Problem 92 - Chemistry The Science in Context

Question

Calculate the \(E_{\text {cell value at } 298} \mathrm{K}\) for the cell based on the reaction $$ mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$$ when \(\left[\mathrm{Ag}^{+}\right]=2.56 \times 10^{-3} \mathrm{Mand}\left[\mathrm{Cu}^{2+}\right]=8.25 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

$$Q = \frac{8.25 \times 10^{-4}}{(2.56 \times 10^{-3})^2} = 0.125$$ #tag_title# Step 5: Calculate the cell potential using the Nernst equation #tag_content# Now, we can plug all the values into the Nernst equation to find the cell potential at the given conditions: $$E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q$$ $$E_{cell} = 0.46\,\text{V} - \frac{(8.314\,\text{J}/\text{K} \cdot \text{mol})(298\,\text{K})}{(2\,\text{mol})(96485\,\text{C}/\text{mol})} \ln (0.125)$$ Calculate the value of \(E_{cell}\):

Step 1: Write the two half-reactions

Here are the two half-reactions for the given redox reaction: 1. Oxidation half-reaction: \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) 2. Reduction half-reaction: \(2 \mathrm{Ag}^{+}(a q) + 2e^- \rightarrow 2\mathrm{Ag}(s)\)

Step 2: Find the standard reduction potentials

We need to find the standard reduction potentials of the two half-reactions. These values are found in tables and are given as: 1. Standard reduction potential for Cu(II) couple: \(E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}= 0.34\,\text{V}\) 2. Standard reduction potential for Ag(I) couple: \(E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}}= 0.80\,\text{V}\)

Step 3: Calculate the standard cell potential

To calculate the standard cell potential, we subtract the oxidation half-reaction potential from the reduction half-reaction potential: $$E_{cell}^{\circ} = E_{\mathrm{Ag}^+/ \mathrm{Ag}}^\circ - E_{\mathrm{Cu}^{2+}/ \mathrm{Cu}}^\circ = 0.80 - 0.34 = 0.46\,\text{V}$$

Step 4: Compute the reaction quotient

From the balanced reaction, the reaction quotient (\(Q\)) is given by: $$Q = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2} = \frac{8.25 \times 10^{-4}}{(2.56 \times 10^{-3})^2}$$ Calculate the value of \(Q\):

Key Concepts

Standard Reduction Potential

Understanding standard reduction potential is crucial for comprehending the spontaneity of redox reactions within an electrochemical cell. Standard reduction potentials, often represented by the symbol \(E^\circ\), are measured under standard conditions (25°C, 1M concentration for each ion, and 1 atm pressure for any gases involved). Each half-reaction in an electrochemical cell has its own standard reduction potential. The more positive the standard reduction potential, the greater the substance's affinity for electrons (i.e., it is reduced more readily).

For example, in the cell reaction given in the exercise, we have two half-reactions involving copper and silver. Copper is being oxidized (loses electrons), and silver is reduced (gains electrons). The standard reduction potential for copper \(\text{Cu}^{2+}/\text{Cu}\) is 0.34 V and for silver \(\text{Ag}^+/\text{Ag}\) is 0.80 V. This means that silver ions have a greater tendency to gain electrons than copper ions do to lose them; thus, silver acts as the cathode and copper as the anode in this cell.

Nernst Equation

The Nernst Equation allows us to calculate the cell potential at any concentrations (not just the standard conditions). It relates the standard cell potential \(E^\circ_{cell}\) to the actual cell potential \(E_{cell}\) taking into account the reaction quotient \(Q\), temperature \(T\), the number of moles of electrons transferred in the reaction \(n\), and the Faraday constant \(F\). The equation can be written as:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log(Q) \]
Where \(\log\) is the base-10 logarithm and 0.0592 V is the result of dividing \(RT/F\) by the natural logarithm of 10 at standard temperature \(T = 298 K\). This equation is a fundamental tool for understanding how cell potential is affected by changing ion concentrations.

Reaction Quotient

The reaction quotient \(Q\) is akin to the equilibrium constant \(K\), but for a system that is not at equilibrium. It is a measure of the relative quantities of reactants and products at a given instant. For the reaction \(\text{aA} + \text{bB} \rightarrow \text{cC} + \text{dD}\), \(Q\) would be expressed as:
\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
For the provided cell reaction, \(Q\) is calculated using the concentrations of \(\text{Cu}^{2+}\) and \(\text{Ag}^+\). Knowing \(Q\) helps us to apply the Nernst Equation accurately to find out the actual cell potential when the solution conditions deviate from standard state.

Galvanic Cell

A Galvanic cell, also known as a voltaic cell, is a device that converts chemical energy into electrical energy through spontaneous redox reactions. It is composed of two half-cells connected by a salt bridge, facilitating the flow of ions. The half-cell where oxidation occurs is the anode, and where reduction occurs is the cathode. Electrons flow from the anode to the cathode through an external circuit, generating an electric current.

In the context of the exercise, the Galvanic cell uses the copper and silver reactions to generate electrical energy. The cell potential produced is the driving force for the flow of electrons and can be calculated using the Nernst Equation if concentrations are not standard.

Redox Reaction

Redox reactions are a family of reactions involving the transfer of electrons between two species. They are composed of two half-reactions: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. The term 'redox' is derived from the combination of two concepts: reduction and oxidation.

In the exercise, we see an example of a redox reaction where solid copper is oxidized to \(\text{Cu}^{2+}\) ions and silver ions are reduced to solid silver. These complementary processes allow for the flow of electrons through an external path in the related Galvanic cell. Understanding the redox mechanism is essential for analyzing and predicting the flow of electrons and hence the potential difference in an electrochemical cell.