How to calculate ∆G (delta G) from Ecell? – (Delta G from Ecell)
∆G stands for the Gibbs free energy change of a chemical reaction. The redox reaction is a type of chemical reaction that facilitates the generation of electrical current in an electrochemical cell.
If we know the electrical potential difference (Ecell), we can easily find ∆G (Delta G) by using the equation ∆G = – nFEcell.
Let us teach you how through the examples given in this article.
What is ∆G (Delta G)?
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
- If ∆G > 0: The reaction is non-spontaneous.
- If ∆G = 0: The reaction stays at
In this way, ∆G (Delta G) not only predicts whether a physical or chemical change is possible but also helps in determining the direction of the reaction.
What is Ecell?
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.
In an electrochemical cell, oxidation occurs at the anode (negatively charged electrode), while reduction takes place at the cathode (positively charged electrode).
The electrons produced at the anode are transferred to the cathode, which generates an electrical current.
Thus, Ecell is calculated as follows:
Electrical potential difference = Ecell = Eϴ (ox) + Eϴ (red)
Where,
- Eϴ (red) = Standard cell potential of the reduction half-cell (cathode)
- Eϴ (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.
How to find ∆G (delta G) from Ecell?
The formula that relates ∆G (delta G) to Ecell is called the cell potential equation:
In ∆ G = -nFE:
- ∆G = Gibbs free energy change (Units: J/mol or kJ/mol).
- n = number of moles of electrons transferred from the anode to the cathode (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).
Solved examples for finding ∆G (Delta G) from Ecell
Example # 1: Use the equation ∆G = -nFE to find the Gibbs free energy change for the redox reaction shown below: Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) |
Solution: To solve this question, use these steps: Step I: From the balanced chemical equation, identify which substance is oxidized and which one is reduced.
This means iron (Fe) is oxidized by losing 2 moles of electrons while copper (Cu) is reduced in this example by gaining these electrons. Step II: Write the two-half-cell reactions, finding the number of moles of electrons transferred. Oxidation half-cell: Fe(s) → Fe2+(aq) + 2e– Reduction half-cell: Cu2+(aq) + 2e–→ Cu(s) Here n = 2. Step III: Find Eϴ (ox) and Eϴ (red) from the electrochemical series. Fe(s) → Fe2+(aq) + 2e– (Eϴ = 0.47 V) Cu2+(aq) + 2e– → Cu(s) (Eϴ = 0.34 V) Step IV: Calculate Ecell Ecell = Eox + Ered Ecell = 0.47 + 0.34 Ecell = 0.81 V Step V: Use the formula ∆G = -nFEcell to find ∆G (Delta G) by substituting the pre-determined Ecell, n, and F= 96,485 C/mol. ⇒ ∆ G = -nFE ∆G = -(2)(96485)(0.81) ∆G = -1.56 x 105 J/mol = -156 kJ/mol Result: The Gibbs free energy change for the above reaction is -156 kJ/mol, which implies that the reaction is spontaneous in the forward direction. |
Example # 2: Find Gibbs free energy change (∆G) for the chemical reaction shown below: 2 Br–(aq) + I2(s) → Br2(l) + 2 I–(aq) |
Solution: The half-cell equations for the reaction are: Reduction half-cell: I2 (aq) + 2e– → 2 I–(aq) (Eϴ = 0.54 V) Oxidation half-cell: 2 Br–(aq) → Br2(l) + 2e– (Eϴ = – 1.09 V) Ecell = Eox + Ered Ecell = 0.54 + (-1.09) Ecell = – 0.55 V Applying the formula: ⇒ ∆ G = -nFE ∆G = -(2)(96485)(-0.55) ∆G = 1.06 x 105 J/mol = 106 kJ/mol Result: The Gibbs free energy change for the above reaction is 106 kJ/mol, which means that the reaction is non-spontaneous in the forward direction. In other words, some external interference is required to propel it forward. |
Example # 3: Find Gibbs free energy change (Delta G) for the chemical reaction shown below: 2 Fe3+(aq) + 2 I–(aq) → 2 Fe2+(aq) + I2(s) |
Solution: The half-cell equations for the reaction are: Reduction half-cell: 2 Fe3+ (aq) + 2 e– → 2 Fe2+(aq) (Eϴ = 0.77 V) Oxidation half-cell: 2 I–(aq) →I2(s) + 2 e– (Eϴ = – 0.54 V) Ecell = Eox + Ered Ecell = -0.54 + 0.77 Ecell = 0.23 V Applying the formula: ⇒ ∆G = -nFE ∆G = -(2)(96485)(0.23) ∆G = -4.44 x 104 J/mol = – 44.4 kJ/mol Result: The Gibbs free energy change for the above reaction is – 44.4 kJ/mol, which implies that the reaction is spontaneous in the forward direction. |
FAQ
How do you find Delta G (∆G) from Ecell? |
Gibbs free energy change (∆G or delta G) can be calculated from electrical cell potential (Ecell) by applying the formula ∆G = -nFEcell where n = number of moles of electrons transferred and F= Faraday’s constant (F= 96,485 C/mol). |
How can Delta G and Ecell predict the spontaneity of a chemical reaction? |
If Ecell > 0 and ∆G (Delta G) < 0, → The reaction is spontaneous. If Ecell < 0 and ∆G (Delta G) > 0, → The reaction is non-spontaneous. |
What is ∆G (Delta G) for the following cell reaction?Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2 Ag(s) + 2OH–(aq) Given that Eϴ Ag+/Ag = +0.80 V and Eϴ Zn2+/Zn = -0.76 V
|
Option B is the correct answer, as shown in the calculation below: The half-cell reactions are: Zn(s) → Zn2+(aq) + 2e– (Eox = 0.76 V) 2 Ag+(aq) + 2e– → 2 Ag(s) (Ered = 0.80 V) Ecell = Eox + Ered = 0.76 + 0.80 = 1.56 V Applying the formula: ⇒ ∆G = -nFEcell ∆G = -(2)(96485)(1.56) ∴ ∆G = – 3.01 x 105 J/mol = -301 kJ/mol |
About the author
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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