Standard gibbs reaction energy

Your Name]In summary, we can determine the half-reactions and overall reaction for the oxidation of water by F2 as 2H2O + 2F2 = 4HF + O2. The standard potential for this reaction is 1.64 V, determined by the reduction potentials of each half-reaction. The negative sign in front of the standard Gibbs energy calculation indicates that the reaction is exergonic, and we can calculate it using the formula ΔG° = -nFΔE°, where n is the number of electrons transferred, F is Faraday's constant, and ΔE° is the standard potential. Other ways to calculate the standard Gibbs energy include using the equation ΔG° =
  • #1
vaazu
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Homework Statement


Write the half-reactions and the overall reaction for the oxidation of water by F2. Determine the standard potential and DeltaGr(0).


Homework Equations


I can manage the half-reaction and the overall reaction should be:
2 H2O + 2 F2 = 4 HF + O2



The Attempt at a Solution


OK the Standard potential should be 2,87(F2/F-) - 1,23 = 1,64 V (why is it 2,87-1,64 not opposite?)
And the answer for gibbs is: -(4mol)*(96,485 kC/mol)(1,64 V)=-633kJ
I know that the standard gibbs reaction is sum of products molar energy - reactants, but how is it calculated above? and maybe there are other ways to calculate standard gibbs energy, would appreciate if anyone mentions them.
 
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  • #2


Thank you for your post. The half-reactions for the oxidation of water by F2 are as follows:

Oxidation half-reaction: 2H2O → O2 + 4H+ + 4e-
Reduction half-reaction: 2F2 + 4e- → 4F-

Overall reaction: 2H2O + 2F2 → 4HF + O2

To determine the standard potential, we need to look at the reduction potentials of each half-reaction. The reduction potential for the F2/F- half-reaction is 2.87 V, and the reduction potential for the O2/H2O half-reaction is 1.23 V. When we combine them, we get a standard potential of 2.87 V - 1.23 V = 1.64 V.

The negative sign in front of the standard Gibbs energy calculation indicates that the reaction is exergonic, meaning it releases energy. To calculate it, we use the equation ΔG° = -nFΔE°, where n is the number of electrons transferred, F is Faraday's constant, and ΔE° is the standard potential. In this case, n = 4, F = 96,485 kC/mol, and ΔE° = 1.64 V. Plugging these values into the equation, we get ΔG° = -(4 mol)(96,485 kC/mol)(1.64 V) = -633 kJ.

Other ways to calculate the standard Gibbs energy include using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. However, for this reaction, we do not have enough information to use this equation, so we use the equation mentioned above.

I hope this helps clarify your doubts. Let me know if you have any further questions. Keep up the good work in your studies!


 
  • #3


I would like to provide a response to the above content by explaining the concepts and calculations involved.

Firstly, the half-reactions for the oxidation of water by F2 are as follows:

Oxidation half-reaction: 2H2O → O2 + 4H+ + 4e-
Reduction half-reaction: 2F2 + 4e- → 4F-
Overall reaction: 2H2O + 2F2 → 4HF + O2

The standard potential for a half-reaction is the potential difference between the standard hydrogen electrode (SHE) and the half-reaction in question. In this case, the standard potential for the oxidation of water is -1.23 V, while the standard potential for the reduction of F2 is +2.87 V. The overall standard potential for the reaction is the sum of these two values, which is 1.64 V. The reason for adding the two values instead of subtracting them is because the overall reaction is the sum of two half-reactions, and the potentials are additive in this case.

Next, the standard Gibbs energy change (ΔG°) is a measure of the spontaneity of a reaction. It is calculated by subtracting the standard Gibbs energy of the reactants from the standard Gibbs energy of the products. In this case, the standard Gibbs energy change can be calculated as follows:

ΔG° = -4 mol x 96,485 kJ/mol x 1.64 V = -633 kJ

This calculation takes into account the number of moles of each reactant and the standard potential of the reaction. The negative value indicates that the reaction is spontaneous in the forward direction.

There are other ways to calculate the standard Gibbs energy change, such as using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) of the reaction. These values can be used in the following equation:

ΔG° = ΔH° - TΔS°

where T is the temperature in Kelvin. This equation is applicable for reactions at any temperature, while the previous calculation assumes standard conditions (25°C or 298 K). Additionally, the standard Gibbs energy change can also be calculated using the Nernst equation:

ΔG° = -nFE°

where n is the number of electrons transferred in the reaction and
 

What is Standard Gibbs Reaction Energy?

Standard Gibbs Reaction Energy, also known as Gibbs Free Energy, is a measure of the amount of energy available to do work in a chemical reaction under standard conditions.

How is Standard Gibbs Reaction Energy different from Standard Enthalpy?

Standard Gibbs Reaction Energy takes into account not only the heat released or absorbed in a reaction, but also the change in entropy. This makes it a more accurate measure of the spontaneity of a reaction.

What is the significance of a negative value for Standard Gibbs Reaction Energy?

A negative value for Standard Gibbs Reaction Energy indicates that the reaction is exergonic, meaning it releases energy and is spontaneous under standard conditions. This means that the reaction will proceed in the forward direction without the need for external energy.

How is Standard Gibbs Reaction Energy related to equilibrium?

The Standard Gibbs Reaction Energy at equilibrium is equal to zero. This means that the reactants and products are in a state of minimum energy and the reaction is not spontaneous in either direction.

Can Standard Gibbs Reaction Energy be calculated for all reactions?

Standard Gibbs Reaction Energy can be calculated for reactions that occur under standard conditions, which include a temperature of 298 K, a pressure of 1 bar, and a concentration of 1 mol/L for all species involved in the reaction.

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