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$$E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 V$$VVV

Short Answer

Expert verified
The standard potential of the cathode is \(0.40 V\).

Step by step solution

01

Understand the Given

This exercise works on the standard potential of a cathode in an electrochemical cell. The standard potential \(E_{cathode}^{\circ}\) is given by the formula \(E_{cathode}^{\circ}=E_{anode}^{0}-E_{cathode}^{0}\), where the standard potential of the anode and cathode are provided.
02

Substitute the Known Values

Substitute the given standard potentials into the standard cathode potential formula: \(E_{cathode}^{\circ} = 2.71 V - 2.310 V\).
03

Perform the Calculation

Perform the subtraction to calculate the standard potential of the cathode: \(E_{cathode}^{\circ} = 2.71 V - 2.310 V = 0.40 V\). The voltage is positive, indicating that the reaction can occur spontaneously under standard conditions.

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Key Concepts

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

Electrochemical Cell
An electrochemical cell is a device that facilitates the conversion of chemical energy into electrical energy through redox reactions. These cells comprise two electrodes immersed in an electrolyte solution. The electrode where oxidation occurs is called the anode, and the electrode where reduction occurs is known as the cathode. A typical electrochemical cell is divided into two half-cells:
  • Anode Half-Cell: The site where the oxidation reaction occurs.
  • Cathode Half-Cell: The site where the reduction reaction occurs.
The electrolyte solution allows ions to move between the two half-cells, maintaining charge balance while electrons flow through an external circuit from the anode to the cathode. This electron flow is what generates a current. If the overall reaction in the cell is spontaneous, the electrochemical cell can function without an external power source, making it practical for energy conversion in remote or standalone devices.
Cathode Potential
Cathode potential refers to the voltage associated with the reduction half-reaction occurring at the cathode. The standard electrode potential of a cathode, denoted as \( E_{cathode}^{\circ} \), is a measure of the tendency of a chemical species to acquire electrons, and thereby, get reduced. It is expressed in volts (V).

Cathode potential is determined using the standard hydrogen electrode (SHE) as a reference and is calculated by comparing the potential difference with a standard electrode under standard conditions (25°C, 1M concentration for each ion, and 1 atm pressure). In the context of the electrochemical cell, a high positive cathode potential indicates a strong tendency for the reduction to occur.
  • A positive cathode potential means the cathode can easily gain electrons.
  • This potential contributes to the overall voltage of the electrochemical cell and indicates how effectively electrons are pulled from the anode to the cathode.
Thus, a large positive \( E_{cathode}^{\circ} \) suggests that the reaction at the cathode is more spontaneous, contributing positively to the feasibility and efficiency of an electrochemical cell.
Anode Potential
Anode potential is the voltage associated with the oxidation half-reaction happening at the anode. In electrochemistry, it is essential as it helps to determine the direction and spontaneity of the electron flow in an electrochemical cell. Like the cathode, the standard electrode potential for an anode, \( E_{anode}^{\circ} \), is also measured against the SHE.

It is expressed in volts and is typically less positive than the potential of the cathode. Since the anode is where oxidation occurs (loss of electrons), it sets the pace for electrons to flow out towards the cathode. Negative or less positive anode potentials indicate a catalyst for electron donation:
  • It doesn't mean the reaction at the anode is not worthwhile; rather, it denotes the need for an electron flow.
  • The lower, more negative anode potential drives the conversion of chemical energy to electrical energy through electron movement.
Understanding anode potential is crucial for optimizing electrochemical cell performance and ensuring the efficiency of the redox reactions involved.

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Most popular questions from this chapter

For the half-reaction \(\mathrm{Hg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(1)\) \(E^{\circ}=0.854 \mathrm{V} .\) This means that \((\mathrm{a}) \mathrm{Hg}(1)\) is more readily oxidized than \(\mathrm{H}_{2}(\mathrm{g}) ;\) (b) \(\mathrm{Hg}^{2+}(\) aq) is more readily reduced than \(\mathrm{H}^{+}(\mathrm{aq}) ;\) (c) \(\mathrm{Hg}(\) l) will dissolve in 1 M HCl; (d) Hg(l) will displace Zn(s) from an aqueous solution of \(\mathrm{Zn}^{2+}\) ion.

Can the displacement of \(\mathrm{Pb}(\mathrm{s})\) from \(1.0 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) be carried to completion by tin metal? Explain.

A quantity of electric charge brings about the deposition of \(3.28 \mathrm{g}\) Cu at a cathode during the electrolysis of a solution containing \(\mathrm{Cu}^{2+}(\text { aq })\). What volume of \(\mathrm{H}_{2}(\mathrm{g}),\) measured at \(28.2^{\circ} \mathrm{C}\) and \(763 \mathrm{mm} \mathrm{Hg},\) would be produced by this same quantity of electric charge in the reduction of \(\mathrm{H}^{+}(\) aq) at a cathode?

The following voltaic cell registers an \(E_{\text {cell }}=0.108 \mathrm{V}\) What is the pH of the unknown solution? $$\operatorname{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(x \mathrm{M}) \| \mathrm{H}^{+}(1.00 \mathrm{M}) |$$ $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) | \mathrm{Pt}$$

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

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