Chapter 19: Problem 71
From the following information, calculate the solubility product of \(\mathrm{AgBr}\) : $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \mathrm{AgBr}(s)+e^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) & E^{\circ}=0.07 \mathrm{~V} \end{array} $$
Short Answer
Step by step solution
Understand the Concepts
Write the Relevant Equations
Determine the Cell Potential
Calculate the Solubility Product
Compute and Verify the Value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electrochemistry
Understanding electrochemistry helps us connect chemical reactions and the electrical energy they produce or consume. This is vital in various real-world applications, including energy storage and electrolysis.
Nernst Equation
- \( E \) is the cell potential under non-standard conditions.
- \( E^{\circ} \) is the standard cell potential.
- \( R \) is the gas constant.
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons.
- \( F \) is Faraday's constant.
- \( Q \) is the reaction quotient.
Half-Reaction
- Oxidation half-reaction: \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \)
- Reduction half-reaction: \( \text{AgBr}(s) + e^- \rightarrow \text{Ag}(s) + \text{Br}^-(aq) \)
Standard Electrode Potential
- \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) is 0.80 V
- \( \text{AgBr}(s) + e^- \rightarrow \text{Ag} + \text{Br}^- \) is 0.07 V