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Chapter 17: Problem 73

Consider the cell described below: $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu} $$ 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 \(T=25^{\circ} \mathrm{C} .\) )

Short Answer

Expert verified
The cell potential after the reaction has changed the concentration of Zn²⁺ by 0.20 mol/L is approximately 1.0861 V.

Step by step solution

01

Write down the balanced redox equation.

The balanced redox equation for a zinc-copper electrochemical cell is given as follows: \( Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \) This gives us the redox reaction for the given electrochemical cell setup.
02

Find the standard cell potential.

To calculate the standard cell potential (\(E°_{cell}\)), we need the standard reduction potentials of Zinc and Copper ions, which are: \(E^{0}_{Zn^{2+}/Zn}\) = -0.76 V and \(E^{0}_{Cu^{2+}/Cu}\) = +0.34 V Now, we can calculate the overall standard cell potential (\(E°_{cell}\)): \(E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode}\) Where the cathode is the reduction half-cell (Copper), and the anode is the oxidation half-cell (Zinc). \(E^{0}_{cell} = (0.34 \ V) -(-0.76 \ V) = 1.10 \ V\) So, the standard cell potential is 1.10 V.
03

Determine the changes in the concentrations of ions in the reaction.

Since the concentration of Zn²⁺ changes by 0.20 mol/L, we have: Initial concentration of Zn²⁺: 1.00 M Final concentration of Zn²⁺: 1.00 + 0.20 = 1.20 M Copper ions are consumed in the process as the reaction proceeds. So the initial concentration of Cu²⁺ is reduced by 0.20 mol/L: Initial concentration of Cu²⁺: 1.00 M Final concentration of Cu²⁺: 1.00 - 0.20 = 0.80 M
04

Apply the Nernst equation to calculate the cell potential after changes in the concentrations of the ions.

To find the cell potential after changes in the ion concentrations, we use the Nernst equation: \(E_{cell} = E^{0}_{cell} - \frac{RT}{nF} \times \ln Q\) Where: \(E_{cell}\) = Cell potential after changes in the concentrations \(E^{0}_{cell}\) = Standard cell potential (1.10 V, as calculated in Step 2) R = Gas constant (\(8.314 \ J/ mol .K\)) T = Temperature (298 K, as given in the problem) n = Number of electrons transferred (2 moles, as seen from the balanced redox equation) F = Faraday's constant (\(96485 \ C/mol\)) Q = Reaction quotient, calculated as \(\frac{[Zn^{2+}]}{[Cu^{2+}]}\) Now applying values to the Nernst equation: \(E_{cell} = 1.10 \ V - \frac{8.314 \ J/mol . K \times 298 \ K}{2 \ mol \times 96485 \ C/mol} \times \ln \frac{1.20 \ mol/L}{0.80 \ mol/L}\) \(E_{cell} = 1.10 \ V - 0.0139 \ V\) \(E_{cell} = 1.0861 \ V\) Thus, the cell potential after the reaction has changed the concentration of Zn²⁺ by 0.20 mol/L is approximately 1.0861 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 formula used to calculate the electrical potential of an electrochemical cell under non-standard conditions. It takes into account the effect of ion concentrations on the overall cell potential. The equation is expressed as:

\[\[\begin{align*}E_{cell} = E^{0}_{cell} - \frac{RT}{nF} \times ln (Q) \end{align*}\]\]

where
  • \(E_{cell}\) is the cell potential under non-standard conditions,
  • \(E^{0}_{cell}\) represents the standard cell potential,
  • \(R\) is the universal gas constant (8.314 J/mol·K),
  • \(T\) is the temperature in Kelvin,
  • \(n\) is the number of moles of electrons exchanged in the redox reaction,
  • \(F\) is Faraday's constant (96485 C/mol), and
  • \(Q\) is the reaction quotient, a dimensionless number representing the ratio of concentrations of products to reactants at any given point in the reaction.
The equation allows us to adjust our expectations for cell potential based on changing conditions like concentration, showing us how the cell's voltage will vary in real-life applications versus idealized scenarios.
Redox Reaction
Redox reactions are a type of chemical reaction that involves the transfer of electrons between two species. These reactions comprise two linked half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained. In the given exercise, the redox reaction occurring in the zinc-copper electrochemical cell is displayed as:

\[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\]

Here, zinc metal (Zn) is oxidized to zinc ions (\(Zn^{2+}\)), losing two electrons in the process, while copper ions (\(Cu^{2+}\)) are reduced to copper metal (Cu), each gaining two electrons. This electron exchange is the fundamental process driving the cell's electrical output.
Standard Reduction Potentials
Standard reduction potentials provide a measure of the tendency of a chemical species to gain electrons and be reduced. These potentials are measured at standard conditions, which are typically 25°C (298 K), a 1 M concentration for each ion, and a pressure of 1 atm for gases. The potentials are displayed in volts and are determined relative to the standard hydrogen electrode, which is assigned a potential of 0 volts. In the exercise provided, the standard reduction potential for zinc (\(E^{0}_{Zn^{2+}/Zn}\)) is -0.76 V and for copper (\(E^{0}_{Cu^{2+}/Cu}\)) is +0.34 V. The more positive the potential, the greater the species' tendency to be reduced. By comparing these values, we understand that copper ions have a greater tendency to gain electrons and be reduced as opposed to zinc ions.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is used in the Nernst equation to assess the progress of the reaction at a particular moment. It considers the concentration of the ions involved in the reaction. For the given zinc-copper cell, \(Q\) is calculated using the relation:

\[Q = \frac{[\text{Products}]}{[\text{Reactants}]}\]

Since our reaction has zinc ions as the product and copper ions as the reactant, the reaction quotient can be written as:\(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\). As the reaction proceeds past the initial state, the concentrations of zinc and copper ions change, affecting the cell potential. The Nernst equation accommodates these concentration changes and thus enables the calculation of the cell potential at any given state of the reaction.

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Most popular questions from this chapter

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

A factory wants to produce \(1.00 \times 10^{3} \mathrm{kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{h}\) to accomplish this?

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{C}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)
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