A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C} :\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu}$$ The mass of each electrode is 200. g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) c. Calculate the mass of each electrode after 10.0 h. d. How long can this battery deliver a current of 10.0 A before it goes dead?

For a zinc-copper battery, the half-reactions are as follows: \(Zn \rightarrow Zn^{2+} + 2e^-\) (anode) \(Cu^{2+} + 2e^- \rightarrow Cu\) (cathode) The standard cell potential (\(E^0\)) for the reaction is found by subtracting the standard potential of the anode (\(Zn\)) from the standard potential of the cathode (\(Cu\)): \(E^0 = E^0_{Cu^{2+}/Cu} - E^0_{Zn^{2+}/Zn}\) The values of the standard reduction potentials are: \(E^0_{Cu^{2+}/Cu} = +0.337 V\) \(E^0_{Zn^{2+}/Zn} = -0.763 V\)

The Nernst equation relates the standard cell potential to the actual cell potential (\(E_{cell}\)): \(E_{cell} = E^0 - \frac{RT}{nF}\ln{Q}\) where: - \(R\) is the universal gas constant, \(8.314 J/(mol \cdot K)\) - \(T\) is the temperature in Kelvin, \(298K\) - \(n\) is the number of electrons transferred, in this case, \(2\) - \(F\) is Faraday's constant, \(9.6485 × 10^4 C/mol\) - \(Q\) is the reaction quotient For the reaction, \(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\). Now we can calculate the initial cell potential: \(E_{cell} = E^0 - \frac{RT}{nF}\ln{Q}\) \(E_{cell} = (+0.337 - (-0.763)) - \frac{8.314 J/(mol \cdot K) \cdot 298K}{2 \cdot 9.6485 × 10^4 C/mol}\ln{\frac{0.10M}{2.50M}}\) This results in \(E_{cell(0)} = 1.065 V\). So, a. the initial cell potential is \(1.065V\).

The charge (in coulombs) that has passed through the battery after 10 hours can be calculated with the formula: \(Q (C) = Current (A) \times Time (s) \) \(Q = 10.0 A \times 10.0 h \times \frac{3600 s}{1 h}\) \(Q = 3.60 \times 10^5 C\) Now, using Faraday's law of electrolysis, we can determine the change in concentration of each ion: \(\Delta [Zn^{2+}] = \frac{Q}{nFV} = \frac{3.60 \times 10^5 C}{2 \cdot (9.6485 × 10^4 C/mol) \cdot 1L}\) \(\Delta [Zn^{2+}] = 0.09351M\) \([Zn^{2+}]_{final} = [Zn^{2+}]_{initial} + \Delta [Zn^{2+}]\) \([Zn^{2+}]_{final} = 0.10 + 0.09351 = 0.19351M\) \([Cu^{2+}]_{final} = 2.50 - 0.09351 = 2.40649M\)

We can now use the Nernst equation as in Step 2 with the final concentrations: \(E_{cell(final)} = E^0 - \frac{RT}{nF}\ln{\frac{0.19351M}{2.40649M}}\) \(E_{cell(final)} = (+0.337 - (-0.763)) - \frac{8.314 J/(mol \cdot K) \cdot 298K}{2 \cdot 9.6485 × 10^4 C/mol}\ln{\frac{0.19351M}{2.40649M}}\) This results in \(E_{cell(final)} = 0.982 V\). So, b. the cell potential after 10 hours is \(0.982V\).

Based on stoichiometry and the fact that current is conserved, the mass loss of the anode (Zn) equals the mass gain of the cathode (Cu). \(m_{Zn} = \Delta_{[Zn^{2+}]} \times M_{Zn}\) \(m_{Cu} = \Delta_{[Cu^{2+}]} \times M_{Cu}\) \(m_{Zn} = 0.09351 mol \times 65.38g/mol = 6.107g\) \(m_{Cu} = 0.09351 mol \times 63.55g/mol = 5.889g\) So, c. the mass of Zn after 10 hours is 200 - 6.107 = 193.893 g, and the mass of Cu after 10 hours is 200 + 5.889 = 205.889 g.

The battery is dead when all the zinc is consumed. The change in zinc concentration can be used to calculate the total time for a complete reaction: \(\Delta t = \frac{[Zn^{2+}]_{final}}{\Delta [Zn^{2+}]/\Delta t}\) \(\Delta t = \frac{[Zn^{2+}]_{final} - [Zn^{2+}]_{initial}}{\Delta [Zn^{2+}]/10h}\) \(\Delta t = \frac{0.19351M - 0.10M}{0.09351M/10h}\) \(\Delta t \approx 10.70h\) So, d. the battery can deliver a current of 10.0 A for approximately 10.70 hours before it goes dead.