Chapter 22: Problem 14
From the standard potentials $$\mathrm{Ag}_{\mathrm{SeO}_{4}(s)}+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{SeO}_{4}^{2-} \quad E^{0}=0.355 \mathrm{V}$$ $$\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightleftharpoons \mathrm{Ag}(s) \quad E^{0}=0.799 \mathrm{V}$$ calculate the solubility product constant for \(\mathrm{Ag}_{2} \mathrm{SeO}_{4}\).
Short Answer
Step by step solution
Write Down the Half-Reactions
Reverse the Ag+ Half-Reaction
Formulate the Solubility Reaction
Calculate the Potential for the Solubility Reaction
Calculate the Equilibrium Constant
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product
- \( \mathrm{Ag}_{2}\mathrm{SeO}_{4}\,(s) \rightleftharpoons 2\mathrm{Ag}^{+}\,(aq) + \mathrm{SeO}_{4}^{2-}\,(aq) \)
- \( K_{sp} = [\mathrm{Ag}^{+}]^2 [\mathrm{SeO}_{4}^{2-}] \)
Nernst Equation
- \( E = E^0 - \frac{RT}{nF} \ln Q \)
- \( E \) is the cell potential at non-standard conditions.
- \( E^0 \) is the standard cell potential.
- \( R \) is the universal gas constant \( 8.314 \, \mathrm{J/mol\,K} \).
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( F \) is the Faraday constant \( 96485 \, \mathrm{C/mol} \).
- \( Q \) is the reaction quotient.
Standard Electrode Potential
- Positive \( E^0 \) values indicate a higher tendency for the species to gain electrons and be reduced.
- Negative \( E^0 \) values suggest the species are more likely to lose electrons and be oxidized.