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(a) What does the term electromotive force mean? (b) What is the definition of the volt? (c) What does the term cell potential mean?

Short Answer

Expert verified
(a) Electromotive force (EMF) is the energy provided by a power source to move electric charge through a circuit. It is measured in volts (V). (b) The volt (V) is the SI unit of electric potential, potential difference, and EMF. It is defined as the electric potential difference across a conductor when a current of one ampere dissipates one watt of power: \[1 V = \dfrac{1 W}{1 A}\] (c) Cell potential, or cell voltage, is the difference in electric potential between the two electrodes of a galvanic cell. It drives the spontaneous flow of electrons through a connected circuit and depends on the reduction potentials of the half-reactions involved.

Step by step solution

01

(a) Electromotive Force Definition

Electromotive force (EMF) is a measure of the energy provided by a power source, such as a battery or generator, to move electric charge through a circuit. It is the force that pushes the electrons to go through the circuit, and it is not an actual physical force. The unit of electromotive force is the volt (V).
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(b) Volt Definition

The volt is the unit of electric potential, electric potential difference, and electromotive force in the International System of Units (SI). It is named after Alessandro Volta, an Italian physicist who invented the first chemical battery. One volt (1V) is defined as the electric potential difference across a conductor when a current of one ampere (1A) dissipates one watt (1W) of power. Mathematically, it can be expressed as: \[1 V = \dfrac{1 W}{1 A}\]
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(c) Cell Potential Definition

Cell potential, also known as cell voltage or cell EMF, refers to the difference in electric potential between the two electrodes of a galvanic (voltaic) cell. It is the driving force behind the spontaneous flow of electrons through a circuit connected to the cell. The cell potential can be positive or negative, depending on the direction of the electron flow. In an electrochemical cell, the cell potential is determined by the reduction potentials of the half-reactions involved in the cell.

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Key Concepts

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

Electrochemistry: Understanding Electromotive Force
Electrochemistry is the branch of chemistry that studies the relationship between electrical energy and chemical changes. A fundamental concept in electrochemistry is the electromotive force (EMF). EMF refers to the energy supplied by a source, such as a battery, to move electric charges through a circuit. Though the name might suggest otherwise, it is not a force in the physical sense. Rather, it is the potential difference generated by the source to drive the flow of electrons.

In simple terms, imagine EMF as the pressure that pushes water through a pipe. Here, the water represents electrons, and the pipe is the circuit. Thus, EMF is crucial for powering devices, as it ensures a continuous flow of electric current. It is measured in volts (V), and higher EMF indicates a stronger ability to cause electron flow.
Getting to Know Volts
Volts are the unit used to measure electric potential, potential difference, and electromotive force. Named after Alessandro Volta, a pioneering physicist, the volt is a central unit in electrical measurements. When discussing volts, it's essential to understand they represent the potential energy per unit charge.

To simplify, think of volts like the height in a water tank; the taller the tank, the more potential energy the water has. Similarly, a higher voltage means more potential energy is available for each charge. The formal definition states that one volt is the potential difference between two points in a circuit when one joule of energy is used to move one coulomb of charge from one point to another. Mathematically, this is represented as: \[1 \, \text{volt} = \frac{1 \, \text{watt}}{1 \, \text{ampere}}\]This helps quantify how much power is being used to drive a current of one ampere between two points.
Cell Potential: The Driving Force of Galvanic Cells
Cell potential, also known as cell voltage or cell EMF, is a pivotal element in studying electrochemical cells, especially galvanic or voltaic cells. It represents the potential difference between the two electrodes when the cell operates. Think of it as the motivational push that encourages electrons to move from the anode to the cathode through an external circuit.

The electrodes' potential difference is crucial as it dictates whether electric energy flows spontaneously or if external energy is needed. Cell potential is influenced by the reduction potentials of the reactions occurring at the electrodes. If the cell potential is positive, the cell can perform work spontaneously, driving a current. In contrast, a negative potential indicates the need for external energy to provide the required current.

In essence, the cell potential enables us to gauge the efficiency and readiness of an electrochemical cell to do work.
Diving into Electric Potential
Electric potential is a key concept that describes the potential energy per unit charge at a point in an electric field. It illustrates how much work is needed to move a charge from one point to another. Often termed potential difference, it helps figure out how much energy is required for charge movement.

Imagine a hill: the higher you go, the more potential energy you have due to gravity. Similarly, in an electric field, the higher the potential, the more energy a charge possesses. Electric potential is crucial in understanding how charges move in fields and circuits. It's measured in volts, emphasizing its role in energy transformation and transfer.
  • Higher electric potential indicates a greater ability to move charges.
  • It's vital for calculating energy transfer in circuits.
Understanding electric potential sets the stage for comprehending how circuits function, and how energy is harnessed and converted in various systems.

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Most popular questions from this chapter

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (b) The platinum electrode in a standard hydrogen electrode is specially prepared to have a large surface area. Why is this important? (c) Sketch a standard hydrogen electrode.

(a) The nonrechargeable lithium batteries used for photography use lithium metal as the anode. What advantages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithium-ion battery does not use lithium metal as an electrode material. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation.

Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are $$ \begin{aligned} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) & E_{\mathrm{red}}^{o}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) & E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{aligned} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+& 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \\ & 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\mathrm{Zn}\) electrodes. The two half-cells have \(\left[\mathrm{Zn}^{2+}\right]=1.8 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=\) \(1.00 \times 10^{-2} M\), respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3},\) and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s)$$ (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 aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

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