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Magazine Chapter 19: Problem 89
\( \quad\) The voltaic cell $$ \operatorname{Cd}(s)\left|\mathrm{Cd}^{2+}(a q) \| \mathrm{Ni}^{2+}(1.0 M)\right| \mathrm{Ni}(s) $$ has a cell potential of \(0.240 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). What is the concentration of cadmium ion? \(\left(E_{\text {cell }}^{\circ}=0.170 \mathrm{~V} .\right)\)
Short Answer
Expert verified
The concentration of cadmium ions is approximately 0.043 M.
Step by step solution
01
Write the Nernst equation
The Nernst equation is used to find the relationship between the cell potential and the concentrations of the ions. The equation is: \[ E = E^{ ext{cell}} - \frac{0.0592}{n} \log \left(\frac{[ ext{products}]}{[ ext{reactants}]}\right) \] where \( E \) is the cell potential at non-standard conditions, \( E^{\text{cell}} \) is the standard cell potential, and \( n \) is the number of moles of electrons exchanged.
02
Identify reactions and calculate n
The redox reaction is as follows: - Oxidation half-equation: \( \text{Cd}(s) \rightarrow \text{Cd}^{2+}(aq) + 2e^- \) - Reduction half-equation: \( \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \) Thus, \( n = 2 \) as 2 moles of electrons are exchanged.
03
Set up the Nernst equation
Substitute the given values and \( n \) into the Nernst equation: \[ 0.240 = 0.170 - \frac{0.0592}{2} \log \left( \frac{[\text{Ni}^{2+}]}{[\text{Cd}^{2+}]} \right) \] Note that the concentration of \( [\text{Ni}^{2+}] \) is 1.0 M.
04
Solve for the concentration of cadmium ions
Rearrange the Nernst equation to isolate the logarithmic term: \[ \frac{0.0592}{2} \log \left( \frac{1.0}{[\text{Cd}^{2+}]} \right) = 0.170 - 0.240 \] \[ \frac{0.0592}{2} \log \left( \frac{1.0}{[\text{Cd}^{2+}]} \right) = -0.070 \] Solve for the logarithm: \[ \log \left( \frac{1.0}{[\text{Cd}^{2+}]} \right) = -\frac{0.070}{0.0296} \] \[ \log \left( \frac{1.0}{[\text{Cd}^{2+}]} \right) = -2.364 \]
05
Calculate cadmium ion concentration
Convert the logarithmic term to concentration: \[ \frac{1.0}{[\text{Cd}^{2+}]} = 10^{-2.364} \] \[ [\text{Cd}^{2+}] = \frac{1.0}{10^{-2.364}} \approx 0.043 \text{ M} \] Thus, the concentration of cadmium ions, \([\text{Cd}^{2+}]\), is approximately \(0.043\) M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst equation
The Nernst equation is an essential part of understanding how electrochemical cells work, especially under non-standard conditions. This equation determines the effect of ion concentration on the cell potential of a voltaic cell. It is expressed as:
The component, \( \log \left(\frac{[\text{products}]}{[\text{reactants}]}\right) \), shows the impact of the concentration of ionic species in the electrochemical reaction. Changes in the ion concentrations shift the cell's potential away from standard conditions. Thus, the Nernst equation is fundamental in predicting the behavior of voltaic cells when concentrations are varied.
- \( E = E^{\text{cell}} - \frac{0.0592}{n} \log \left(\frac{[\text{products}]}{[\text{reactants}]}\right) \)
The
Voltaic cell
A voltaic cell is a fundamental element of electrochemistry, often referred to as a galvanic cell. It converts chemical energy into electrical energy through spontaneous redox reactions. These cells harness such reactions to produce an electric current.In a voltaic cell, two half-cells are connected:
- Each half-cell contains a metallic electrode and an electrolyte solution where redox reactions occur.
- An example setup might be \(\operatorname{Cd}(s) | \text{Cd}^{2+}(aq) || \text{Ni}^{2+}(aq) | \text{Ni}(s) \).
Redox reactions
Redox reactions are chemical processes in which oxidation and reduction occur simultaneously. These reactions involve the transfer of electrons between two species. Within an electrochemical cell, redox reactions are responsible for generating electricity.
Let's break down the components:
- Oxidation: This is the loss of electrons by a molecule, atom, or ion. At the anode, oxidation occurs, and electrons are supplied to the external circuit.
- Reduction: This involves the gain of electrons. At the cathode, reduction takes place, completing the circuit as electrons are received from the anode.
Cell potential
The cell potential, also known as electromotive force (EMF), is a measure of the voltage produced by a voltaic cell. It reflects the cell's ability to drive an electron flow from the anode to the cathode due to redox reactions.Several factors influence cell potential:
- Standard Cell Potential, \( E^{\text{cell}} \): This is the potential difference when the cell operates under standard conditions (1 M concentrations, 1 atm pressure, room temperature 25掳C).
- Concentration: As ions are consumed or produced, their concentrations change. The Nernst equation explains how these changes affect the cell potential.
