Question

A voltaic cell is constructed with two \(\mathrm{Cu}^{2+}-\mathrm{Cu}\) electrodes. The two half-cells have \(\left[\mathrm{Cu}^{2+}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \times 10^{-4} \mathrm{M}\), respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cu}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

Short answer

a) The cathode is the half-cell with \(\left[\mathrm{Cu}^{2+}\right]=0.100\,\mathrm{M}\). b) The standard emf of the cell is \(0\,V\). c) The cell emf for the given concentrations is approximately \(0.089\,V\). d) The \(\left[\mathrm{Cu}^{2+}\right]\) will decrease at the cathode and increase at the anode as the cell operates.

Step-by-step solution

Identify the half-reactions

We have a voltaic cell with two Cu-Cu(II) half-cells. In a voltaic cell, the reduction half-reaction takes place at the cathode. The reduction half-reaction for copper ions is: \[ \mathrm{Cu^{2+}}(aq) + 2e^- → \mathrm{Cu}(s) \]

Determine which electrode is the cathode

The electrode where reduction occurs is the cathode. In this cell, the cathode will be the half-cell with the higher concentration of \(\mathrm{Cu}^{2+}\), because this will favor the reduction of \(\mathrm{Cu}^{2+}\) ions. Therefore, the cathode is the half-cell with \(\left[\mathrm{Cu}^{2+}\right]=0.100\,\mathrm{M}\).

Find the standard emf of the cell

For a \( Cu^{2+} - Cu \) redox couple, the standard reduction potential, \( E^0 \), is equal to \( +0.337\,V \) . Since both half-cells have the same redox couple, the standard emf of this cell is \(0\,V\), because the potential difference between them is: \[ E_\text{cell}^0 = E_\text{cathode}^0 - E_\text{anode}^0 = 0.337\,V - 0.337\,V = 0\,V \]

Calculate the cell emf using Nernst equation

Now we will find the cell emf for the given concentrations using the Nernst equation, which is: \[ E_\text{cell} = E_\text{cell}^0 - \frac{RT}{nF} \ln \frac{(Q)} \] In our case, since the temperature is not given, we will assume it's 25°C (298.15 K) for a standard condition. Also, the number of electrons exchanged in the reaction (n) is 2. R is the gas constant, which is 8.314 J/(mol·K), and F is the Faraday constant, 96485 C/mol. Q is the reaction quotient given by: \[ Q = \frac{[\mathrm{Cu}^{2+}]_\text{anode}}{[\mathrm{Cu}^{2+}]_\text{cathode}} \] Now we can calculate the cell emf: \[ E_\text{cell} = 0\,V - \frac{8.314 \mathrm{\cancel{J/m.K}} \times 298.15 \mathrm{\cancel{K}}}{2 \times 96485 \mathrm{C} \times \mathrm{mol^{-1}} } \ln \frac{1.00 \times 10^{-4} \mathrm{\cancel{M}}}{0.100\,\mathrm{\cancel{M}}} \]

Evaluate the cell emf

\[ E_\text{cell} = - \frac{8.314 \times 298.15}{2 \times 96485} \ln (10^{-3}) \approx - \frac{2472.9347}{192970} \times (-6.9078) \approx 0.089\,V \] The cell emf for the given concentrations is about \( 0.089\,V \).

Predict changes in Cu(II) concentration

As the cell operates, reduction occurs at the cathode (higher concentration of \(\mathrm{Cu}^{2+}\)) while oxidation occurs at the anode (lower concentration of \(\mathrm{Cu}^{2+}\)). Accordingly: a) At the cathode, the \(\mathrm{Cu^{2+}}\) concentration will decrease, as they are being reduced to form solid \(\mathrm{Cu}\). b) At the anode, the \(\mathrm{Cu^{2+}}\) concentration will increase, since solid copper is being oxidized to form \(\mathrm{Cu^{2+}}\) ions. To summarize the answers: a) The electrode with \(\left[\mathrm{Cu}^{2+}\right]=0.100\,\mathrm{M}\) is the cathode. b) The standard emf of the cell is \(0\,V\). c) The cell emf for the given concentrations is approximately \(0.089\,V\). d) The \(\left[\mathrm{Cu}^{2+}\right]\) will decrease at the cathode and increase at the anode as the cell operates.

