Determine the potential of each of the following cells: (a) \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H} \mathrm{H}_{2}(\mathrm{~g}, 1.0\) bar \()\left|\mathrm{HCl}(\mathrm{aq}, 0.075 \mathrm{M}) \| \mathrm{HCl}\left(\mathrm{aq}, 1.0 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right)\right|\) \(\mathrm{H}_{2}(\mathrm{~g}, 1.0\) bar \() \mathrm{Pt}(\mathrm{s})\) (b) \(\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}\left(\mathrm{aq}, 0.37 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right) \| \mathrm{Ni}^{2+}\left(\mathrm{aq}, 0.059 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right)\right| \mathrm{Ni}(\mathrm{s})\) (c) \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{Cl}_{2}(\mathrm{~g}, 2.50\) Torr \()|\mathrm{HCl}(\mathrm{aq}, 1.0 \mathrm{M}) \| \mathrm{HCl}(\mathrm{aq}, 0.85 \mathrm{M})| \mathrm{H}_{2}(\mathrm{~g}\), 125 Torr \() \mid \mathrm{Pt}(\mathrm{s})\) (d) \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}\left(\mathrm{aq}, 0.277 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right)\right| \mathrm{Sn}^{4+}\left(\mathrm{aq}, 0.867 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right), \mathrm{Sn}^{2+}\) \(\left(\mathrm{aq}, 0.55 \mathrm{~mol} \cdot \mathrm{L}^{-1}\right) \mid \mathrm{Pt}(\mathrm{s})\)

Short Answer

Expert verified
The calculations provided are general formulas and guidelines to calculate the potential for each cell using the Nernst equation and standard reduction potentials where appropriate. The actual numerical values can only be provided if standard reduction potentials and reaction conditions are known or assumed.

Step by step solution

01

- Calculate the cell potential for part (a)

The reaction is a hydrogen electrode, so we can use the Nernst equation: \( E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log \frac{[\text{acid}]}{[\text{base}]} \). For a standard hydrogen electrode, \( E^\circ_{cell} = 0 \) V. Since the pressures of hydrogen gas are both 1 bar (standard conditions) and do not affect potential, we only consider the concentration of HCl. \( n = 2 \) for the \( 2H^+ + 2e^- \rightarrow H_2 \) reaction. Now plug in the concentrations: \( E_{cell} = -(0.0592/2) \log(0.075) \).
02

- Calculate the cell potential for part (b)

Use the standard reduction potentials from a table (assumed to be available). The reaction is \( Zn^{2+} + 2e^- \rightarrow Zn(s) \) and \( Ni^{2+} + 2e^- \rightarrow Ni(s) \). \( E_{cell} = E^\circ_{cathode} - E^\circ_{anode} - (0.0592/n) \log(\frac{[Zn^{2+}]^{n1}_{anode}}{[Ni^{2+}]^{n2}_{cathode}}) \). You would use the standard potentials for \( Zn^{2+}/Zn \) and \( Ni^{2+}/Ni \) and the concentrations provided to calculate the potential.
03

- Calculate the cell potential for part (c)

This cell involves a gas (Cl2), so we use the gas constant (R), temperature (T), Faraday's constant (F), and the pressure of Cl2 to calculate the potential using a modified Nernst equation incorporating pressures: \( E_{cell} = E^\circ_{cell} - (RT/nF) \ln(P) \). The overall n is 2 for the reaction involving chlorine gas. The pressures of \( H_2 \) and the concentrations of HCl also play a role in the calculation.
04

- Calculate the cell potential for part (d)

The reaction for this cell involves tin at two different oxidation states. The Nernst equation applies here as well, considering the concentrations of \( Sn^{2+} \) and \( Sn^{4+} \) ions. The reaction can be written as \( Sn^{4+} + 2e^- \rightarrow Sn^{2+} \), with \( n = 2 \). Use the relevant standard reduction potentials for the half-reactions and apply the Nernst equation: \( E_{cell} = E^\circ_{cell} - (0.0592/2) \log(\frac{[Sn^{2+}]_{product}}{[Sn^{4+}]_{reactant}}) \).

