# Find the equilibrium constant for a 3-electron transfer process, whose standard emf is "0.59 V" at "298.15 K"? F = "96485.33 C/mol e"^(-)

## a) ${10}^{15}$ b) ${10}^{20}$ c) ${10}^{25}$ d) ${10}^{30}$

Jun 28, 2016

The standard EMF (electromotive "force") is ${E}_{\text{cell}}^{\circ}$. So, you should relate the following two equations:

$\setminus m a t h b f \left(\Delta G = \Delta {G}^{\circ} + R T \ln Q\right)$

$\setminus m a t h b f \left(\Delta {G}^{\circ} = - n F {E}_{\text{cell}}^{\circ}\right)$

where:

• $\Delta G$ is the Gibbs' free energy for the process.
• $\Delta {G}^{\circ}$ is the Gibbs' free energy for the process at $m a t h b f \left({25}^{\circ} \text{C}\right)$ and $\setminus m a t h b f \left(\text{1 bar}\right)$ pressure.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature at which the process occurs, in $\text{K}$. For a standard thermodynamic process we assume it is $\text{298.15 K}$.
• $n$ is the $\text{mol}$s of electrons transferred.
• $\text{F}$ is Faraday's constant, which is approximately ${\text{96485.33 C/mol e}}^{-}$.
• ${E}_{\text{cell}}^{\circ}$ is in $\text{V}$, and $\text{1 V"cdot"C" = "1 J}$.

Since we are finding ${K}_{\text{eq}}$, we know that $\Delta G = 0$ and $Q = K$ at equilibrium. Therefore:

$\Delta {G}^{\circ} = - R T \ln {K}_{\text{eq" = -nFE_"cell}}^{\circ}$

$\implies \textcolor{b l u e}{{K}_{\text{eq") = e^(nFE_"cell"^@"/}} R T}$

= e^(("3 mols e"^(-)"/1 mol atom"cdot"96485.33 C/mol e"^(-)cdot"0.59 V")"/"("8.314472 J/mol"cdot"K"cdot"298.15 K")

$= {e}^{68.89}$

$\approx \textcolor{b l u e}{8.3 \times {10}^{29}}$

I found the actual answer to this elsewhere, and all they give is ${10}^{30}$, which is close enough, since the other given answer choices were ${10}^{15}$, ${10}^{20}$, and ${10}^{25}$.