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.

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: \[ \mathrm{Cu^{2+}}(aq) + 2e^- → \mathrm{Cu}(s) \]

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 \(\left[\mathrm{Cu}^{2+}\right]=0.100\,\mathrm{M}\).

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: \[ E_\text{cell}^0 = E_\text{cathode}^0 - E_\text{anode}^0 = 0.337\,V - 0.337\,V = 0\,V \]

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: \[ E_\text{cell} = 0\,V - \frac{8.314 \mathrm{\cancel{J/m.K}} \times 298.15 \mathrm{\cancel{K}}}{2 \times 96485 \mathrm{C} \times \mathrm{mol^{-1}} } \ln \frac{1.00 \times 10^{-4} \mathrm{\cancel{M}}}{0.100\,\mathrm{\cancel{M}}} \]

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

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.