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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 \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M},\) \([\mathrm{H}]=0.050 M,\) and \(P_{\mathrm{H}}=1.0 \mathrm{~atm} ?\)

Short Answer

Expert verified
The emf of the cell is \(-0.207 \, \text{V}\).

Step by step solution

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01

Write the Half Reactions

The two half-cells involved in the electrochemical cell are: 1. \( \mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb} \) 2. \( \mathrm{H}^+ + e^- \rightarrow \frac{1}{2}\mathrm{H}_2 \). The reaction for the hydrogen half-cell must be adjusted to match the electrons in the \(\mathrm{Pb}^{2+}/\mathrm{Pb}\) reaction, giving \(2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2\).
02

Identify Standard Electrode Potentials

The standard electrode potential for the lead half-reaction \( \mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb} \) is \( E^\circ = -0.13 \mathrm{~V} \). The standard hydrogen electrode (SHE) potential \( 2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2 \) is defined as \( E^\circ = 0.00 \mathrm{~V} \).
03

Calculate the Standard Cell Potential

The standard cell potential \( E^\circ_{\text{cell}} \) is calculated by subtracting the standard electrode potential of the anode reaction from the cathode reaction. Here, the hydrogen half-cell is the anode and lead half-cell is the cathode:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.13 \, \text{V}) - (0.00 \, \text{V}) = -0.13 \, \text{V}. \]
04

Use the Nernst Equation to Find the Cell Potential

The Nernst equation is used to calculate the cell potential under non-standard conditions:\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q. \]Here, \( n = 2 \) (electrons transferred). The reaction quotient \( Q \) can be found with the equation:\[ Q = \frac{1}{ [\text{Pb}^{2+}] \cdot ([\text{H}^+]^2 / P_{\text{H}_2}) } = \frac{1}{ [0.10] \cdot (0.050^2 / 1.0) } = 400. \] Inserting these values into the Nernst equation we get:\[ E_{\text{cell}} = -0.13 \text{ V} - \frac{0.0592}{2} \log 400 = -0.13 \text{ V} - 0.0592 \times 1.30103 = -0.207 \text{ V}. \]
05

Conclusion

The emf of the cell under the given conditions is \(-0.207 \mathrm{~V}\). The negative sign indicates that the direction chosen for the reactions does not occur spontaneously as a galvanic cell.

Key Concepts

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

Electrode Potential
In electrochemistry, **electrode potential** represents the voltage difference between an electrode and its surrounding solution. It can be understood as the tendency of a chemical species to be reduced, which is the gain of electrons. This potential is defined under standard conditions and is termed as Standard Electrode Potential, denoted as \(E^\circ\). Each half-reaction in electrochemistry has an associated standard electrode potential.

  • When a half-reaction occurs in a cell, the species with the higher electrode potential typically acts as the cathode (where reduction occurs).
  • The species with the lower electrode potential acts as the anode (where oxidation occurs).

For example, in the original exercise, the lead and hydrogen half-cells have specific electrode potentials. Identifying these potentials helps in calculating the overall cell potential, which determines the voltage or emf of the cell.
Nernst Equation
The **Nernst Equation** gives us a way to calculate the potential of an electrochemical cell under non-standard conditions. It extends the concept of electrode potential to real-life scenarios where standard conditions (such as concentration) are not met. The equation is:\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \]where:
  • \(E_{\text{cell}}\) is the cell potential under non-standard conditions.
  • \(E^\circ_{\text{cell}}\) is the standard cell potential.
  • \(n\) is the number of electrons involved per reaction.
  • \(Q\) is the reaction quotient.

The Nernst Equation indicates how the cell potential changes with temperature, pressure, and concentration. In practice, it's invaluable for adjusting theoretical predictions of cell voltage to match experimental conditions.
Reaction Quotient
The **reaction quotient**, denoted as \(Q\), is a measure of the relative amounts of products and reactants present during a reaction at any given time. It's similar to the equilibrium constant but is used for non-equilibrium conditions.

The formula for calculating \(Q\) is expressed as:\[ Q = \frac{\text{[products]}}{\text{ [reactants]}} \]This ratio is used in the Nernst Equation to adjust the potential calculations based on actual concentrations and pressures during the reaction. In the exercise, it's calculated using the concentrations of Pb\(^{2+}\) and H\(^+\) ions, as well as the pressure of H\(_2\) gas.

Understanding \(Q\) helps predict if the reaction will proceed towards products or reactants to reach equilibrium. If \(Q\) is less than the equilibrium constant (\(K\)), the reaction will proceed in the forward direction, and vice versa.
Standard Hydrogen Electrode
The **Standard Hydrogen Electrode** (SHE) is a universally used reference electrode in electrochemistry. It consists of a platinum electrode in contact with 1 M H\(^+\) ions and H\(_2\) gas at 1 atmosphere. By convention, the electrode potential of SHE is set to zero volts under standard conditions, providing a baseline or reference point:
  • It allows comparison of the electrode potentials of all other electrodes.
  • It defines the absolute measure of electrode potential.

The SHE is crucial for determining standard electrode potentials of other half-reactions. In this context, it offers a consistent benchmark for computing standard cell potentials. Understanding SHE simplifies the process of comparing and predicting cell reactions.

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Most popular questions from this chapter

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 \(1.00-\mathrm{g}\) sample of plant matter requires \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) 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?

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

What is the hourly production rate of chlorine gas (in \(\mathrm{kg})\) from an electrolytic cell using aqueous \(\mathrm{NaCl}\) electrolyte and carrying a current of \(1.500 \times 10^{3} \mathrm{~A} ?\) The anode efficiency for the oxidation of \(\mathrm{Cl}^{-}\) is 93.0 percent.

Discuss the advantages and disadvantages of fuel cells over conventional power plants in producing electricity.

Balance the following redox equations by the half-reaction 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) (e) \(\mathrm{Mn}^{2+}+\mathrm{BiO}_{3}^{-} \longrightarrow \mathrm{Bi}^{3+}+\mathrm{MnO}_{4}^{-}\) (in acidic solution)

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