Question

An electrochemical cell consists of a silver metal electrode immersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.00 M\) separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00\(M \mathrm{NH}_{3}\) that also contains \(2.4 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) The equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\) is: $$\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad K=1.0 \times 10^{13}$$ and the two cell half-reactions are: $$\begin{array}{rl}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} & {\mathscr{E}^{\circ}=0.80 \mathrm{V}} \\ {\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}} & {\mathscr{E}^{\circ}=0.34 \mathrm{V}}\end{array}$$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C} ?\)

Short answer

The cell potential at \(25^{\circ}C\) with the given concentrations and half-cell potentials is approximately \(0.10\; \mathrm{V}\).

Step-by-step solution

Write the half-cell reactions and the overall cell reaction

The given half-cell reactions are: $$ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \qquad (\mathscr{E}^{\circ}=0.80 \;\mathrm{V}) $$ $$ \mathrm{Cu}^{2+}+2\mathrm{e}^{-} \longrightarrow \mathrm{Cu} \qquad (\mathscr{E}^{\circ}=0.34 \;\mathrm{V}) $$ We are given that \(\mathrm{Ag}^{+}\) is being reduced, and the second half-cell reaction must, therefore, be the reverse of the Cu one: $$ \mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2\mathrm{e}^{-} $$ Now, we can write the overall cell reaction: $$ \mathrm{Ag}^{+} + \mathrm{Cu} \longrightarrow \mathrm{Ag} + \mathrm{Cu}^{2+} $$

Find the equilibrium constant (K) for the overall cell reaction

The given equilibrium constant (K) for the formation of the copper-ammonia complex: $$ \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad (K=1.0 \times 10^{13}) $$ This equilibrium constant will not affect the overall cell potential directly, so we don't need to calculate the equilibrium constant (K) for the overall cell reaction.

Calculate the cell potential using the Nernst equation

Now, we can calculate the cell potential using the Nernst equation: $$ E_{cell}=E_{cathode}^{\circ}-E_{anode}^{\circ}-\frac{RT}{nF}\ln Q $$ Here, \(E_{cathode}^{\circ}\) is the standard potential of the reduction half-cell reaction (Ag), \(E_{anode}^{\circ}\) is the standard potential of the oxidation half-cell reaction (Cu), \(Q\) is the reaction quotient, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, and F is the Faraday constant. The given values are: $$ E_{cathode}^{\circ} = 0.80 \;\mathrm{V} $$ $$ E_{anode}^{\circ} = 0.34 \;\mathrm{V} $$ $$ [\mathrm{Ag}^{+}] = 1.00\;\mathrm{M} $$ $$ [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}] = 2.4 \times 10^{-3} \;\mathrm{M} $$ $$ R = 8.314 \;\mathrm{J \; mol^{-1}\; K^{-1}} $$ $$ T = 25^{\circ} \mathrm{C} + 273.15 \;\mathrm{K} = 298.15 \;\mathrm{K} $$ $$ n = 1 \;\mathrm{mol \; e^{-}} $$ $$ F = 96485 \;\mathrm{C \; mol^{-1}} $$ Substituting the values into the Nernst equation, we get: $$ E_{cell} = 0.80 - 0.34 - \frac{8.314 * 298.15}{1 * 96485} \ln \frac{1}{2.4 \times 10^{-3}} $$ Simplify: $$ E_{cell} = 0.46 - 0.0257 \ln \frac{1}{2.4 \times 10^{-3}} $$ Now, we can calculate the cell potential: $$ E_{cell} = 0.46 - 0.0257 \ln (416.67) \approx 0.10 \;\mathrm{V} $$ The cell potential at \(25^{\circ}C\) with the given concentrations and half-cell potentials is approximately \(\boxed{0.10\; \mathrm{V}}\).

Key concepts

Nernst Equation

The Nernst Equation is a fundamental concept in electrochemistry that relates the cell potential of an electrochemical cell under non-standard conditions to the concentrations of the involved substances. It helps predict how the potential changes when concentrations differ from standard conditions.

The equation is given by:\[E_{cell} = E_{cell}^\circ - \frac{RT}{nF}\ln Q\]Where:

• \(E_{cell}\) is the cell potential under current conditions.
• \(E_{cell}^\circ\) is the standard cell potential.
• \(R\) is the universal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is the Faraday constant (96485 C/mol).
• \(Q\) is the reaction quotient, which is the ratio of the concentrations of the products to the reactants.

The Nernst Equation adjusts the standard cell potential (\(E_{cell}^\circ\)) by accounting for differences in concentration, pressure, and temperature, making it more relevant to real-world conditions. It's useful in predicting cell behavior and calculating the potential when concentrations are not at equilibrium.

Half-cell Reactions

In electrochemical cells, reactions occur in two parts called half-cell reactions. One occurs at the anode and the other at the cathode. Each half-reaction involves a transfer of electrons, with one side gaining electrons (reduction) and the other losing them (oxidation). These reactions are crucial in determining the overall function and efficiency of the cell.

For this electrochemical cell, we have:

• The reduction half-reaction at the cathode: \( \mathrm{Ag}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}\) with a standard potential \(\mathscr{E}^{\circ}=0.80 \;\mathrm{V}\).
• The oxidation half-reaction at the anode: \( \mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+} + 2\mathrm{e}^{-}\) with a standard potential \(\mathscr{E}^{\circ}=0.34 \;\mathrm{V}\).

Half-cell reactions are written separately to help identify which species are oxidized and reduced, and to compute the cell potential. Understanding each half-reaction also allows chemists to balance the overall cell reaction by ensuring the electrons lost in oxidation are equal to those gained in reduction.

Cell Potential Calculation

Calculating the cell potential is a key step in evaluating the feasibility and direction of an electrochemical reaction. It combines the individual potentials of the cathode and anode reactions, considering both concentration and standard conditions.

To compute the cell potential at a specific temperature and concentration, the standard potentials of both half-reactions are used:

• Cathode potential: \(E_{cathode}^{\circ} = 0.80 \;\mathrm{V}\)
• Anode potential: \(E_{anode}^{\circ} = 0.34 \;\mathrm{V}\)

The cell potential formula, adjusted by the Nernst Equation, is:\[E_{cell}=E_{cathode}^{\circ}-E_{anode}^{\circ}-\frac{RT}{nF}\ln Q\]Inserting the values from the exercise:

• \(R = 8.314 \;\mathrm{J \; mol^{-1}\; K^{-1}}\)
• \(T = 298.15 \;\mathrm{K}\)
• \(n = 1\) since one mole of electrons is transferred in the silver half-cell reaction
• \(F = 96485 \;\mathrm{C \; mol^{-1}}\)
• \(Q = \frac{1}{2.4 \times 10^{-3}}\)

Substituting these into the Nernst Equation provides a measure of how the concentration affects the potential, resulting in:\[E_{cell} = 0.46 - 0.0257 \ln(416.67) \approx 0.10 \;\mathrm{V}\]Thus, the cell potential at 25°C is approximately 0.10 V, showing the potential difference driving the electrochemical reaction.