Problem 57
Question
A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?\) (c) if \(n=3 ?\)
Step-by-Step Solution
Verified Answer
For n=1, \(K \approx 2.74 \times 10^4\); for n=2, \(K \approx 7.51 \times 10^7\); for n=3, \(K \approx 2.06 \times 10^{11}\).
1Step 1: Understanding the Nernst Equation
The Nernst equation relates the standard cell potential \(E^°\) to the equilibrium constant \(K\). The equation is given by \[ E^° = \frac{0.0592}{n} \log K \] at standard temperature (\(298\, \text{K}\)).
2Step 2: Rearrange the Equation for K
To find \(K\), rearrange the Nernst equation to:\[ \log K = \frac{nE^°}{0.0592} \] Then solve for \(K\):\[ K = 10^{\left(\frac{nE^°}{0.0592}\right)} \].
3Step 3: Substitute Values and Solve for n=1
For \(n=1\), substitute \(E^° = 0.257 \mathrm{~V}\) into the rearranged equation:\[ \log K = \frac{1 \times 0.257}{0.0592} \] Calculate this to find \(K\). \[ K = 10^{\left(\frac{0.257}{0.0592}\right)} \approx 2.74 \times 10^4 \].
4Step 4: Substitute Values and Solve for n=2
For \(n=2\), the equation becomes:\[ \log K = \frac{2 \times 0.257}{0.0592} \] Calculate to determine \(K\).\[ K = 10^{\left(\frac{0.514}{0.0592}\right)} \approx 7.51 \times 10^7 \].
5Step 5: Substitute Values and Solve for n=3
For \(n=3\), the equation becomes:\[ \log K = \frac{3 \times 0.257}{0.0592} \] Calculate to find \(K\).\[ K = 10^{\left(\frac{0.771}{0.0592}\right)} \approx 2.06 \times 10^{11} \].
Key Concepts
Standard Cell PotentialNernst EquationElectrochemistry
Standard Cell Potential
The standard cell potential, represented as \(E^°\), is a crucial concept in electrochemistry. It measures the driving force behind a chemical reaction occurring in an electrochemical cell under standard conditions. Standard conditions generally mean a temperature of 298 K, a pressure of 1 atm, and concentrations of 1 M for any solutions involved.
The standard cell potential is determined by the difference in potentials between the two electrodes in a galvanic cell. It's expressed in volts (V), and it's calculated by subtracting the standard reduction potential of the anode reaction from that of the cathode reaction. An essential detail is that a positive \(E^°\) suggests the reactants spontaneously convert to products, which is a favorable and forward reaction.
To sum it up:
The standard cell potential is determined by the difference in potentials between the two electrodes in a galvanic cell. It's expressed in volts (V), and it's calculated by subtracting the standard reduction potential of the anode reaction from that of the cathode reaction. An essential detail is that a positive \(E^°\) suggests the reactants spontaneously convert to products, which is a favorable and forward reaction.
To sum it up:
- Standard conditions: 298 K, 1 atm, 1 M concentrations
- Positive \(E^°\) implies a spontaneous reaction
- Measured in volts
Nernst Equation
The Nernst Equation is a fundamental formula in electrochemistry that relates the standard cell potential to the concentration of the involved species. It provides a way to calculate the actual cell potential at any given concentrations, not just under standard conditions. The Nernst Equation is represented as: \[E = E^° - \frac{RT}{nF} \ln{Q}\]where \(E\) is the cell potential at any conditions, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant.
However, at standard temperature (298 K), this equation simplifies to: \[E = E^° - \frac{0.0592}{n} \log Q\]When a reaction reaches equilibrium, \(E\) becomes zero, and \(Q\) equals the equilibrium constant \(K\). By rearranging the equation, you can solve for \(K\) to understand how far the reaction will proceed.
To highlight its use:
However, at standard temperature (298 K), this equation simplifies to: \[E = E^° - \frac{0.0592}{n} \log Q\]When a reaction reaches equilibrium, \(E\) becomes zero, and \(Q\) equals the equilibrium constant \(K\). By rearranging the equation, you can solve for \(K\) to understand how far the reaction will proceed.
To highlight its use:
- Predicts cell potential under non-standard conditions
- Links standard potential to equilibrium constant \(K\)
- Helps understand reaction progress
Electrochemistry
Electrochemistry is the branch of chemistry that explores the relationship between electricity and chemical reactions. It involves studying reactions where electrons are transferred, called redox reactions, and these usually occur in electrochemical cells.
Electrochemical cells are devices that convert chemical energy into electrical energy (as in galvanic cells) or use electrical energy to provoke chemical changes (as in electrolytic cells). These principles power various everyday applications—from batteries that power devices to processes like electroplating and electrolysis.
In a practical sense:
Electrochemical cells are devices that convert chemical energy into electrical energy (as in galvanic cells) or use electrical energy to provoke chemical changes (as in electrolytic cells). These principles power various everyday applications—from batteries that power devices to processes like electroplating and electrolysis.
In a practical sense:
- Galvanic cells transform chemical energy into electrical energy
- Electrolytic cells drive chemical changes with electrical energy
- Key reactions involve electron transfer, known as redox reactions
Other exercises in this chapter
Problem 53
If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{-5},\) calculate the corresponding \(\Delta G^{\circ}\)
View solution Problem 54
If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{5}\), calculate the corresponding \(\Delta G^{\circ}\)
View solution Problem 58
At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.63 \mathrm{~V}\). The equilibrium constant for the reaction is \(3.8 \times 10^{10}
View solution Problem 60
Consider the voltaic cell illustrated in Figure \(20.5,\) which is based on the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn
View solution