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Consider a redox reaction for which \(E^{b}\) is a negative number. (a) What is the sign of \(\Delta G^{\text {e }}\) 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 ΔG is positive, as ΔG = -nFE and E is negative. (b) The equilibrium constant (K) for the reaction is smaller than 1 since ln(K) < 0 when ΔG > 0. (c) No, an electrochemical cell based on this reaction cannot accomplish work on its surroundings, as it has a negative cell potential (E).

Step by step solution

01

Determine the sign of ΔG based on E

The relationship between Gibbs free energy (ΔG) and cell potential (E) is given by the formula: ΔG = -nFE where n is the number of moles of electrons transferred, F is the Faraday constant (96,485 C/mol), and E is the cell potential. If E is a negative number, the product of -nFE will be positive. So: ΔG > 0 (a) The sign of ΔG is positive.
02

Determine the relationship between equilibrium constant (K) and ΔG

We can use the relationship between ΔG and the equilibrium constant (K) to determine whether K is larger or smaller than 1. That relationship is given by the formula: ΔG = -RTln(K) where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant. Since we already know that ΔG is positive: 0 < -RTln(K) Therefore, ln(K) < 0 which implies: K < 1 (b) The equilibrium constant for the reaction is smaller than 1.
03

Determine if the cell can accomplish work on its surroundings

An electrochemical cell can accomplish work on its surroundings when it has a positive cell potential (E). Since E is given to be negative in this problem, the cell cannot accomplish work on its surroundings. (c) No, 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 (\(\Delta G\)) is a crucial thermodynamic quantity used to determine the spontaneity of a chemical reaction. Simply put, it tells us whether a reaction can occur by itself. For a reaction to be spontaneous, \(\Delta G\) must be negative, indicating that the process can release free energy and proceed without any external input.
In contrast, when \(\Delta G\) is positive, as derived from \(\Delta G = -nFE\), the reaction is non-spontaneous under standard conditions. Here, the negative sign before the equation suggests that when the cell potential \((E)\) is negative, \(\Delta G\) becomes positive. Hence, the reaction requires energy input to proceed, making it less favorable. Understanding Gibbs Free Energy helps predict the direction and feasibility of chemical processes.
Equilibrium Constant
The Equilibrium Constant (\(K\)) plays a fundamental role in predicting the extent of a reaction at equilibrium. It is a ratio that compares the concentrations of products to reactants for a reaction at equilibrium. A key equation linking Gibbs Free Energy to \(K\) is \(\Delta G = -RT\ln(K)\), where \(R\) is the gas constant and \(T\) is temperature in Kelvin.
When \(\Delta G\) is positive, as discovered, this leads to \(\ln(K) < 0\), implying that \(K < 1\). Meaning, at equilibrium, reactant concentrations are higher than those of the products. This tells us that the reaction does not go significantly towards the products, aligning with its non-spontaneous nature. The equilibrium constant thus provides valuable insight into the equilibrium position and overall reaction tendencies.
Cell Potential
Cell Potential (\(E\)), measured in volts, is a measure of the electromotive force of an electrochemical cell. It determines the ability of a redox reaction to push electrons through a circuit, doing useful work. The expression \(\Delta G = -nFE\) links cell potential with Gibbs Free Energy, showing how a positive or negative \(E\) influences the reaction's spontaneity.
A negative cell potential, as given, implies that the reaction is not self-sustaining—the system must receive energy input rather than providing energy. Consequently, this contributes to a positive \(\Delta G\), further affirming that the cell cannot effectively perform its intended task of generating electricity or doing mechanical work. Understanding cell potential helps in designing and predicting the performance of electrochemical cells.
Electrochemical Cell
An Electrochemical Cell is a device that converts chemical energy into electrical energy through redox reactions. It consists of two half-cells connected by a salt bridge and an external circuit. The cell potential (\(E\)) is crucial for determining the cell's ability to perform work.
Naturally, when \(E\) is positive, the electrochemical cell can perform work, such as powering electronic devices or triggering chemical reactions. However, with a negative cell potential as given in this exercise, it is concluded that the cell is inefficient in producing work. Instead, it might require energy input to function, highlighting its limitation in this particular reaction. Understanding the workings of electrochemical cells is essential for applications in batteries and fuel cells.
Thermodynamics
Thermodynamics is the branch of chemistry that deals with energy changes, particularly focusing on heat and work involved during chemical reactions and physical processes. It is built upon laws that predict the direction and feasibility of a process.
The link between thermodynamics and Gibbs Free Energy in electrochemical cells is central. The relationship expressed in \(\Delta G = -nFE\) provides a thermodynamic view of how energy transformations occur in electrochemical reactions. When cell potential is negative—and thus thermodynamically unfavorable—a deeper thermo understanding sheds light on why such cells need an external energy source.
  • First Law: Energy can't be created or destroyed.
  • Second Law: Every energy transfer leads to increased entropy.
These laws among others guide the study and application of electrochemical cells and their energy-related behaviors.

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Most popular questions from this chapter

(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal hydride batteries over nickel-cadmium batteries?

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Sn}\) to \(\mathrm{Sn}^{2+} b_{y} \mathrm{I}_{2}\) (to form IT), (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}\) (to form \(\mathrm{I}_{2}\) ), (c) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\), (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cuby} \mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{47}\) ).

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{~S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{~S}(s)+\) \(2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)\) (b) \(4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\) \(5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)\) (c) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\). (a) Use data in Appendix C to calculate \(\Delta H^{t}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

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