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Chapter 18: Problem 69

Consider the cell described below: $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M) \| \mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{~mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

Short Answer

Expert verified
The cell potential after the reaction has operated long enough for the [Zn虏鈦篯 to have changed by 0.20 mol/L is approximately \(1.079 V\).

Step by step solution

01

Identify the half-reactions

To calculate the cell potential, we need to identify the half-reactions occurring at the anode (Zn) and cathode (Cu). The balanced half-reactions are given below: At the anode (oxidation): \(Zn \rightarrow Zn^{2+} + 2e^-\) At the cathode (reduction): \(Cu^{2+} + 2e^- \rightarrow Cu\)
02

Write the overall cell reaction

The overall cell reaction is obtained by adding the balanced oxidation and reduction half-reactions: \(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu\)
03

Calculate the standard cell potential

To calculate the standard cell potential, we need to consult a table of standard electrode potentials and find the values for Zn虏鈦�/Zn and Cu虏鈦�/Cu half-cells: The standard reduction potentials are: \(E^0_{Zn^{2+}/Zn} = -0.76 V\) \(E^0_{Cu^{2+}/Cu} = +0.34 V\) Now, we can calculate the standard cell potential (\(E^0_{cell}\)): \(E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 V - (-0.76 V) = 1.10 V\)
04

Apply the Nernst equation

The Nernst equation relates the cell potential, standard cell potential, temperature, and the concentrations of the species involved in the reaction: \(E_{cell} = E^0_{cell} - \frac{0.05916}{n} \cdot \log{Q}\) where: - \(E_{cell}\) is the cell potential - \(E^0_{cell}\) is the standard cell potential - n is the number of moles of electrons transferred - Q is the reaction quotient For this particular cell, the reaction quotient Q is: \(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\) Since [Zn虏鈦篯 has changed by 0.20 mol/L and [Cu虏鈦篯 has changed by the same value but in the opposite direction, we can calculate the new concentrations: \([\mathrm{Zn}^{2+}]_\mathrm{final} \ = \ [\mathrm{Zn}^{2+}]_\mathrm{initial} \ + 0.20 \mathrm{~mol/L} \ = \ 1.00 \mathrm{~mol/L} \ + 0.20 \mathrm{~mol/L} \ = \ 1.20 \mathrm{~mol/L}\) \([\mathrm{Cu}^{2+}]_\mathrm{final} \ = \ [\mathrm{Cu}^{2+}]_\mathrm{initial} \ - 0.20 \mathrm{~mol/L} \ = \ 1.00 \mathrm{~mol/L} \ - 0.20 \mathrm{~mol/L} \ = \ 0.80 \mathrm{~mol/L}\) Now, we can substitute the values into the Nernst equation: \(E_{cell} = 1.10 V - \frac{0.05916}{2} \cdot \log{\frac{1.20}{0.80}}\)
05

Calculate the cell potential

Now we can calculate the cell potential for the new concentrations using the Nernst equation: \(E_{cell} \approx 1.10 V - \frac{0.05916}{2} \cdot \log{1.5}\) \(E_{cell} \approx 1.10 V - 0.02117 V \approx 1.07883 V\) The cell potential after the reaction has operated long enough for the [Zn虏鈦篯 to have changed by 0.20 mol/L is approximately \(1.079 V\).

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Key Concepts

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

Nernst Equation
The Nernst Equation is a powerful tool that helps us calculate the cell potential under non-standard conditions, offering a way to understand how the potential changes when the concentration of reactants and products vary. It is derived from the Gibbs free energy equation and is particularly useful for determining how electrochemical cells behave over time. The equation is written as:\[ E_{cell} = E^0_{cell} - \frac{0.05916}{n} \cdot \log{Q} \]Here's what the components mean:
  • \(E_{cell}\): The actual cell potential under current conditions.
  • \(E^0_{cell}\): The standard cell potential, a reference point measured under standard conditions (1 M concentration, 25掳C, 1 atm pressure).
  • \(n\): Number of electrons transferred in the reaction.
  • \(Q\): Reaction quotient, a snapshot of the ratio of the concentrations of products over reactants.
By substituting the known values into the equation, we can determine the real-time potential of an electrochemical cell, illustrating the dynamic aspect of redox reactions.
Standard Electrode Potential
Standard Electrode Potential (SEP) is a critical concept in electrochemistry, providing a measure of the individual potential of a reversible electrode at standard state conditions, which include a concentration of 1 M, a temperature of 25掳C, and a pressure of 1 atm.
  • It's denoted by \( E^0 \) and reported in volts (V).
  • The value is derived from a hydrogen reference electrode, which is universally assigned a value of 0.00 V.
  • Standard electrode potentials help us predict the direction of redox reactions and determine which species will be oxidized or reduced.
For our electrochemical cell including the zinc and copper reactions, the standard electrode potentials are:- \( E^0_{Zn^{2+}/Zn} = -0.76 \, V \) indicating that zinc is likely to oxidize, being a more negative potential.- \( E^0_{Cu^{2+}/Cu} = +0.34 \, V \) suggesting that copper will likely be reduced, as it is more positive.The difference between these potential values, \( E^0_{cell} = 1.10 \, V \), represents the voltage the cell can produce under standard conditions.
Redox Reactions
Redox reactions are at the heart of electrochemistry, involving the transfer of electrons between two substances. These reactions comprise two key processes: oxidation and reduction.- Oxidation is the loss of electrons, where the oxidizing agent supports this process.- Reduction is the gain of electrons, facilitated by reducing agents.In our cell:
  • Zinc (Zn) undergoes oxidation, shifting to zinc ions \( (Zn^{2+}) \), while losing electrons.
  • Copper ions \( (Cu^{2+}) \) undergo reduction, gaining electrons to form copper metal.
The balance of these reactions is crucial to maintaining the flow of electricity in the cell. By understanding redox reactions, one can predict and calculate the behavior of electrochemical cells. It forms the foundation for applications in batteries, corrosion, electroplating, and much more.

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Most popular questions from this chapter

Consider the following half-reactions: $$\begin{array}{cc}\mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 \mathrm{M}\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution
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