Delta G = -nFE, How to find n, F?

The equation ∆G = -nFE represents Gibbs free energy change for an electrochemical cell. It is also sometimes known as the cell potential equation.

Let’s find out in this article what is Delta G = -nFE, where and how to use it to solve a diversity of numerical problems.

What are the components of ∆G (Delta G) = -nFE?

In ∆ G = -nFE:

• ∆G (Delta G) = Gibbs free energy change (Units: J/mol or kJ/mol).
• n = number of moles of electrons transferred in the reaction (Units: mol).
• F= Faraday’s constant. It denotes the electrical charge carried by one mole of electrons. The value of F stays constant at F = 96,485 C/mol.
• E= Electrical potential difference or cell potential across an electrochemical cell (Unit: V).

Where and how to use Delta G = -nFE? – Examples  

The equation ∆G = -nFE is most applicable for redox reactions.

A redox reaction is a chemical change in which oxidation and reduction take place simultaneously.

Oxidation refers to the loss of electrons, which leads to an increase in the oxidation state.

In contrast, reduction denotes the gain of electrons that decreases the oxidation state of an element in a molecule or a compound.

For example, in the reaction given below:

H_{2 (g)} + F_{2 (g)} → 2 HF _(g)….Reaction (i)

In this example, hydrogen (H) is oxidized, while fluorine (F) is reduced. The oxidation state of H increases from 0 (in H₂) to +1 (in HF). Contrarily, the oxidation state of F reduces from 0 (in F₂) to -1 (in HF).

Reaction (i) can be represented as two half-cell reactions shown below:

H_{2 (g)} → 2 H⁺ _(g) + 2 e^– (Oxidation half-cell)

F_{2 (g)} + 2 e^– → 2 F^–_(g) (Reduction half-cell)

In the above example, 2 moles of electrons are transferred from H₂ to F₂ for the redox reaction to take place. Thus, n = 2 in the equation ∆G = -nFE.

In an electrochemical cell, oxidation occurs at the anode (negatively charged cell/electrode), while reduction takes place at the cathode (positively charged cell/electrode).

The electrons produced at the anode get transferred to the cathode, which generates an electrical current. 

Electricity generated as a result of a chemical reaction is the basic principle of electrochemistry.

Electrical potential difference = E^ϴ_cell = E^ϴ (ox) + E^ϴ (red)

Where,

• E^ϴ (red) = Standard cell potential of the reduction half-cell (cathode)
• E^ϴ (ox) = Standard cell potential of the oxidation half-cell (anode)

Standard cell potential refers to the maximum electrical potential difference of a given half-reaction when all the components are in their standard physical states.

If the standard cell potentials are given, we can find the value of E_cell and substitute it in the formula ∆G = -nFE to find the Gibbs free energy change (∆G) for the reaction.

By convention, the hydrogen half-cell is called the reference electrode. Its E^ϴ is given an arbitrary value of zero. E^ϴ for all other reactions are thus marked w.r.t E^ϴ (H₂).

Gibbs free energy change (∆G) refers to reversible work done on a system at constant temperature and pressure. In other words, ∆G quantifies the energy that is converted to useful work during a chemical process.

• If ∆G < 0: The reaction is spontaneous
• If ∆G > 0: The reaction is non-spontaneous.
• If ∆G = 0: The reaction stays at equilibrium.

In this way, ∆G not only predicts whether a physical or chemical change is possible but also helps in determining the directionality of the reaction.

Now, let us find out ∆G for H_{2 (g)} + F_{2 (g)} → 2 HF _(g) from the information given below:

H_{2 (g)} → 2 H⁺ _(g) + 2 e^– (E^ϴ = 0)

F_{2 (g)} + 2 e^– → 2 F^–_(g) (E^ϴ = + 2.86 V)

As per the data given above,

E_cell = 2.86 + 0 = 2.86 Volts

We already know that F = 96,485 C/mol and n = 2.

Applying the formula (Delta G = -nFE) and substituting all the known values gives us:

⇒ ∆G = -nFE

∆G = – (2) (96,485) (2.86)

∆G = – 5.52 x 10⁵ J/mol

∴ ∆G = – 551.9 = 552 kJ/mol

∆G < 0 implies that the reaction H_{2 (g)} + F_{2 (g)} → 2 HF _(g) is spontaneous in the forward direction.

Another example is- The redox reaction given below can be split into two half-cell reactions with their standard electrode potentials, as shown:

Balanced chemical equation:

Ni_(s) + Pb (NO₃)_{2 (aq)} → Ni (NO₃)_{2 (aq)} + Pb_(s)……Reaction (ii)

Half-cell reactions:

Oxidation half-cell: Ni _(s) →Ni²⁺ _(aq) + 2e^–    (E^ϴ = + 0.26 V)

Reduction half-cell: Pb²⁺ _(aq) + 2e^– → Pb_(s) (E^ϴ = -0.13 V)

Use this information to find the Gibbs free energy change for reaction ii.

Solution

Step i) Determine the number of moles of electrons transferred (n) from the half-cell reactions.

Here n = 2

Step ii) Find E_cell

E^ϴ_cell = E^ϴ (ox) + E^ϴ (red)

E^ϴ_cell = 0.26 + (-0.13)

E^ϴ_cell = 0.13 V

Step iii) Find ∆G by applying the formula ∆G = -nFE and substituting F = 96,485 C/mol

⇒ ∆G = -nFE

∆G = – (2) (96485) (0.13)

∴ ∆G = -25086.1 J/mol = – 25.1 kJ/mol.

Result: The Gibbs free energy change for the given reaction is -25.1 kJ/mol.

FAQ

What is represented by the equation ∆G (Delta G) = -nFE?

The equation ∆G = -nFE represents Gibbs free energy change for an electrochemical reaction. In ∆G = -nFE: ∆G = Gibbs free energy change n = number of moles of electrons transferred F = Faraday’s constant E = Electrochemical cell potential/ potential difference

Is the value of n in ∆G = -nFE ever negative?

No. n denotes the number of moles of electrons transferred during an electrochemical reaction, which represents a positive integer. Therefore, the value of n is always positive. If no electrochemical reaction occurred, then n = 0, but it is never negative.

How can the two equations ∆G = -nFE and ∆G = ∆H – T∆S be used to derive an expression for energy (E) as a function of temperature (T)?

The common entity between ∆G = -nFE and ∆G = ∆H – T∆S is Gibbs free energy change (Delta G); therefore, equating the two expressions and making E the subject of the formula gives us: -nFE = ∆H – T∆S