# Cell Potential Relationships

A cell potential can be related to Gibbs free energy (ΔG) and an equilibrium constant, K.

## E°cell and ΔG°

A cell potential can be directly related to a Gibbs free energy change such that

$\Delta G^{\circ} = -nFE^{\circ}_{\mathrm{cell}}$

where

• ΔG° = free energy (in kJ mol–1)
• n = moles of e transferred in redox reaction
• F = Faraday’s constant = 96,485 C/mol e (or J/V mol e)
• E°cell = cell potential (in V or J C–1)

Important Features

• A positive voltage (E°cell) gives a negative ΔG° (spontaneous)
• A negative voltage (E°cell) gives a positive ΔG° (nonspontaneous)

Faraday’s constant, F, is a measure of the electric charge per 1 mole of electrons.

Since

$\mathrm{1~mol~of~electrons = 6.022 \times 10^{23}~electrons}$ and the charge of one electron is approximately 1.602 × 10–19 C, Faraday’s constant is simply

\begin{align*} F &= \dfrac{(1.602\times 10^{-19}~\mathrm{C})(6.022\times 10^{23})}{\mathrm{1~mol~e^-}}\\ &= 96,485~\mathrm{C/mol}~e^{-} \end{align*}

If V = J C–1, then C = J V–1. Faraday’s constant can be expressed in J/V mol e.

$\dfrac{\mathrm{C}}{\mathrm{mol}~e^-} \rightarrow \dfrac{\mathrm{J}}{\mathrm{V~mol}~e^-}$

## E°cell and K

A cell potential can be directly related to an equilibrium constant such that

$E_{\mathrm{cell}}^{\circ} = \left (\dfrac{RT}{nF}\right )\ln K$

where

• E°cell = cell potential (in V or J C–1)
• R = gas constant = 8.315 J mol–1 K–1
• T = temperature (in K)
• n = moles e transferred
• F = Faraday’s constant = 96,485 C/mol e (or J/V mol e)
• K = equilibrium constant = (unitless; “products over reactants”)

Important Features

• A positive voltage (E°cell) gives a K > 1 (product favored)
• A negative voltage (E°cell) gives a K < 1 (reactant favored)

## ΔG°cell and K

As demonstrated in Chapter 15, free energy is related to an equilibrium constant

$\Delta G^{\circ} = -RT\ln K$

where

• ΔG° = free energy (in kJ mol–1)
• R = gas constant = 8.315 J mol–1 K–1
• T = temperature (in K)
• K = equliibrium constant = (unitless; “products over reactants”)

Important Features

• A negative ΔG° gives a K > 1 (product favored)
• A positive ΔG° gives a K < 1 (reactant favored)

## The Golden Triangle

The golden triangle is not a real name (I just made that up for fun). However, it does show, visually, the relationships we’ve introduced here between ΔG°, E°cell, and K. (Image from openStax)

Given that ΔG°, E°cell, and K are all related, we can make the following qualitative conclusions.

K ΔG° E°cell Outcome under standard conditions
> 1 < 0 > 0 Reaction is spontaneous; product favored
< 1 > 0 < 0 Reaction is non-spontaneous; reactant favored
= 1 = 0 = 0 Reaction is at equilibrium; reactants and products equally abundant

### Example

Determine the standard free energy (in kJ mol–1) and equilibrium constant for the following galvanic cell at 25 °C.

$\mathrm{Cr}(s) ~\rvert ~\mathrm{Cr^{2+}}(aq) ~\rvert\lvert~ \mathrm{Pb}^{2+}(aq) ~ \lvert ~\mathrm{Pb}(s)$

Find the standard cell potential.

\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}\\ &= E_{\mathrm{Pb^{2+}/Pb}}^{\circ} - E_{\mathrm{Cr^{2+}/Cr}}^{\circ}\\ &= (-0.13~\mathrm{V}) - (-0.913~\mathrm{V})\\ &= 0.78~\mathrm{V}\\ &= 0.78~\mathrm{J~C^{-1}} \end{align*}

Convert the standard cell potential into a standard free energy.

\begin{align*} \Delta G^{\circ} &= -nFE^{\circ}_{\mathrm{cell}}\\ &= -(2~\mathrm{mol}~e^-)(96,485~\mathrm{C~mol^{-1}}~e^-)(0.78~\mathrm{J~C^{-1}})\left( \dfrac{1~\mathrm{kJ}}{\mathrm{10^3~J}} \right ) \\ &= -150.5~\mathrm{kJ~mol^{-1}} \end{align*}

Determine K from the standard free energy.

\begin{align*} \Delta G^{\circ} &= -RT\ln K \\ K &= e^{-\dfrac{\Delta G^{\circ}}{RT}}\\ &= e^{-\dfrac{-150.5~\mathrm{kJ~mol^{-1}}}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}} \\ &= 2.32\times 10^{26} \end{align*}

Concept Check: We expect a large K given the very negative ΔG° value. A very spontaneous reaction is very product favored.

Alternative: You could also determine K from the standard cell potential. Be sure to convert E°cell from J C–1 (which is V) into kJ C–1.

\begin{align*} E_{\mathrm{cell}}^{\circ} &= \left (\dfrac{RT}{nF}\right )\ln K \\ K &= e^{\dfrac{nFE_{\mathrm{cell}}^{\circ}}{RT}}\\ &= e^{\dfrac{(2~\mathrm{mol}~e^-)(96,485~\mathrm{C~mol^{-1}}~e^-)(0.00078~\mathrm{kJ~C^{-1}})}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}}\\ &= 2.33\times 10^{26} \end{align*}

Despite which way you solve for K, you will notice that the answers match to one decimal place.

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