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Chapter 16: Problem 598

Given \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{+2}+2 \mathrm{e}^{-}\) with \(\mathrm{E}^{\circ}=+.763\), calculate \(\mathrm{E}\) for a Zn electrode in which \(\mathrm{Zn}^{+2}=.025 \mathrm{M}\)

Short Answer

Expert verified
The actual cell potential (E) for the Zn electrode in a Zn虏鈦� solution with a concentration of 0.025 M is approximately 0.803 V.

Step by step solution

01

Understand the Nernst equation

The Nernst equation enables us to calculate the cell potential (E) based on the standard cell potential (E掳), concentrations, and the number of electrons exchanged in the redox reaction. The Nernst equation is given as: \[E = E掳 - \frac{RT}{nF} \ln Q\] Where: - E is the cell potential to be calculated - E掳 is the standard cell potential (+0.763 V) - R is the gas constant (8.314 J/mol路K) - T is the temperature in Kelvin (assume 298 K for room temperature) - n is the number of electrons transferred in the redox reaction (2 for Zn) - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient, calculated using the given concentrations (0.025 M for Zn虏鈦�)
02

Calculate the reaction quotient (Q)

As the given reaction is a reduction half-reaction, the reaction quotient (Q) can be calculated only using the concentration of the zinc ion (Zn虏鈦�). The balanced reaction is: \[\mathrm{Zn^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Zn}\] Hence, Q becomes: \[Q = [\mathrm{Zn}^{2+}]\] From the exercise, the given concentration of Zn虏鈦� is 0.025 M: \[Q = 0.025\]
03

Apply the Nernst equation

With the computed value of Q, all parameters are now available to be plugged into the Nernst equation: \[E = E掳 - \frac{RT}{nF} \ln Q\] \[E = 0.763 - \frac{(8.314)(298)}{(2)(96485)} \ln 0.025\] Now, you can calculate the value of E: \[E \approx 0.803\, V\]
04

State the final answer

The actual cell potential (E) for the Zn electrode in a Zn虏鈦� solution with a concentration of 0.025 M is approximately 0.803 V.

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Key Concepts

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

Understanding Cell Potential
The concept of cell potential is crucial in electrochemistry, as it determines the electrical energy a cell can produce. The cell potential, often denoted as \( E \), can be considered as a measure of the driving force behind an electrochemical reaction. When a redox reaction takes place, electrons transfer from one species to another. This electron flow, which occurs due to potential difference, can do work鈥攍ike powering a device. The standard cell potential, \( E^\circ \), measures this potential when each reactant and product is in its standard state. It provides a reference point, typically at concentrations of 1 M, pressure of 1 atm, and temperature of 298 K (25掳C). However, under non-standard conditions, you need tools like the Nernst equation to find the cell potential.
The Mechanism of Redox Reactions
Redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons between two species. One species undergoes oxidation by losing electrons, while the other is reduced by gaining those electrons. It is essential to identify the two half-reactions in any redox process鈥攏amely, oxidation and reduction.
For instance, zinc (Zn) in the reaction \( \mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2 \mathrm{e^-} \) undergoes oxidation as it loses electrons. Meanwhile, in return, it can also gain electrons in a reduction process involving zinc ions: \( \mathrm{Zn^{2+}} + 2 \mathrm{e^-} \rightarrow \mathrm{Zn} \). This electron exchange is what contributes to the electrochemical cell potential.
The Role of Reaction Quotient in Calculating Potential
The reaction quotient, \( Q \), is vital for understanding how a reaction proceeds towards equilibrium under non-standard conditions. It reflects the ratio of the concentrations of reactants to products at any point in the reaction. Calculating \( Q \) is essential for applying the Nernst equation effectively, as it allows us to adjust the standard cell potential to the actual conditions of the reaction.
In the given exercise, \( Q \) was calculated using only the concentration of zinc ions, yielding \( Q = 0.025 \). This was plugged into the Nernst equation for adjustments due to the deviation from standard conditions. With \( Q \) and other constants, you can find the actual cell potential, reflecting real-world scenarios where concentrations rarely remain at ideal states.

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Most popular questions from this chapter

When a current of \(5.36 \mathrm{~A}\) is passed for \(30 \mathrm{~min}\). through an electrolytic cell containing a solution of a zinc salt, it is found that \(3.27 g\) of zinc metal is deposited at the cathode. What is the equivalent weight of zinc in the salt?

You have the following cell process: \(\mathrm{Fe}(\mathrm{s})+\mathrm{Co}^{2+}(.5 \mathrm{M}) \rightarrow \mathrm{Fe}^{2+}(1.0 \mathrm{M})+\mathrm{Co}(\mathrm{s})\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \leftrightarrows \mathrm{Fe}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.44 \mathrm{e}\) and \(\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \leftrightharpoons \mathrm{Co}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.28\), find the standard cell potential \(\Delta \mathrm{E}\), the cell potential \(\Delta \mathrm{E}\) and the concentration ratio at which the potential generated by the cell is exactly zero. which the potential generated by the cell is exactly zero.

Calculate the voltage (E) of a cell with \(\mathrm{E}^{\circ}=1.1\) volts, If the copper half-cell is at standard conditions but the zinc ion concentration is only \(.001\) molar. Temperature is \(25^{\circ} \mathrm{c}\). The overall reaction is \(\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Cu}+\mathrm{Zn}^{+2}\)

You pass a \(1.0\) amp current through an electrolytic cell for \(1.0 \mathrm{hr}\). There are 96,500 coul in a Faraday (F). Calculate the number of grams of each of the following that would be deposited at the cathode: (2) Cu from a \(\mathrm{Cu}^{+2}\) solution and (1) Ag from an Ag ' solution, (3) A.1 from an \(\mathrm{Al}^{3+}\) solution.

A meter reads that a battery is putting out 450 amp in the external circuit of a cell. The cell is involved in the electrolysis of a copper sulfate solution. During the \(30.0 \mathrm{~min}\) that the current was allowed to flow, a total of \(30 \mathrm{~g}\) of copper metal was deposited at the cell's cathode. Was the meter an accurate measurement of current? \((\mathrm{F}=96,500\) coul. \()\)
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