Problem 56 - Chemistry The Science in Context

Question

The standard potential of the Cu-Zn cell reaction, $$\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)$$ is \(1.10 \mathrm{V} .\) Would the potential of the Cu-Zn cell differ from \(1.10 \mathrm{V}\) if the concentrations of both \(\mathrm{Cu}^{2+}\) and \(\mathrm{Zn}^{2+}\) were \(0.25 M ?\)

Step-by-Step Solution

Answer

Answer: No, the potential of the Cu-Zn cell does not differ from the standard potential when both Cu²⁺ and Zn²⁺ ions have concentrations of 0.25 M.

Step 1: Write down the cell reaction

The Cu-Zn cell reaction is given by: $$\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)$$

Step 2: Recall the Nernst equation

The Nernst equation is used to calculate the cell potential at non-standard conditions and is given by: $$E=E^{\circ}-\frac{RT}{nF} \ln Q$$ where: - \(E\) is the cell potential under the given conditions, - \(E^{\circ}\) is the standard cell potential, - \(R\) is the gas constant (\(8.314\ \mathrm{J/(mol\ K)} \)), - \(T\) is the temperature (assume \(298\ \mathrm{K}\) for room temperature), - \(n\) is the number of electrons transferred in the redox reaction, - \(F\) is the Faraday constant (\(96485\ \mathrm{C/mol}\)), and - \(Q\) is the reaction quotient.

Step 3: Calculate the number of electrons transferred (n)

In the reaction of Zn (s) and Cu²⁺, the Zn (s) loses 2 electrons to form Zn²⁺ and Cu²⁺ gains 2 electrons to form Cu (s). So, the total number of electrons exchanged in the reaction is 2.

Step 4: Determine the reaction quotient (Q)

The reaction quotient (Q) for the given reaction can be expressed as: $$Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}$$ Since both \(\mathrm{Cu}^{2+}\) and \(\mathrm{Zn}^{2+}\) concentrations are \(0.25\ \mathrm{M}\), we have: $$Q = \frac{0.25}{0.25} = 1$$

Step 5: Calculate the cell potential (E) using the Nernst equation

By substituting the given values and the calculated Q value in the Nernst equation, we get: $$E=1.10\ \mathrm{V}-\frac{8.314\ \mathrm{J/(mol\ K)}\times 298\ \mathrm{K}}{2\times 96485\ \mathrm{C/mol}} \ln 1$$ As \(\ln 1 = 0\), we get: $$E = 1.10\ \mathrm{V}$$

Step 6: Compare the calculated potential with the given standard potential

The calculated potential (\(E = 1.10\ \mathrm{V}\)) is equal to the given standard potential (\(E^{\circ} = 1.10\ \mathrm{V}\)). Therefore, the potential of the Cu-Zn cell does not differ from the given standard potential when the concentrations of both \(\mathrm{Cu}^{2+}\) and \(\mathrm{Zn}^{2+}\) ions are \(0.25\ \mathrm{M}\).

Key Concepts

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry. It helps us calculate the electromotive force (EMF) or cell potential at conditions that are not standard, when concentrations are different from the default 1 M. Typically shown as:

• \( E = E^{\circ} - \frac{RT}{nF} \ln Q \)

Here are the components:

• \( E \) is the cell potential under specific conditions.
• \( E^{\circ} \) stands for standard cell potential.
• \( R \), the gas constant, is \( 8.314 \, \text{J/(mol K)} \).
• \( T \) represents the temperature in Kelvin; often, room temperature is \( 298 \, \text{K} \).
• \( n \) is the number of moles of electrons transferred.
• \( F \), Faraday’s constant, is \( 96485 \, \text{C/mol} \).
• \( Q \) is the reaction quotient.

In essence, the Nernst Equation adjusts the standard potential \( E^{\circ} \) based on the actual, non-standard conditions of the reaction. It's crucial because real-world reactions rarely happen in ideal conditions.

Standard Cell Potential

The standard cell potential, represented by \( E^{\circ} \), is the voltage (or electric potential difference) of a cell under standard conditions, which are:

• A pressure of \( 1 \, ext{atm} \) for any gases present.
• Concentrations of \( 1 \, ext{M} \) for all solutions.
• A temperature of \( 298 \, ext{K} \) (about \( 25 \, ^{\circ}\text{C} \)).

This potential is essentially the measure of how effectively a current can move from the anode to the cathode under these defined conditions. For the reaction \( \mathrm{Zn}(s) + \mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q) + \mathrm{Cu}(s) \), the standard cell potential is known to be \( 1.10 \, \text{V} \). This helps us understand how favorable the reaction is; the greater the potential, the more spontaneous the reaction.

Reaction Quotient

The Reaction Quotient \( Q \) is a parameter that gives the ratio of the concentrations of products to reactants at any point in the reaction, just like the equilibrium constant, but not necessarily at equilibrium. It is calculated using the formula:

• \( Q = \frac{[\text{products}]}{[\text{reactants}]} \)

For the given reaction, the quotient is formulated as:

• \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \)

In our scenario, both \( \mathrm{Cu}^{2+} \) and \( \mathrm{Zn}^{2+} \) ions have a concentration of \( 0.25 \mathrm{M} \), so \( Q \) becomes \( 1 \). This indicates the quotient is at equilibrium; hence, the potential calculated (\( E \)) using the Nernst Equation remains the same as the standard potential \( E^{\circ} \), because \( \ln 1 = 0 \). Understanding \( Q \) helps us determine how the reaction is proceeding and how it influences the cell potential away from standard conditions.