The Eᵒcell and the reaction are given below. What is the Gibbs free energy of the system at standard conditions? Pb²+ (aq) + Mg(s) = Pb(s) + Mg2+ (aq) Ecell = +2.24 V Eº AG° = [?] kJ

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### Calculating Gibbs Free Energy from Cell Potential

Below we explore the relationship between the standard electrode potential (\(E^\circ_{cell}\)) and Gibbs free energy (\(\Delta G^\circ\)) for the following redox reaction at standard conditions.

#### Reaction Details:
\[ \text{Pb}^{2+}(\text{aq}) + \text{Mg}(\text{s}) \rightleftharpoons \text{Pb}(\text{s}) + \text{Mg}^{2+}(\text{aq}) \]

Given:
\[ E^\circ_{cell} = +2.24 \, \text{V} \]

#### Problem Statement:
What is the Gibbs free energy of the system at standard conditions?

#### Formula for Gibbs Free Energy:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]

where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( n \) is the number of moles of electrons transferred,
- \( F \) is the Faraday constant (\(96,485 \, \text{C/mol}\)),
- \( E^\circ_{cell} \) is the standard cell potential.

#### Calculation Placeholder:
\[ \Delta G^\circ = [ \ \boxed{?} \ ] \, \text{kJ} \]

**Input Field:**
Please enter either a \( + \) or \( - \) sign and the magnitude of your answer.

**Submission Box:**
- **Enter your value for Free Energy, kJ:**
  \[ \boxed{\phantom{Enter your value here}} \]

- [ **Enter** ] (button to submit the calculated value)
Transcribed Image Text:### Calculating Gibbs Free Energy from Cell Potential Below we explore the relationship between the standard electrode potential (\(E^\circ_{cell}\)) and Gibbs free energy (\(\Delta G^\circ\)) for the following redox reaction at standard conditions. #### Reaction Details: \[ \text{Pb}^{2+}(\text{aq}) + \text{Mg}(\text{s}) \rightleftharpoons \text{Pb}(\text{s}) + \text{Mg}^{2+}(\text{aq}) \] Given: \[ E^\circ_{cell} = +2.24 \, \text{V} \] #### Problem Statement: What is the Gibbs free energy of the system at standard conditions? #### Formula for Gibbs Free Energy: \[ \Delta G^\circ = -nFE^\circ_{cell} \] where: - \( \Delta G^\circ \) is the standard Gibbs free energy change, - \( n \) is the number of moles of electrons transferred, - \( F \) is the Faraday constant (\(96,485 \, \text{C/mol}\)), - \( E^\circ_{cell} \) is the standard cell potential. #### Calculation Placeholder: \[ \Delta G^\circ = [ \ \boxed{?} \ ] \, \text{kJ} \] **Input Field:** Please enter either a \( + \) or \( - \) sign and the magnitude of your answer. **Submission Box:** - **Enter your value for Free Energy, kJ:** \[ \boxed{\phantom{Enter your value here}} \] - [ **Enter** ] (button to submit the calculated value)
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