Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write the equations relating \(\Delta G^{\circ}\) and \(K\) to the standard emf of a cell. Define all the terms.

Short Answer

Expert verified
ΔG° = -nFE° and ΔG° = -RT ln K relate ΔG° to E° and K. E° = (RT/nF) ln K relates E° to K.

Step by step solution

01

Understand the Relationship Between ΔG° and E°

The relationship between the standard free energy change (ΔG°) and the standard cell potential (E°) is given by the equation:\[ΔG° = -nFE°\]where ΔG° is the standard free energy change, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (approximately 96,485 C/mol), and E° is the standard cell potential in volts (V).
02

Relate ΔG° to the Equilibrium Constant (K)

The relationship between the standard free energy change (ΔG°) and the equilibrium constant (K) is expressed as:\[ΔG° = -RT \, ext{ln} \, K\]where R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant of the reaction.
03

Relate E° to the Equilibrium Constant (K)

By combining the relationship between ΔG° and E° with that of ΔG° and K, we can derive a direct relationship between E° and K:\[E° = \frac{RT}{nF} \ln K\]This equation links the standard cell potential with the equilibrium constant. To convert this to base 10 logarithms, the equation often appears as:\[E° = \frac{0.0592}{n} \log K \quad (at \ 298 \ K)\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation Relations
Understanding the equations that connect Gibbs Free Energy and other factors is key in chemistry and thermodynamics. The relationships form a fundamental basis for predicting and explaining chemical reactions.
\(\Delta G^{\circ}\) (standard free energy change) is linked to two important quantities through equations. First, it relates to the standard emf (electromotive force, E°) of a cell via the equation:
  • \(\Delta G^{\circ} = -nFE^{\circ}\)
    • \(n\) stands for the number of moles of electrons in the redox reaction.
    • \(F\) is Faraday's constant, approximately 96,485 C/mol.
This formula shows the energy change in a reaction and how it relates to the cell's potential difference. The more negative \(\Delta G^{\circ}\), the more likely the reaction is spontaneous.
Another important equation is:
  • \(\Delta G^{\circ} = -RT \ln K\)
    • \(R\) is the universal gas constant (8.314 J/mol·K).
    • \(T\) represents the temperature in Kelvin.
    • \(K\) is the equilibrium constant of the reaction.
This relationship tells us how the spontaneity expressed by \(\Delta G^{\circ}\) links to the extent of a reaction at equilibrium.
Standard Emf
The standard emf, or standard cell potential (E°), measures the voltage or electrical potential difference of a cell under standard conditions. This is a way to gauge the tendency of a redox reaction to occur.
When calculating E°, remember it reflects the potential energy difference between the cathode and anode in an electrochemical cell. The sign of E° offers insight:
  • Positive E° indicates a spontaneous reaction under standard conditions.
  • Negative E° suggests a non-spontaneous reaction unless additional energy is provided.
To solidify this understanding, look at the equation relating \(\Delta G^{\circ}\) and E°:
  • \(\Delta G^{\circ} = -nFE^{\circ}\)
Here, a positive E° contributes to a negative \(\Delta G^{\circ}\), reinforcing the notion of a spontaneous process. This connection helps in predicting the feasibility of various electrochemical reactions.
Equilibrium Constant
The equilibrium constant (K) plays a vital role in chemistry, reflecting the balance of a reaction at equilibrium. It gives a quantitative idea of the position of equilibrium and tells how far a reaction progresses.
In the relationship \(\Delta G^{\circ} = -RT \ln K\), we can see:
  • A large K value (>1) corresponds to a negative \(\Delta G^{\circ}\), suggesting the reaction favors products and is spontaneous.
  • A small K value (<1) means positive \(\Delta G^{\circ}\), where the reactants are favored.
This explains whether reactions are product or reactant-favored. Importantly, under standard conditions, we can also relate K to E°:
  • \(E^{\circ} = \frac{RT}{nF} \ln K\)
This equation bridges E° and K, allowing understanding of how electrical potential relates to the chemical equilibrium.
Standard Cell Potential
The standard cell potential (E°) is a significant measure in electrochemistry since it indicates the voltage a cell can supply when operating reversibly under standard conditions (298 K, 1 atm pressure, and 1 M concentration solutions).
It is calculated using the standard reduction potentials of the cathode and anode:
  • \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\)
This potential drives the movement of electrons from the anode to the cathode. When considering equilibrium, the equation \(E^{\circ} = \frac{0.0592}{n} \log K\) at 298 K is especially useful. It relates the cell potential to the equilibrium constant accounting for the number of electrons (n) transferred.
Understanding E° helps predict the energetics and directionality of electrochemical reactions. Its integration with \(\Delta G^{\circ}\) and K makes it a central concept in understanding both thermodynamic stability and chemical equilibrium of reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the Nernst equation for the following processes at some temperature \(T\) : (a) \(\mathrm{Mg}(s)+\mathrm{Sn}^{2+}(a q) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\operatorname{Sn}(s)\) (b) \(2 \mathrm{Cr}(s)+3 \mathrm{~Pb}^{2+}(a q) \rightleftarrows 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{~Pb}(s)\)

An aqueous solution of a platinum salt is electrolyzed at a current of \(2.50 \mathrm{~A}\) for \(2.00 \mathrm{~h}\). As a result, \(9.09 \mathrm{~g}\) of metallic Pt is formed at the cathode. Calculate the charge on the Pt ions in this solution.

The nitrite ion \(\left(\mathrm{NO}_{2}^{-}\right)\) in soil is oxidized to the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right)\) by the bacterium Nitrobacter agilis in the presence of oxygen. The half-reactions are: \(\mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{NO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=0.42 \mathrm{~V}\) $$\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=1.23 \mathrm{~V}$$ Calculate the yield of ATP synthesis per mole of nitrite oxidized.

Given the following standard reduction potentials, calculate the ion-product, \(K_{\mathrm{w}},\) for water at \(25^{\circ} \mathrm{C}:\) $$ \begin{array}{ll} 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(\mathrm{~g}) & E^{\circ}=0.00 \mathrm{~V} \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q) & E^{\circ}=-0.83 \mathrm{~V} \end{array} $$

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M},\) \([\mathrm{H}]=0.050 M,\) and \(P_{\mathrm{H}}=1.0 \mathrm{~atm} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free