# Cell Potential Relationships

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

*E*°_{cell} and Δ*G*°

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

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

where

- Δ
*G*° = 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 Δ*G*° (spontaneous) - A negative voltage (
*E*°_{cell}) gives a positive Δ*G*° (nonspontaneous)

**Faraday’s Constant**

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)

## Δ*G*°_{cell} and *K*

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

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

where

- Δ
*G*° = 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 Δ
*G*° gives a*K*> 1 (product favored) - A positive Δ
*G*° 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 Δ*G*°, *E*°_{cell}, and *K*.

(Image from openStax)

Given that Δ*G*°, *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 Δ*G*° 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.