Chapter 19: Problem 110
Determine the cell potential of the following cell. $$\mathrm{Pt}\left|\mathrm{H}_{2}(1.0 \mathrm{~atm})\right| \mathrm{H}^{+}(1.0 \mathrm{M}) \| \mathrm{Cl}^{-}(1.0 \mathrm{M}), \mathrm{AgCl}(s) \mid \mathrm{Ag}$$ The cathode is essentially a silver electrode, \(\mathrm{Ag}^{+}(a q) \mid \mathrm{Ag}\). However, the cathode solution is saturated with silver chloride, so that the silver-ion concentration is determined by the solubility product of \(\operatorname{AgCl}\left(K_{s p}=1.8 \times 10^{-10}\right)\).
Short Answer
Step by step solution
Determine Standard Reduction Potentials
Calculate Concentration of Silver Ions
Use Nernst Equation to Find Potential at Cathode
Calculate Cell Potential
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Reduction Potential
For example, the \( ext{Ag}^+ + e^- ightarrow ext{Ag} \) reaction has a standard reduction potential of +0.80 V. This indicates that silver ions are good at gaining electrons (reduction). Similarly, the hydrogen electrode reaction, \( ext{H}_2 ightarrow 2 ext{H}^+ + 2e^- \), has a potential of 0.00 V, serving as the reference point for all reduction potentials.
These values help in calculating the overall cell potential by combining the half-reaction potentials. You'll need to subtract the anode potential from the cathode potential to find the total cell voltage.
Nernst Equation
The equation is:\[E = E^0 - \frac{RT}{nF} \ln Q\]Here's a breakdown of each part:
- \( E \) is the cell potential under non-standard conditions.
- \( E^0 \) is the standard potential.
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons exchanged in the reaction.
- \( F \) is Faraday's constant (96485 C/mol).
- \( Q \) is the reaction quotient, calculated from the concentrations of the reactants and products.
Solubility Product
In the case of silver chloride (AgCl), the \( K_{sp} \) is \( 1.8 \times 10^{-10} \). This extremely low value indicates that AgCl is not very soluble in water. As a result, in any given solution, only a small amount of Ag can dissociate into \( ext{Ag}^+ \) ions. From the relation \( K_{sp} = [ ext{Ag}^+][ ext{Cl}^-] \), with \( [ ext{Cl}^-] \) known, we can precisely calculate \( [ ext{Ag}^+] \).
This determined ion concentration is vital when using the Nernst Equation, as it serves as the \( Q \) value for calculating cell potential.
Electrochemical Cell
The main components include:
- **Anode**: The site of oxidation and the electrode where electrons are released. In this exercise, it is the hydrogen electrode.
- **Cathode**: The site of reduction, where electrons are gained. The silver chloride electrode acts as the cathode.
- **Electrolyte**: The medium that allows ions to move between the electrodes.
Silver Ion Concentration
In a saturated solution, where maximum dissolution takes place, few silver ions are present because the solubility is low. The ion concentration can be found using:\[[\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]}\]Given \( K_{sp} \) of AgCl is extremely low, once the chloride ion concentration is known (1 M in our case), the concentration of \( [\text{Ag}^+] \) can be accurately calculated.
With \( [\text{Ag}^+] = 1.8 \times 10^{-10} \text{ M} \), and using this value in the Nernst Equation, we determine the real-world potential of the silver electrode under non-standard conditions.