What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\) \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(\mathrm{P}_{\mathrm{H}_{2}}=1.0 \mathrm{~atm} ?\)

Short Answer

Expert verified
The emf of the cell is approximately 0.043 V.

Step by step solution

01

Identify the half-reactions

First, write out the half-reactions for each individual cell. For the \(\mathrm{Pb}^{2+} / \(\mathrm{Pb}\) half-cell, the half-reaction is \(\mathrm{Pb}^{2+} + 2e^- ➔ \(\mathrm{Pb}\). For the \(\mathrm{Pt} / \mathrm{H}+\) / \(\mathrm{H}_{2}\) half-cell, the half-reaction is 2\(\mathrm{H}^{+} + 2e^- ➔ \(\mathrm{H}_{2}\).
02

Identify the half-cell potentials

Next, the half-cell potential (E°) values need to be looked up in a table. For the \(\mathrm{Pb}^{2+} + 2e^- ➔ \(\mathrm{Pb}\) half-cell, E° = -0.13 V. For the 2\(\mathrm{H}^{+} + 2e^- ➔ \(\mathrm{H}_{2}\) half-cell, E° = 0 V (since it is the standard hydrogen electrode).
03

Apply the Nernst equation

Substituting the half-cell potential and concentrations into Nernst equation gives: E = E° - RT/nF * lnQ. Here, E° is the cell potential (E° = E°cathode – E°anode), n is the number of moles of electrons transferred in the balanced chemical reaction (n = 2 for both of the half-reactions), Q is the reaction quotient, R is the universal gas constant and T is the temperature in Kelvin (assume standard room temperature of 298.15 K for this problem). Substituting into the equation, we get: E = -0.13 - (8.314 J/mol·K * 298.15 K / 2 * 96485 C/mol) * ln(1/(0.10)). The value for F, Faraday’s constant is 96485 C/mol.
04

Calculate the emf

Evaluate the above expression to get the emf (E). The emf of the cell will be the solution.

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Key Concepts

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

Nernst Equation
Understanding electrochemical cells requires a good grasp of the Nernst equation, which is vital for calculating the cell potential at any given concentration of reactants and products. The Nernst equation gives us the relationship between the electromotive force (emf) of a cell and the concentration of ions involved in the cell reactions.

At its core, the Nernst equation adjusts the standard half-cell potential to account for non-standard conditions. The standard equation is expressed as \[\begin{equation} E = E^° - \frac{RT}{nF} \ln(Q)\text{,}\end{equation}\] where \(E\) is the cell potential, \(E^°\) 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 in the reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

Through this equation, students can see the direct relationship between emf and the reaction's driving forces. The equation tells us that, as the reaction quotient increases, signifying a higher concentration of products to reactants, the emf of the cell decreases, and vice versa.
Standard Half-Cell Potential
The standard half-cell potential, denoted as \(E^°\), is a measure of the intrinsic tendency of a species to gain or lose electrons compared to the standard hydrogen electrode, which is arbitrarily set to 0 volts. These values are determined under standard conditions, which means all solutes are at 1 M concentration, and all gases are at 1 atm pressure.

When dealing with standard half-cell potentials, remember it's about comparing electrodes. For example, in the textbook exercise, the half-cell potential for a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) electrode is given as -0.13 V. This negativity indicates that lead is less eager to gain electrons than hydrogen. Knowing these potentials allows students to predict the direction of electron flow and the cell’s natural tendency to undergo the reaction.
Reaction Quotient
The reaction quotient, \(Q\), is an expression that uses the concentrations of the reactants and products to predict the direction in which a reaction will proceed to achieve equilibrium. It is similar in form to the equilibrium constant but can be applied at any moment during the reaction, not just at equilibrium.

In the electrochemical cell context, \(Q\) helps us to understand how the emf will change as the reaction progresses. The reaction quotient is calculated by taking the product of the concentrations of the products, each raised to the power of its stoichiometric coefficient, and dividing that by the product of the concentrations of the reactants, also raised to their stoichiometric coefficients. In the given cell, this would involve the concentration of \(\mathrm{H}^+\) ions and the partial pressure of \(\mathrm{H}_2\) gas. Since \(Q\) is in the denominator of the Nernst equation, a higher concentration of \(\mathrm{Pb}^{2+}\) ions would increase \(Q\) and reduce the emf.
Galvanic Cell
A galvanic cell, also known as a voltaic cell, is an electrochemical cell that derives electrical energy from spontaneous oxidation-reduction reactions taking place within the cell. It has two half-cells, each consisting of a metal electrode submerged in a solution containing metal cations. The half-cells are connected by a salt bridge that allows the flow of ions, while electrons flow through an external circuit from the anode to the cathode.

The galvanic cell functions thanks to the different standard half-cell potentials of its two electrodes, resulting in a potential difference and the flow of electrons. In the exercise, the cell consists of a lead electrode and a hydrogen electrode, and the spontaneous reaction is driven by the potential difference created between these two electrodes. By studying galvanic cells, students learn how chemical energy is converted into electrical energy, which is the principle behind batteries and many types of portable power sources.

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Most popular questions from this chapter

How does the tendency of iron to rust depend on the \(\mathrm{pH}\) of solution?

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$ \mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2} $$ (b) If a 1.00 -g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

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Balance the following redox equations by the halfreaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

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