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Key Concepts

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

Nernst Equation
Understanding the Nernst equation is crucial for calculating the potential of electrochemical cells under non-standard conditions. It allows us to account for changes in concentration or partial pressure of the reactants and products involved in the cell reactions. The Nernst equation is given by:
\[\begin{equation} E_{cell} = E^\circ_{cell} - \dfrac{0.0592}{n} \log\dfrac{[\text{products}]}{[\text{reactants}]} \end{equation}\]
where:
  • E_{cell} is the cell potential under non-standard conditions,
  • E^\circ_{cell} is the standard cell potential,
  • n is the number of moles of electrons transferred in the reaction, and
  • The term \log\dfrac{[\text{products}]}{[\text{reactants}]} represents the reaction quotient, Q, which is a measure of the driving force of the reaction.
When the concentrations of the reactants and products are at their standard states (1M for solutions, 1atm for gases), the cell operates under standard conditions, and E_{cell} equals E^\circ_{cell}. The Nernst equation becomes particularly handy in problems like those presented in the exercise, as it allows the determination of cell potentials for each specific scenario, considering the deviations from standard conditions.
Standard Reduction Potential
The standard reduction potential (E^\circ) is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. Each half-cell has a characteristic standard reduction potential, which can be found in electrochemical series tables. The standard cell potential (E^\circ_{cell}) is calculated by subtracting the potential of the anode (where oxidation occurs) from that of the cathode (where reduction occurs):
\[\begin{equation} E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \end{equation}\]
This value is central to predicting the direction of electron flow and the feasibility of a reaction occurring spontaneously as a galvanic cell. Positive values suggest a spontaneous reaction under standard conditions. For instance, in part (b) of the exercise, the difference in standard reduction potentials between zinc and nickel can be used to determine the baseline voltage before considering concentration effects.
Galvanic Cell Reactions
Galvanic cells, also known as voltaic cells, operate on the principle of spontaneous redox reactions, producing electrical energy as a result. The key components of a galvanic cell are the anode and cathode, which are often composed of different materials allowing for different standard reduction potentials. The electrochemical reactions in these cells are divided into two half-reactions, one representing oxidation and the other reduction. The flow of electrons from the anode to the cathode through an external circuit generates an electric current.

Importance of Standard Conditions

The operation under standard conditions assumes activities of reactants and products to be unity, which simplifies the calculation of cell potentials as showcased in part (a). But real-world scenarios often deviate from these conditions, implementing the Nernst equation for accurate potential calculation.

Relevance of Concentrations and Pressures

The exercise solutions exemplify how fluctuations in concentrations, like in part (d), or partial pressures, as in part (c), influence the potential of the cell. Such variations impact the driving force of the reaction, altering the cell's potential through the reaction quotient in the Nernst equation. These principles encapsulate the fundamental nature of galvanic cell reactions and embody the dynamic interplay between chemistry and electricity, enabling practical applications like batteries and various other electrochemical devices.

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Most popular questions from this chapter

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) \(2 \mathrm{Ce}^{++}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Ce}^{3+}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq})\) \(E_{\mathrm{cell}^{\circ}}=+1.08 \mathrm{~V}\) (b) \(6 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 6 \mathrm{Fe}^{2+}(\mathrm{aq})+\) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq}), E_{\text {cell }}{ }^{\circ}=-1.29 \mathrm{~V}\)

Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: (a) \(\mathrm{AgBr}(\mathrm{s})=\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\), a solubility equilibrium (b) \(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}\)(aq) \(\longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), the Bronsted neutralization reaction (c) \(\mathrm{Cd}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_{3}(\mathrm{~s}) \rightarrow \mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{~s})\), the reaction in the nickel-cadmium cell

The density of the electrolyte in a lead-acid battery is measured to assess its state of charge. Explain how the density indicates the state of charge of the battery.

Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations: (a) \(\mathrm{Ni}^{2+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Ni}(\mathrm{s})+\mathrm{Zn}^{2+}\) (aq) (b) \(\mathrm{Ce}^{4+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{I}_{2}(\mathrm{~s})+\mathrm{Ce}^{3+}(\mathrm{aq})\) (c) \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{aq})\) (d) \(\mathrm{Au}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Au}(\mathrm{s})+\mathrm{Au}^{3+}(\mathrm{aq})\)

When a pH meter was standardized with a boric acid-borate buffer with a pH of \(9.40\), the cell was \(+0.060 \mathrm{~V}\). When the buffer was replaced with a solution of unknown hydronium ion concentration, the cell potential was \(+0.22 \mathrm{~V}\). What is the \(\mathrm{pH}\) of the solution?

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