilibrium:

In a Galvanic Cell, the Gibbs free energy is related to the potential by:

$\overline{){\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{{\mathbf{nFE}}}_{{\mathbf{cell}}}^{{\mathbf{o}}}{\mathbf{}}}$

If the E^{o}_{cell }> 0, then the process is spontaneous (galvanic cell)

If the E^{o}_{cell }< 0, then the process is nonspontaneous (electrolytic cell)

ΔG^{o} can also be expressed by the equilibrium constant Keq of the reaction.

$\overline{){\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{RT}}{\mathbf{}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{eq}}}{\mathbf{}}}$

Combining the two equations since both are equal to ΔG^{o} gives:

$\overline{){\mathbf{-}}{{\mathbf{nFE}}}_{{\mathbf{cell}}}^{{\mathbf{o}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{RT}}{\mathbf{}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{eq}}}{\mathbf{}}}$

where

n = # of e^{-} transferred

F = Faraday’s constant = 96485 J/(mol e^{-})

E°_{cell} = standard cell potential, V

R = gas constant, 8.314 J/mol e^{- }K

T = temperature in K

K_{eq} = equilibrium constant of the reaction

In the Gibbs Equation, we have to determine the E°_{cell} first, as well as the **anode (oxidation) **and the **cathode (reduction)** in the concentration cell and the **number of electrons transferred (n)**.

Compute the equilibrium constant for the spontaneous reaction between Ni^{2+}(aq) and Cd(s).

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.

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