# D41.1 Relationships among ΔG°, K°, and E°

The thermodynamics of a redox reaction can be described in terms of Δ_{r}*G*° and *K°*, and these can be further related to *E*°_{cell} because they all describe whether the redox reaction would be reactant-favored or product-favored at equilibrium.

In a voltaic cell, the difference in Gibbs free energy between products and reactants allows the cell to do electrical work. We represent electrical work done by the cell as *w*_{elec}.

*w*

_{elec}(J) = charge transferred (C) × potential difference (V)

The charge of 1 mole of electrons is known as the **Faraday’s constant** (*F*):

Hence, the total quantity of charge transferred per mole of a redox reaction is:

In this equation, ** n** is

*the number of electrons transferred in the balanced redox reaction*, which can be obtained from the balanced half-reactions that are added to produce the overall redox reaction. For example, in the reaction:

^{3+}(aq) + 3 Zn(s) ⟶ 3 Zn

^{2+}(aq) + 2 Au(s)

6 e^{−} are transferred because each half-reaction, after multiplying by a factor to balance electrons, involves 6 e^{−}:

Oxidation: | 3 × [Zn(s) | ⟶ | Zn^{2+}(aq) + 2 e^{−}] |

Reduction: | 2 × [Au^{3+}(aq) + 3 e^{−} |
⟶ | Au(s)] |

Overall: | 2 Au^{3+}(aq) + 3 Zn(s) |
⟶ | 3 Zn^{2+}(aq) + 2 Au(s) |

When this redox reaction happens once, 6 e^{−} are transferred. When a mole of this reaction takes place, 6 moles of e^{−} are transferred; that is, 6 mol* × *96485 C/mol = 578910 C are transferred. Hence, we have:

w_{elec} |
= | charge transferred × potential difference |

w_{elec} |
= | (nF) × (E_{cell}) |

If we operate a voltaic cell such that the maximum possible electrical work, *w*_{elec}, is done and the only work done is electrical work, then:

_{r}

*G*= −

*w*

_{max}= −

*w*

_{elec}= −nFE

_{cell}

The negative sign makes sense because a positive *E*_{cell} indicates a spontaneous redox reaction while for Δ_{r}*G* a negative value indicates a spontaneous reaction.

If all the reactants and products are in their standard states, then the equation becomes:

_{r}

*G°*= –

*nFE°*

_{cell}

This equation also links standard cell potentials to equilibrium constants, since:

*nFE°*

_{cell}= Δ

_{r}

*G*° = –

*RT*ln

*K*°

Therefore:

Thus, if any one of Δ_{r}*G*°, *K°*, or *E*°_{cell} is known or can be calculated for a redox reaction, the other two quantities can be determined using the relationships shown below. Moreover, any of the three quantities can be used to determine whether a reaction would be product-favored at equilibrium.

**Exercise: Standard Gibbs Free Energy Change for a Redox Reaction
**

**Exercise: Gibbs Free Energy Change and Equilibrium Constant for a Redox Reaction
**

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