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 \(10.00 \mathrm{M} \mathrm{NH}_{3}\) 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: \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{13}\) and the two cell half-reactions are: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C}\) ?

Short answer

The cell potential at 25°C can be calculated using the Nernst equation and the given standard electrode potentials. First, find the concentration of \(\mathrm{Cu}^{2+}\) using the equilibrium constant and given concentrations. Then, calculate the electrode potential for the copper half-reaction using the Nernst equation. Finally, calculate the cell potential by subtracting the copper electrode potential from the given standard electrode potential of the silver half-reaction. The cell potential at 25°C is: \[\mathscr{E}_\text{cell} = 0.80 - \mathscr{E}_\mathrm{Cu}\]

Step-by-step solution

Write the Nernst equation for the copper half-reaction

The Nernst equation is used to calculate the electrode potential for half-reactions at non-standard conditions. The general Nernst equation for a half-reaction looks like this: \[\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF}\ln{Q_c}\] where \(\mathscr{E}\) is the electrode potential at non-standard conditions, \(\mathscr{E}^\circ\) is the standard electrode potential, \(R\) is the gas constant (\(8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\)), \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred in the half-reaction, \(F\) is the Faraday constant (\(96485 \frac{\mathrm{C}}{\mathrm{mol}}\)), and \(Q_c\) is the reaction quotient. We will apply the Nernst equation to the copper half-reaction: \(\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \longrightarrow \mathrm{Cu}\)

Calculate the reaction quotient for the copper half-reaction

First, we need to find the concentration of \(\mathrm{Cu}^{2+}\). We are given the equilibrium constant and concentrations of other species involved in the equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\). That equilibrium reaction is as follows: \(\mathrm{Cu}^{2+}(aq)+4 \mathrm{NH}_{3}(aq) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(aq) \quad K=1.0 \times 10^{13}\) Using the equilibrium constant expression, we can solve for the concentration of \(\mathrm{Cu}^{2+}\): \[K = \frac{[\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4}\] Substituting the given values: \[1.0 \times 10^{13} = \frac{2.4 \times 10^{-3}}{[\mathrm{Cu}^{2+}](10.00)^4}\] Now, solve for the concentration of \(\mathrm{Cu}^{2+}\): \[[\mathrm{Cu}^{2+}] = \frac{2.4 \times 10^{-3}}{(1.0 \times 10^{13})(10.00)^4}\]

Calculate the electrode potential for the copper half-reaction using the Nernst equation

Now we are ready to plug in the values into the Nernst equation for the copper half-reaction: \[\mathscr{E}_\mathrm{Cu} = \mathscr{E}_\mathrm{Cu}^\circ - \frac{RT}{nF}\ln{Q_c}\] Substituting the given values and the temperature of 25°C (298.15 K): \[\mathscr{E}_\mathrm{Cu} = 0.34 - \frac{8.314 \times 298.15}{2 \times 96485}\ln{\frac{[\mathrm{Cu}^{2+}]}{2.4 \times 10^{-3}}}\]

Calculate the cell potential

We are given the standard electrode potential for the silver half-reaction: \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \quad \mathscr{E}^\circ_\mathrm{Ag}=0.80 \,\mathrm{V}\) Since we assumed that \(\mathrm{Ag}^{+}\) is reduced, and it has a standard electrode potential of 0.80 V, the cell potential can be found by subtracting the copper electrode potential: \[\mathscr{E}_\text{cell} = \mathscr{E}_\mathrm{Ag} - \mathscr{E}_\mathrm{Cu}\]

Finalize the solution

Substitute the calculated electrode potentials: \[\mathscr{E}_\text{cell} = 0.80 - \mathscr{E}_\mathrm{Cu}\] Plugging in the value for \(\mathscr{E}_\mathrm{Cu}\) from Step 3, perform the calculations to get the final value for the cell potential at 25°C.

Key concepts

Nernst Equation

The Nernst Equation helps us find the cell potential when conditions aren't standard. It's like a tool that adjusts the potential based on temperature, concentration, and more. The equation is:\[ \mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF}\ln{Q} \]This formula uses several constants:

• \(R\): Gas constant \(8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\).
• \(T\): Temperature in Kelvin.
• \(n\): Number of electrons transferred.
• \(F\): Faraday constant \(96485 \frac{\mathrm{C}}{\mathrm{mol}}\).

Application:
Consider the copper half-reaction given in the exercise. By plugging in the respective values including concentration terms (from the Reaction Quotient), you can find the non-standard potential of the reaction.

Standard Electrode Potential

The Standard Electrode Potential \(\mathscr{E}^\circ\) is like a reference point showing the tendency of a chemical species to be reduced. It's usually measured under standard conditions: 1 M concentration, 1 atm pressure, and 25°C.For the silver and copper reactions:

• Silver: \( \mathrm{Ag}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{Ag} \quad \mathscr{E}^\circ = 0.80 \, \mathrm{V} \)
• Copper: \( \mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \quad \mathscr{E}^\circ = 0.34 \, \mathrm{V} \)

These values indicate how likely the ions are to accept electrons. In general, higher values suggest a higher likelihood of reduction. These potentials are essential for calculating cell potentials using the Nernst Equation.

Reaction Quotient

The Reaction Quotient \(Q\) helps us understand the state of reactions at any given time. It's similar to the equilibrium constant \(K\), but \(Q\) can be evaluated at any point in time, not just at equilibrium.For a reaction:\[ \mathrm{aA} + \mathrm{bB} \rightleftharpoons \mathrm{cC} + \mathrm{dD} \]The expression for \(Q\) is:\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]In the exercise, the complexation of copper with ammonia affects \(Q\). Given the high equilibrium constant \(K\) for forming \(\mathrm{Cu(NH}_3)_4^{2+}\), knowing \(Q\) allows us to plug it into the Nernst Equation to find non-standard conditions.

Half-Reaction

Half-reactions break down the oxidation and reduction parts of a redox process. They show the electron transfer more clearly and help in calculating potentials.The given half-reactions are:

• Silver: \( \mathrm{Ag}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag} \)
• Copper: \( \mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \)

Why Half-Reactions Matter:

They help in finding the number of moles of electrons transferred, crucial for using the Nernst Equation. By splitting the entire cell reaction, you gain insights about which species are getting oxidized and reduced. This step is key when calculating final cell potentials and understanding how energy is harnessed from chemical reactions.