A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{c}{\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q)}\end{array} $$ The cell is operated at 298 \(\mathrm{K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 M\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential?

The given half-cell reactions are: 1) $$ \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) $$ 2) $$ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q) $$

For each half-cell reaction, we need to find the standard electrode potential (also known as standard reduction potential): $$ E_{\mathrm{Cu}^{+} \mathrm{/Cu}}^{\circ} = +0.52\ \mathrm{V} $$ $$ E_{\mathrm{I_{2}/I^{-}} }^{\circ} = +0.54\ \mathrm{V} $$ We can now find the electromotive force (E°) under standard conditions for the cell by using the formula: $$ E^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} $$ Since the standard reduction potential of I2/I- is greater than that of Cu+/Cu, I2/I- acts as the cathode and Cu+/Cu acts as the anode. $$ E^{\circ} = E_{\mathrm{I_{2}/I^{-}}}^{\circ} - E_{\mathrm{Cu}^{+} \mathrm{/Cu}}^{\circ} = 0.54\ \mathrm{V} - 0.52\ \mathrm{V} = 0.02\ \mathrm{V} $$

The Nernst equation is given by: $$ E = E^{\circ} - \frac{RT}{nF} \ln Q_c $$ where: E = cell potential at non-standard conditions R = gas constant (8.314 J/mol K) T = temperature (298 K) n = number of electrons transferred F = Faraday's constant (96485 C/mol) Qc = reaction quotient We will calculate the cell potential for both half-cells using the Nernst equation. For Cu+/Cu: $$ E_{\mathrm{Cu}^{+} \mathrm{/Cu}} = E_{\mathrm{Cu}^{+} \mathrm{/Cu}}^{\circ} - \frac{RT}{F} \ln \left[\frac{a(\mathrm{Cu}^{+})}{1}\right] $$ For I2/I-: $$ E_{\mathrm{I_{2}/I^{-}}} = E_{\mathrm{I_{2}/I^{-}} }^{\circ} - \frac{RT}{2F} \ln \left[\frac{1}{a(\mathrm{I}^{-})^2}\right] $$ Substituting the known values, we get: $$ E_{\mathrm{Cu}^{+} \mathrm{/Cu}} = +0.52\ \mathrm{V} - \frac{(8.314\ \mathrm{J/mol\ K})(298\ \mathrm{K})}{(96485\ \mathrm{C/mol})} * \ln \left(\frac{0.25\ \mathrm{M}}{1}\right) $$ $$ E_{\mathrm{I_{2}/I^{-}}} = +0.54\ \mathrm{V} - \frac{(8.314\ \mathrm{J/mol\ K})(298\ \mathrm{K})}{(2)(96485\ \mathrm{C/mol})} * \ln \left(\frac{1}{(0.035\ \mathrm{M})^2}\right) $$ Now we can find E (a) as: $$ E = E_{\mathrm{I_{2}/I^{-}}} - E_{\mathrm{Cu}^{+} \mathrm{/Cu}} $$ After calculating all the values, we get: $$ E \approx +0.021\ \mathrm{V} $$ (a) The cell potential E is approximately 0.021 V.

As we mentioned earlier, the anode is the Cu+/Cu half-cell, which has lower standard reduction potential. (b) The anode of the cell is Cu+/Cu.

The results obtained in Step 2 tell us that under standard conditions, Cu+/Cu should still be the anode. (c) Yes, the answer to part (b) is the same as it would be if the cell were operated under standard conditions.

For a cell with zero potential: $$ E = 0 $$ First, we can calculate E for the Cu+/Cu half-cell when [Cu+] = 0.15 M: $$ E_{\mathrm{Cu}^{+} \mathrm{/Cu}} = +0.52\ \mathrm{V} - \frac{(8.314\ \mathrm{J/mol\ K})(298\ \mathrm{K})}{(96485\ \mathrm{C/mol})} * \ln \left(\frac{0.15\ \mathrm{M}}{1}\right) $$ Now, we will calculate the concentration of I- when E = 0 using the Nernst equation for I2/I-: $$ 0 = E_{\mathrm{I_{2}/I^{-}}} - E_{\mathrm{Cu}^{+} \mathrm{/Cu}} $$ $$ 0 = \left[E_{\mathrm{I_{2}/I^{-}} }^{\circ} - \frac{RT}{2F} \ln \left(\frac{1}{a(\mathrm{I}^{-})^2}\right)\right] - E_{\mathrm{Cu}^{+} \mathrm{/Cu}} $$ After rearranging and solving for a(I-), we will get the concentration of I- for zero cell potential (d): $$ a(\mathrm{I}^{-}) \approx 0.026\ \mathrm{M} $$ (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential? The cell would have zero potential when [I-] is approximately 0.026 M.