Consider an electrochemical cell: \(\mathrm{A}(\mathrm{s})\left|\mathrm{A}^{\mathrm{n}+}(\mathrm{aq}, 2 \mathrm{M}) \| \mathrm{B}^{2 \mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{B}(\mathrm{s})\). The value of \(\Delta H^{\ominus}\) for the cell reaction is twice that of \(\Delta G^{\theta}\) at \(300 \mathrm{~K}\). If the emf of the cell is zero, the \(\Delta S^{\theta}\) (in \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) of the cell reaction per mole of \(\mathrm{B}\) formed at \(300 \mathrm{~K}\) is (Given: \(\ln (2)=0.7, R\) (universal gas constant) \(=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \cdot H, S\) and \(G\) are enthalpy, entropy and Gibbs energy, respectively.)

Short Answer

Expert verified
-11.62 \(\mathrm{J K^{-1} mol^{-1}}\)

Step by step solution

01

Understanding the Information Given

From the information provided, we understand that the cell operates at 300 K, the standard enthalpy change \( \Delta H^{\ominus} \) is twice the standard Gibbs free energy change \( \Delta G^{\theta} \) for the cell reaction, and the electromotive force (EMF) of the cell is zero. We are asked to find the standard entropy change \( \Delta S^{\theta} \) for the reaction when one mole of B is formed. Additionally, we know that at equilibrium, \( \Delta G = 0 \) and the Nernst equation can be used.
02

Relating Enthalpy (\( \Delta H^{\ominus} \)) and Gibbs Free Energy (\( \Delta G^{\theta} \))

We use the relationship between Gibbs free energy, enthalpy, and entropy, which is given by \( \Delta G^{\theta} = \Delta H^{\ominus} - T \Delta S^{\theta} \) at constant temperature. We also know that \( \Delta G^{\theta} = \frac{\Delta H^{\ominus}}{2} \) from the problem statement.
03

Setting up the Equation for Entropy Change

Since the EMF is zero, \( \Delta G^{\theta} = 0 \) at equilibrium, which implies \( 0 = \Delta H^{\ominus} - T \Delta S^{\theta} \) or \( \Delta S^{\theta} = \frac{\Delta H^{\ominus}}{T} \). Substituting \( \Delta H^{\ominus} = 2 \Delta G^{\theta} \) gives us \( \Delta S^{\theta} = \frac{2 \Delta G^{\theta}}{T} \).
04

Calculating the Value of \( \Delta G^{\theta} \)

We can use the Nernst equation to find \( \Delta G^{\theta} \). Nernst equation at equilibrium where EMF is zero: \( 0 = -\frac{\Delta G^{\theta}}{nF} + RT\ln Q \) where \( Q \) is the reaction quotient. Since the reaction is at standard conditions (concentration of 1 M), \( Q = \frac{[A^n+]}{[B^{2n+}]} = \frac{2M}{1M} = 2 \) and \(\Delta G^{\theta} = -RT\ln Q = -8.3 \times 300 \times 0.7 = -1743 J/mol \).
05

Calculating the Entropy Change \( \Delta S^{\theta} \)

Using the value of \( \Delta G^{\theta} \) calculated and substituting it into the equation for \( \Delta S^{\theta} \), we get \( \Delta S^{\theta} = \frac{2 \times (-1743)}{300} = \frac{-3486}{300} = -11.62 \mathrm{J K^{-1} mol^{-1}} \).

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Key Concepts

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

Thermodynamics in Electrochemistry
In electrochemistry, understanding thermodynamics is crucial because it reveals how energy changes during chemical reactions in an electrochemical cell. An electrochemical cell uses chemical reactions to produce electrical energy, and the thermodynamics of these reactions tell us whether the process is spontaneous and how efficient it can be.

Thermodynamics ties into electrochemistry primarily through the study of energy changes involving Gibbs free energy (\( \triangle G \)), enthalpy (\( \triangle H \bar{\theta} \bar{\theta}= -RT\text{ln}Q \triangle S \theta \triangle H \ominus \triangle S \theta \triangle H \ominus \triangle G \theta \triangle G \theta \triangle S \theta \bar{\theta} \bar{\theta} \bar{\theta} \triangle S \theta \bar{\theta} \bar{\theta} \)
Gibbs Free Energy
At the heart of predicting spontaneity in electrochemical reactions is Gibbs free energy (\( \triangle G \bar{\theta} = \triangle H \ominus - T\triangle S \theta \triangle G \theta \triangle H \ominus = \frac{\triangle H \ominus}{2} \bar{\theta} \triangle G \theta \bar{\theta} \ominus \ominus \theta \triangle S \theta \triangle S \theta = \frac{\triangle H \ominus}{T} \bar{\theta} = \frac{2\triangle G \theta}{T} \triangle G \theta \)
Nernst Equation
A powerful tool in the study of electrochemistry is the Nernst equation, vital for calculating the electromotive force (EMF) of a cell under non-standard conditions. However, it can also inform us about the change in Gibbs free energy (\( \triangle G \theta = -RT\text{ln}Q \bar{\theta} = -RT\text{ln}Q \triangle G \theta \theta = -RT\text{ln}Q \bar{\theta} = -8.3 \times 300 \times 0.7 = -1743 \text{J/mol} \ominus \ominus \ominus \bar{\theta} \)
Entropy and Enthalpy Relationship
Entropy (\( \triangle S \triangle H \triangle S \triangle G = \triangle H - T\triangle S \triangle H \ominus \triangle G \theta \triangle S \theta \theta \triangle S \theta \ominus \ominus \ominus \triangle S \theta = \frac{2 \times (-1743)}{300} = \frac{-3486}{300} = -11.62 \text{J K}^{-1} \text{mol}^{-1} \ominus \ominus \ominus \bar{\theta} \triangle S \theta \bar{\theta} = \frac{\triangle H \ominus}{T} \)

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Most popular questions from this chapter

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