Chapter 21: Problem 69
A voltaic cell consists of an \(\mathrm{Mn} / \mathrm{Mn}^{2+}\) half-cell and a \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. Calculate \(\left[\mathrm{Pb}^{2+}\right]\) when \(\left[\mathrm{Mn}^{2+}\right]\) is \(1.4 M\) and \(E_{\text {cell }}\) is \(0.44 \mathrm{~V}\).
Short Answer
Expert verified
[\text{Pb}^{2+}] \thickapprox 2.7 \times 10^{-21} \text{ M}
Step by step solution
01
Write down the half-reactions
First, identify the half-reactions for both \( \text{Mn/Mn}^{2+} \) and \( \text{Pb/Pb}^{2+} \): \text{Oxidation (anode)}: \[ \text{Mn} \rightarrow \text{Mn}^{2+} + 2e^{-} \] \text{Reduction (cathode)}: \[ \text{Pb}^{2+} + 2e^{-} \rightarrow \text{Pb} \]
02
Write the Nernst equation
The Nernst equation for the cell is: \( E_{\text{cell}} = E^{\text{o}}_{\text{cell}} - \frac{0.0592}{n} \text{log} \frac{[\text{Mn}^{2+}]}{[\text{Pb}^{2+}]} \) Here, \( n = 2 \) because each half-reaction involves two electrons.
03
Find the standard cell potential
Use the standard reduction potentials: \[ E^{\text{o}}_{\text{Mn}^{2+}/\text{Mn}} = -1.18 \text{ V} \] \[ E^{\text{o}}_{\text{Pb}^{2+}/\text{Pb}} = -0.13 \text{ V} \] The standard cell potential is: \[ E^{\text{o}}_{\text{cell}} = E^{\text{o}}_{\text{cathode}} - E^{\text{o}}_{\text{anode}} \] \[ E^{\text{o}}_{\text{cell}} = -0.13 \text{ V} - (-1.18 \text{ V}) = 1.05 \text{ V} \]
04
Plug in the known values into the Nernst equation
Substitute the given values and the standard cell potential into the Nernst equation: \[ 0.44 \text{ V} = 1.05 \text{ V} - \frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \]
05
Solve for the concentration of \( \text{Pb}^{2+} \)
First, isolate the logarithmic term: \[ 0.44 \text{ V} - 1.05 \text{ V} = -\frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \] \[ -0.61 \text{ V} = -\frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \] Next, eliminate the negative sign and solve for the log term: \[ 0.61 = \frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \] \[ \text{log} \frac{1.4}{[\text{Pb}^{2+}]} = \frac{0.61 \times 2}{0.0592} \] \[ \text{log} \frac{1.4}{[\text{Pb}^{2+}]} = 20.61 \]
06
Exponentiate both sides
To solve for the concentration of \( \text{Pb}^{2+} \), exponentiate both sides: \[ \frac{1.4}{[\text{Pb}^{2+}]} = 10^{20.61} \]\[ [\text{Pb}^{2+}] = \frac{1.4}{10^{20.61}} \]\[ [\text{Pb}^{2+}] \thickapprox 2.7 \times 10^{-21} \text{ M} \]
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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 fundamental in electrochemistry. It provides a way to calculate the cell potential under non-standard conditions, which is essential for understanding how real-world cells operate. The equation is: \[ E_{\text{cell}} = E^{\text{o}}_{\text{cell}} - \frac{0.0592}{n} \text{log} \frac{[\text{Oxidant}]}{[\text{Reductant}]} \] This formula tells us how to adjust the standard cell potential based on the concentrations of the reactants and products. Here’s a breakdown of the variables:
- E_{\text{cell}}: The cell potential under non-standard conditions.
- E^{\text{o}}_{\text{cell}}: The standard cell potential.
- n: The number of moles of electrons transferred in the reaction.
- [\text{Oxidant}]: Concentration of the oxidized species.
- [\text{Reductant}]: Concentration of the reduced species.
Standard reduction potential
Standard reduction potential is a measure of the tendency of a chemical species to be reduced, and it’s measured under standard conditions (1 M concentration, 1 atm pressure, 25°C). Reduction potentials are often tabulated for various half-reactions. For example:
- Mn^{2+}/Mn has a standard reduction potential of -1.18 V.
- Pb^{2+}/Pb has a standard reduction potential of -0.13 V.
Concentration calculation
In electrochemistry, concentrations of species play a critical role in determining cell potential, especially under non-standard conditions. Here’s a focused example on how to calculate the concentration. We start from the Nernst equation: \[ E_{\text{cell}} = E^{\text{o}}_{\text{cell}} - \frac{0.0592}{n} \text{log} \frac{[\text{Mn}^{2+}]}{[\text{Pb}^{2+}]} \] Plug the given values into the equation: \[ 0.44 \text{ V} = 1.05 \text{ V} - \frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \] Solving for [\text{Pb}^{2+}], we'll isolate the logarithmic term:\[ 0.44 \text{ V} - 1.05 \text{ V} = -\frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \]\[ -0.61 \text{ V} = -\frac{0.0592}{2} \text{log} \frac{1.4}{[\text{Pb}^{2+}]} \]Then solve for the log term and exponentiate both sides to find:\[ [\text{Pb}^{2+}] = \frac{1.4}{10^{20.61}} \thickapprox 2.7 \times 10^{-21} \text{ M} \] This shows how significant the influence of concentration is on the cell potential.
Cell potential
The cell potential, \( E_{\text{cell}} \), is the difference in potential energy between the two electrodes of an electrochemical cell. It drives the flow of electrons from the anode to the cathode and is measured in volts (V). The standard cell potential, \( E^{\text{o}}_{\text{cell}} \), is calculated using the standard reduction potentials of the half-reactions. Here it’s computed as: \[ E^{\text{o}}_{\text{cell}} = E^{\text{o}}_{\text{cathode}} - E^{\text{o}}_{\text{anode}} \]For our Mn/Pb cell, this results in:\[ E^{\text{o}}_{\text{cell}} = -0.13 \text{ V} - (-1.18 \text{ V}) = 1.05 \text{ V} \] Under non-standard conditions, the actual cell potential is adjusted using the Nernst equation. Hence, understanding both standard and actual cell potentials are vital for accurately predicting the cell’s behavior in various operating conditions.