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One half-cell in a voltaic cell is constructed from a copper wire dipped into a \(4.8 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) The other half-cell consists of a zinc electrode in a \(0.40 \mathrm{M}\) solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the cell potential.

Short Answer

Expert verified
The cell potential is approximately 0.9625 V.

Step by step solution

01

Identify the half-cell reactions

The half-cell reactions for a copper-zinc voltaic cell are: For the copper half-cell: \[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\]For the zinc half-cell: \[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^-\]
02

Determine standard electrode potentials

The standard reduction potential for copper is \( E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \) and for zinc it is \( E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \).
03

Calculate the standard cell potential

The standard cell potential \(E^0_{\text{cell}}\) is given by the equation: \[ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \]Plug in the values:\[ E^0_{\text{cell}} = (+0.34 \text{ V}) - (-0.76 \text{ V}) = +1.10 \text{ V}\]
04

Use the Nernst Equation to find the actual cell potential

The Nernst Equation is:\[ E_{\text{cell}} = E^0_{\text{cell}} - \left( \frac{RT}{nF} \right) \ln Q \]Where:- \( R = 8.314 J/mol \cdot K\) (universal gas constant)- \( T = 298 K\) (assumed standard temperature)- \( F = 96485 C/mol \) (Faraday's constant)- \( n = 2 \) (number of moles of electrons exchanged)- \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \) (reaction quotient)Inserting the values and simplifying, we get:\[ E_{\text{cell}} = 1.10 \text{ V} - \left( \frac{(8.314)(298)}{2(96485)} \right) \ln \left( \frac{0.40}{4.8 \times 10^{-3}} \right) \]
05

Simplify the Nernst Equation result

The term \( \frac{(8.314)(298)}{2(96485)} \approx 0.0257 \text{ V} \) which simplifies our equation to:\[ E_{\text{cell}} = 1.10 \text{ V} - 0.0257 \ln \left( \frac{0.40}{4.8 \times 10^{-3}} \right) \]Evaluating \( \ln \left( \frac{0.40}{4.8 \times 10^{-3}} \right) \) gives approximately 5.35.So, \[ E_{\text{cell}} = 1.10 \text{ V} - 0.0257 \times 5.35 \approx 1.10 \text{ V} - 0.1375 \text{ V} \approx 0.9625 \text{ V} \]
06

Conclusion on the cell potential

The calculated cell potential for the voltaic cell is approximately \( 0.9625 \text{ V} \).

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Key Concepts

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

Voltaic Cell
A voltaic cell is a fascinating device that generates electrical energy from chemical reactions. It consists of two distinct half-cells, each with a metal electrode immersed in an electrolyte solution. One half-cell undergoes oxidation, losing electrons, while the other undergoes reduction, gaining electrons.
The external circuit allows the electrons to flow from one electrode to the other, creating an electric current. This flow is crucial for devices like batteries, where energy conversion takes place.
Each half-cell has a specific reaction and potential, represented by its electrode. Thus, the overall cell potential is the difference between the potentials of the two half-cells. In the case of a copper-zinc voltaic cell:
  • Zinc acts as the anode, where oxidation occurs.
  • Copper acts as the cathode, where reduction occurs.
By applying the Nernst Equation along with the known concentration of ions, the actual cell potential can be calculated.
Nernst Equation
The Nernst Equation is a powerful tool for predicting the cell potential under non-standard conditions. It relates the concentration of reactants and products to the electromotive force (EMF) of a cell.
This equation is crucial when the concentrations of the substances involved are not in their standard states. The formula is given by:
\[E_{\text{cell}} = E^0_{\text{cell}} - \left( \frac{RT}{nF} \right) \ln Q \] where:
  • \(E_{\text{cell}}\) is the cell potential.
  • \(E^0_{\text{cell}}\) is the standard cell potential.
  • \(R\) is the universal gas constant.
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons transferred.
  • \(F\) is Faraday's constant.
  • \(Q\) is the reaction quotient, reflecting the concentrations of reactants and products.
Using this equation, we can see that changes in concentration affect the cell's voltage, providing practical insights for electrochemical applications.
Standard Electrode Potentials
Standard electrode potentials are fundamental in estimating the voltage a voltaic cell can produce. They measure the tendency of a metal ion to gain electrons, hence dictating whether it will act as an oxidizing or reducing agent.
The standard electrode potential values are determined under standard conditions, typically at 1 M concentration, 1 atm pressure, and 25°C. These values are referenced against the standard hydrogen electrode, which has a potential set to zero.
The copper half-cell in our example has a standard reduction potential of \( +0.34 \text{ V} \), indicating a strong tendency to gain electrons compared to zinc, which has \( -0.76 \text{ V} \).
The difference in these potentials, \(E^0_{\text{cell}}\), is a key factor in determining the spontaneity and potential difference in a voltaic cell. The positive sign of total cell potential signifies a spontaneous reaction, allowing the generation of electrical energy.
Copper-Zinc Reaction
The copper-zinc reaction exemplifies a classic voltaic cell mechanism. In this reaction, zinc undergoes oxidation, losing electrons and forming zinc ions:
\[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \]
Meanwhile, copper ions gain electrons through reduction to form solid copper:
\[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \]
The stark difference in their standard electrode potentials results in a noticeable voltage.
  • The zinc half-cell acts as the anode, losing electrons.
  • The copper half-cell, the cathode, receives these electrons.
The electrons travel through the external circuit from zinc to copper, creating an electric current at a measurable cell potential.
This specific configuration also informs battery design, underlying mechanisms in devices that power various technologies.

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Most popular questions from this chapter

Chlorine gas is obtained commercially by electrolysis of brine (a concentrated aqueous solution of \(\mathrm{NaCl}\) ). If the electrolysis cells operate at \(4.6 \mathrm{V}\) and \(3.0 \times 10^{5} \mathrm{A},\) what mass of chlorine can be produced in a 24 -h day?

The standard potential for the reaction of \(\mathrm{Mg}(\mathrm{s})\) with \(\mathrm{I}_{2}(\mathrm{s})\) is \(+2.91 \mathrm{V} .\) What is the standard free energy change, \(\Delta G^{\circ}\) for the reaction?

A hydrogen-oxygen fuel cell operates on the simple reaction $$\mathbf{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)$$ If the cell is designed to produce 1.5 A of current, and if the hydrogen is contained in a 1.0 -L tank at \(200 .\) atm pressure at \(25^{\circ} \mathrm{C},\) how long can the fuel cell operate before the hydrogen runs out? (Assume there is an unlimited supply of \(\left.\mathbf{O}_{2 \cdot}\right)\)

An aqueous solution of \(\mathrm{Na}_{2} \mathrm{S}\) is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external battery, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the half-reaction that occurs at this electrode. (b) Sulfur is the primary product at the anode. Write an equation for its formation.

Which of the following ions is most easily reduced? (a) \(\mathrm{Cu}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Zn}^{2+}(\mathrm{aq})\) (c) \(\mathrm{Fe}^{2+}(\mathrm{aq})\) (d) \(\mathrm{Ag}^{+}(\mathrm{aq})\) (e) \(\mathrm{Al}^{3+}(\mathrm{aq})\)

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