### Calculating Gibbs Free Energy from Cell PotentialBelow 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)

Answered: Image /qna-images/answer/2aa475db-3344-4aa5-b75a-4c55a80585fd.jpg

Answered: The Eᵒcell and the reaction are given…

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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Question

### 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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