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Chapter 20: Problem 7

Consider a redox reaction for which \(E^{\circ}\) is a negative number. (a) What is the sign of \(\Delta G^{\circ}\) for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than \(1 ?\) (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings? [Section 20.5\(]\)

Short Answer

Expert verified
(a) The sign of \(\Delta G^{\circ}\) is positive since \(E^{\circ}\) is negative. (b) The equilibrium constant, K, is smaller than 1. (c) An electrochemical cell based on this reaction cannot accomplish work on its surroundings.

Step by step solution

01

Calculating the sign of 螖G掳

Using the equation: \(\Delta G^\circ = -nFE^\circ\) Since \(E^\circ\) is a negative number, and both n (number of moles of electrons transferred) and F (Faraday's constant) are positive numbers, the product of -nF and the negative \(E^\circ\) will be a positive number. Thus, the sign of \(\Delta G^\circ\) is positive.
02

Determining the equilibrium constant comparison to 1

Next, we will use the equation: \(\Delta G^\circ = -RT\ln K\) Since \(\Delta G^\circ\) is positive, then the \(\ln K\) must be a negative number, because R (Ideal Gas Constant) and T (Temperature in Kelvin) are both positive numbers. Recall that the natural logarithm of a number between 0 and 1 is negative. Hence, the equilibrium constant K is smaller than 1.
03

Evaluating if the electrochemical cell can perform work on its surroundings

In an electrochemical cell, if the cell potential, \(E^\circ\), is negative, it indicates the reaction occurring is a non-spontaneous process. A non-spontaneous redox reaction will not be able to accomplish work on its surroundings under standard conditions. Therefore, an electrochemical cell based on this reaction cannot accomplish work on its surroundings.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G \), is a measure of the maximum reversible work that a thermodynamic system can perform at a constant temperature and pressure. It helps predict whether a reaction will be spontaneous or non-spontaneous.
For redox reactions, the change in Gibbs free energy \( \Delta G^{\circ} \) can be calculated using the equation:
  • \( \Delta G^{\circ} = -nFE^{\circ} \)
Where:
  • \( n \) is the number of moles of electrons transferred in the reaction.
  • \( F \) is Faraday's constant \((96485 \, C/mol)\).
  • \( E^{\circ} \) is the standard cell potential.
If \( E^{\circ} \) is negative, it indicates that \( \Delta G^{\circ} \) is positive, meaning the reaction is non-spontaneous. This implies that external energy would be needed for the reaction to proceed.
Equilibrium Constant
The equilibrium constant, \( K \), gives an idea of the ratio of products to reactants at equilibrium for a given reaction at a certain temperature. It can be related to Gibbs free energy change by the equation:
  • \( \Delta G^{\circ} = -RT\ln K \)
Where:
  • \( R \) is the ideal gas constant \((8.314 \, J/mol\cdot K)\).
  • \( T \) is the temperature in Kelvin.
When \( \Delta G^{\circ} \) is positive, it indicates that \( \ln K \) is negative, which means the value of \( K \) is less than 1. This suggests that reactants are favored at equilibrium over the products, reflecting a non-spontaneous process where equilibrium lies towards the reactants.
Electrochemical Cell
An electrochemical cell is a device that can convert chemical energy into electrical energy through redox reactions or vice versa. It consists of two half-cells, each containing an electrode and an electrolyte.
For an electrochemical cell to do work, the cell potential \( E^{\circ} \) should be positive, indicating that the redox reaction is spontaneous and can provide electrical energy.
  • A negative \( E^{\circ} \) value signals that the reaction is non-spontaneous, meaning the cell can't do work on its surroundings without an external power source.
Thus, in a redox reaction where \( E^{\circ} \) is negative, the electrochemical cell it forms cannot accomplish work spontaneously, and external input would be required to drive the reaction.

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Most popular questions from this chapter

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ, \(1973,\) Vol. 50,61 ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten \(\mathrm{NaCl} ?\) (d) Why is sodium metal not obtained when an aqueous solution of \(\mathrm{NaCl}\) undergoes electrolysis?

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$\mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s)$$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\).

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)\) (acidic solution) (b) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q)\) (acidic solution) (c) \(\mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q)\) (acidic solution) (e) \(\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g)\) (basic solution) (f \(\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (basic solution)
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