Key concepts

Electrode Reactions

In a voltaic cell, chemical reactions occur at electrodes that involve the transfer of electrons. These reactions are crucial in generating electric current. A voltaic cell comprises two half-cells, each containing different metals or metal ions. When an electrode undergoes reduction (gain of electrons), it is termed a cathode. Conversely, when an electrode undergoes oxidation (loss of electrons), it is termed an anode.

In the case of the copper voltaic cell discussed, each half-cell includes a copper electrode submerged in a solution containing copper ions (\(\mathrm{Cu^{2+}}\)). The reaction at the cathode involves the reduction of copper ions:\[\mathrm{Cu^{2+}}(aq) + 2e^- \rightarrow \mathrm{Cu}(s)\]This equation shows that copper ions in the aqueous solution gain electrons to become solid copper on the electrode's surface.

Understanding these reactions is key to predicting the behavior of the voltaic cell and the direction of electron flow in external circuits.

Standard Electrode Potential

Each material has a standardized measure called the standard electrode potential, \(E^0\), which reflects how likely it is for a half-reaction to occur as a reduction. The potential is measured under standard conditions (1 M concentration, 1 atm pressure, 25°C temperature) and compared to a standard hydrogen electrode. A more positive \(E^0\) indicates a greater tendency to gain electrons, making it a better oxidizing agent.

For copper, the standard electrode potential is \(+0.337 \, V\). In this scenario, since both half-cells use copper, the potential difference doesn't create a standard voltage but instead focuses on concentration differences. Hence, the standard emf of our described copper cell (where both electrodes are copper) becomes \(0 \, V\), as their respective standard potentials cancel each other out, \(0.337 \, V - 0.337 \, V\).

Knowing these potentials helps predict and explain the electron flow direction in various electrochemical processes.

Nernst Equation

The Nernst equation becomes crucial when calculating the cell potential under non-standard conditions, such as varied ion concentrations. It relates the actual cell emf to the standard emf and the reaction quotient (Q), giving insights on how concentration changes impact cell potential.

The equation is formulated as: \[E_{cell} = E_{cell}^0 - \frac{RT}{nF}\ln(Q)\]where:

• \(E_{cell}\) is the actual emf.
• \(E_{cell}^0\) is the standard emf.
• \(R\) is the ideal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of electrons transferred.
• \(F\) is Faraday's constant (96485 C/mol).
• \(Q\) is the reaction quotient.

The reaction quotient (\(Q\)) for this particular cell reflects the ratio of \([\mathrm{Cu^{2+}}]_{anode}\) to \([\mathrm{Cu^{2+}}]_{cathode}\), capturing how concentration differences drive the reaction. Thus, the calculated \(E_{cell}\) of \(0.089 \, V\) gives a real measure of the electromotive force given current conditions.

Concentration Changes in Electrodes

As the voltaic cell functions, concentration changes occur at the electrodes due to ongoing electrode reactions. Specifically, these changes are associated with the dynamics of reduction and oxidation processes at the cathode and anode, respectively.

At the cathode, the concentration of \(\mathrm{Cu^{2+}}\) decreases because the copper ions are converted into solid copper by electron gain. Hence, there are fewer copper ions in solution over time as the cell is in operation.

Conversely, at the anode, the reaction involves the oxidation of solid copper, which increases the \(\mathrm{Cu^{2+}}\) concentration in solution as copper atoms lose electrons to become copper ions. As a result, the number of copper ions in solution increases.

Monitoring these changes highlights the importance of electron transfer in determining the direction and extent of reaction progression in electrochemical cells.