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Chapter 17: Problem 33

The negative sign in Equation \(17.3\left(w_{\text {clec }}=-Q E_{\text {cell }}\right)\) seems to indicate that a voltaic cell with a positive cell potential does negative electrical work. How is this possible?

Short Answer

Expert verified
Answer: In a voltaic cell with a positive cell potential, negative electrical work occurs. This is because the flow of electrons is spontaneous from the anode to the cathode, generating electrical energy without the need for external energy input. The cell is converting chemical energy to electrical energy as the electrons flow through the circuit.

Step by step solution

01

(Step 1: Understanding Cell Potential)

(Cell potential, also known as electromotive force (EMF) or cell voltage, is the electric potential difference between two half-cells of a voltaic cell. It represents the driving force that causes the flow of electrons from one half-cell to another. A positive cell potential indicates that electrons will flow spontaneously from the anode (where oxidation occurs) to the cathode (where reduction occurs), generating electrical energy in the process.)
02

(Step 2: Understanding Electrical Work)

(Electrical work is the amount of energy that is converted into other forms of energy due to the flow of charge. The electrical work can be either positive or negative, depending on the direction of the flow of charge. Positive electrical work occurs when external energy is used to force charges through an electric field against a voltage potential. Negative electrical work occurs when charges flow spontaneously through an electric field, releasing energy in the process.)
03

(Step 3: Positive and Negative Electrical Work)

(Positive electrical work requires energy from an external source to move charges against the electric field, like in the case of charging a battery. On the other hand, negative electrical work occurs when charges move spontaneously in the electric field without requiring external energy, like in the case of discharging a battery.)
04

(Step 4: Explanation of Negative Electrical Work in a Voltaic Cell)

(In a voltaic cell with a positive cell potential, the flow of electrons is spontaneous from the anode to the cathode. This means that the electrons move naturally through the external circuit, releasing electrical energy as they go. As they do not require external energy to move, the electrical work done by the cell is considered negative. The negative sign in the equation \(w_{\text {clec }}=-Q E_{\text {cell }}\) indicates that the electrical work done by the cell is negative, which means the voltaic cell is converting chemical energy to electrical energy as the electrons flow through the circuit.)

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Key Concepts

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

Cell Potential
In the realm of electrochemistry, cell potential, also known as electromotive force (EMF), is crucial. It is the voltage difference between two half-cells within a voltaic cell. This voltage drives the electron flow from one electrode to the other. Imagine it as the pushing force that makes electrons move. A positive cell potential means that electrons will naturally and spontaneously flow from the anode, where oxidation happens, to the cathode, where reduction occurs. This spontaneous flow is what generates electrical energy. Hence, a positive cell potential is fundamental for a voltaic cell to work effectively.
Electrical Work
Electrical work involves the energy transformation that happens when charges move through an electric field. It can either be positive or negative based on the direction and nature of this movement. Positive electrical work occurs when external energy is used to move charges, like charging a battery. Negative electrical work, however, happens when charges move on their own through the electric field, releasing energy in the process. This distinction is crucial for understanding how energy is managed in electrochemical reactions.
Electron Flow
Electron flow in a voltaic cell is a fundamental concept. Electrons move from the anode to the cathode through an external circuit. This movement is driven by the cell potential, which provides the necessary push. As electrons travel, they carry energy, enabling it to be harnessed for electrical work. This spontaneous movement is what makes voltaic cells so efficient at converting stored chemical energy into usable electrical energy. Understanding electron flow helps in grasping how electric circuits are powered by chemical reactions.
Chemical to Electrical Energy Conversion
Voltaic cells brilliantly illustrate the conversion of chemical energy into electrical energy. This transformation occurs when a chemical reaction induces the flow of electrons. As these electrons pass through the external circuit, they release energy, which can be used to perform electrical work. This process encapsulates the essence of how batteries operate, converting the stored chemical potential into a useful electrical form. Recognizing this conversion is key to understanding how everyday devices power up using chemical reactions.

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Most popular questions from this chapter

In a voltaic cell, why is the cathode labeled the positive terminal and the anode the negative terminal?

Laptops The first generation of laptop computers was powered by nickcl-cadmium (NiCad) batterics, which generated \(1.20 \mathrm{V}\) based on the following cell reaction: \(\mathrm{Cd}(s)+2 \mathrm{NiO}(\mathrm{OH})(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{Ni}(\mathrm{OH})_{2}(s) $$ What is the value of \(\Delta G_{\text {cell }} ?\)

What is the value of \(E^{\circ}\) for the following reaction? $$ 2 \mathrm{AgCl}(s)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{HCl}(a q) $$

Quantitative Analysis Electrolysis can be used to determine the concentration of \(\mathrm{Cu}^{2+}\) in a given volume of solution by electrolyzing the solution in a cell equipped with a platinum cathode. If all of the \(\mathrm{Cu}^{2+}\) is reduced to \(\mathrm{Cu}\) metal at the cathode, the increase in mass of the electrode provides a measure of the concentration of \(\mathrm{Cu}^{2+}\) in the original solution. To ensure the complete (99.9996) removal of the \(\mathrm{Cu}^{2+}\) from a solution in which \(\left[\mathrm{Cu}^{2+}\right]\) is initially about \(1.0 M,\) will the potential of the cathode (versus SHE) have to be more negative or less negative than \(0.34 \mathrm{V}\) (the standard potential for \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) )?

From the table of standard reduction potentials in Appendix 6 a. Select an oxidizing agent that will oxidize \(\mathrm{Cr}(s)\) to \(\mathrm{Cr}^{3+}(a q)\) but not \(\mathrm{Cd}(s)\) to \(\mathrm{Cd}^{2+}(a q)\) b. Select a reducing agent that will reduce Br \(_{2}(\ell)\) to \(\mathrm{Br}^{-}(a q)\) but not \(\mathrm{I}_{2}(s)\) to \(1^{-}(a q).\)
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