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The standard EMF of a galvanic cell can be calculated from (a) the size of the electrode (b) the \(\mathrm{pH}\) of the solution (c) the amount of metal in the anode (d) the \(E^{\circ}\) values of the half-cells

Short Answer

Expert verified
The standard EMF of a galvanic cell is calculated using \(d\) the \(E^{\circ}\) values of the half-cells.

Step by step solution

01

Understand the concept of EMF in a galvanic cell

The Electromotive Force (EMF) of a galvanic cell reflects the voltage produced by the cell when no current is flowing. It is determined by the difference in potential energy of the electrochemical reactions at the anode and cathode, not by the physical size or amount of material present in the cell.
02

Identify factors affecting EMF

The EMF of a galvanic cell is affected by the intrinsic properties of the chemicals involved in the half-cell reactions. The standard EMF \(E^{\circ}\) can be calculated using the standard electrode potentials of the two half-cells involved. The pH can also affect EMF if it changes the effective concentration of ions in solution, but it is not the direct method of calculating standard EMF.
03

Choose the correct answer

Since the EMF of a galvanic cell depends on the potentials of the half-cell reactions and not on the physical characteristics like size of electrode or amount of metal, the correct answer is \(d\) the \(E^{\circ}\) values of the half-cells.

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Key Concepts

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

Standard Electrode Potentials
Understanding standard electrode potentials is crucial when studying galvanic cells and their operation. These potentials, denoted as \( E^{\circ} \), are measures of the individual potential of a reversible electrode at standard state, which includes a concentration of 1 molar solution at 1 atmosphere of pressure and a temperature of 298 K (25°C). These values are determined relative to the standard hydrogen electrode (SHE), which is assigned a potential of 0 volts.

Standard electrode potentials allow us to predict the direction of the electrochemical reactions that will occur within a cell and to determine the cell's standard electromotive force (EMF). By convention, electrodes with higher standard potentials will act as cathodes, where reduction occurs, while those with lower potentials will serve as anodes, where oxidation happens.

When constructing a galvanic cell, the difference in standard electrode potentials between the two half-cells dictates the voltage that the cell can theoretically supply. This difference is calculated by subtracting the potential of the anode from that of the cathode. It is also important to note that the potentials are intensive properties; they do not depend on the size of the electrode or the amount of substance present.
Electromotive Force
Electromotive Force, commonly referred to as EMF, represents the energy provided by a power source, such as a galvanic cell. Measured in volts, EMF is essentially the maximum potential difference between two electrodes when no current is being drawn from the cell. The concept of EMF is central to understanding how batteries and similar devices generate electrical energy.

Contrary to what the term 'force' might suggest, EMF is not actually a force; it is a potential, which explains why it is measured in voltage. The value of EMF is directly linked to the spontaneous electrochemical reactions taking place within the cell. When these reactions naturally tend toward equilibrium, they generate a voltage that can be harnessed.

The standard EMF of a cell can be calculated by using the standard electrode potentials of the two half-reactions. The correct application of the Nernst equation can also help you to find the EMF under non-standard conditions, such as different temperatures or concentrations.
Electrochemical Reactions
Electrochemical reactions form the basis of any galvanic cell's function. These are the chemical reactions that cause electrons to move through an external circuit, creating an electric current. Such reactions involve oxidation and reduction processes, often abbreviated as redox reactions.

In the context of a galvanic cell, the anode experiences oxidation; it is where the loss of electrons occurs. Electrons then flow from the anode to the cathode through the external circuit. At the cathode, reduction takes place as these electrons are gained. The substances that undergo these redox reactions typically involve metals and their ions, but other materials and electrolytes can also participate.

Electrochemical reactions are fascinating in that they convert chemical energy directly into electrical energy. The specific nature of the reactions within a cell determines the amount of voltage produced, which is why understanding these reactions is key to mastering the principles behind galvanic cells and their EMFs. The predictability of these reactions is derived from the standard electrode potentials and the thermodynamic likelihood of electrons moving from the higher energy anode to the lower energy cathode.

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Most popular questions from this chapter

Use of lithium metal as an electrode in high energy density batteries is due to (a) lithium is the lightest element. (b) lithium has the highest oxidation potential. (c) lithium is quite reactive. (d) lithium does not corrode readily.

The preparation of \(\mathrm{LiOH}\) by the electrolysis of a \(35 \%\) solution of \(\mathrm{LiCl}\) using a platinum anode led to a current efficiency of \(80 \%\). What weight of \(\mathrm{LiOH}\) was formed by the passage of \(2.5 \mathrm{~A}\) for \(4825 \mathrm{~s} ?\) (a) \(1.92 \mathrm{~g}\) (b) \(2.40 \mathrm{~g}\) (c) \(0.96 \mathrm{~g}\) (d) \(0.672 \mathrm{~g}\)

What is the equilibrium constant of the reaction: \(2 \mathrm{Fe}^{3+}+\mathrm{Au}^{+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Au}^{3+} ?\) Given \(E_{\mathrm{Au}^{+} \mid \mathrm{Au}}^{\circ}=1.68 \mathrm{~V}, E_{\mathrm{Au}^{3 \cdot} \mid \mathrm{Au}}^{\circ}=1.50 \mathrm{~V}\), \(E_{\mathrm{Fe}^{3 *} \mid \mathrm{Fe}^{\mathrm{e}}^{\circ}}=0.75 \mathrm{~V}\) and \(2.303 R T / F=0.06\) (a) \(1 \times 10^{22}\) (b) \(1 \times 10^{-22}\) (c) \(1 \times 10^{-11}\) (d) \(1 \times 10^{-72}\)

During the electrolysis of \(0.1 \mathrm{M}-\mathrm{CuSO}_{4}\) solution using copper electrodes, a depletion of \(\mathrm{Cu}^{2+}\) occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the local concentration of \(\mathrm{Cu}^{2+}\) near the anode and cathode are, respectively, \(0.12 \mathrm{M}\) and \(0.08 \mathrm{M}\), the back EMF developed at \(298 \mathrm{~K}\) is \((\log 1.5\) \(=0.18,2.303 R T / F=0.06)\) (a) \(0.33 \mathrm{~V}\) (b) \(5.4 \mathrm{mV}\) (c) \(2.7 \mathrm{mV}\) (d) \(10.8 \mathrm{mV}\)

For a cell reaction involving a twoelectron change, the standard EMF of the cell is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction at \(25^{\circ} \mathrm{C}\) will be (a) \(1 \times 10^{10}\) (b) \(1 \times 10^{-10}\) (c) \(29.5 \times 10^{-2}\) (d) \(2 \times 10^{10}\)

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