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 \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

To find the initial cell potential, we will start by determining the standard electrode potential for each electrode. These values can usually be found in chemistry textbooks or online resources. We will denote the standard electrode potential of Zn as \(E^0_{Zn}\) and for Cu as \(E^0_{Cu}\). \(E^0_{Zn} = -0.76V\) \(E^0_{Cu} = +0.34V\)

Now, we will use the Nernst equation: \(E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10}\frac{Q}{1}\) where: \(E_{cell}\) = cell potential at the given concentrations \(E^0_{cell}\) = standard cell potential (\(E^0_{cell} = E^0_{Cu} - E^0_{Zn}\)) n = number of electrons transferred (for Zn - Cu system, n = 2) Q = reaction quotient Using the given concentrations: Q = \(\frac{[Cu^{2+}]}{[Zn^{2+}]}\) = \(\frac{2.50}{0.10}\) = 25 Now, plug in the values and solve for \(E_{cell}\): \(E_{cell} = (0.34 - (-0.76)) - \frac{0.0592}{2} \log_{10}(25)\) \(E_{cell} = 1.10 - 0.0442 \times 1.398\) \(E_{cell} = 1.10 - 0.0617\) \(E_{cell} =1.0383V\) So, the initial cell potential is 1.0383V. Step 2: Calculate the cell potential after 10.0A of current has flowed for 10.0h

We will use Faraday's Law of electrolysis to determine the number of moles of electrons transferred: n = \(\frac{It}{F}\) where: I = current in Amperes (A) t = time in seconds (s) F = Faraday's constant (\(96485C/mol\)) First, convert the time to seconds: t = 10.0h × 3600s/h = 36000s Now calculate the moles of electrons transferred: n = \(\frac{10.0A \times 36000s}{96485C/mol}\) = 3.732 mol

Using stoichiometry, we can now update the concentrations of \(Zn^{2+}\) and \(Cu^{2+}\): Zn: \(0.10 M + \frac{3.732 mol}{1.00 L} = 0.10 M + 3.732 M = 3.832 M\) Cu: \(2.50 M - \frac{3.732 mol}{1.00 L} = 2.50 M - 3.732 M = -1.232 M\) Since the concentration of \(Cu^{2+}\) cannot be negative, we have reached the battery's dead state and the cell potential will be zero. Step 3: Calculate the mass of each electrode after 10.0h The mass change in the electrodes will be due to the stoichiometric conversion of metal ions from \(Zn(s)\) to \(Zn^{2+}\) and \(Cu^{2+}\) to \(Cu(s)\). We know that the number of moles of electrons transferred is 3.732 mol, and the molar mass of Zn and Cu are 65.38 g/mol and 63.55 g/mol, respectively.

For the Zn electrode: \(∆m_{Zn} = n \times M_{Zn}\) \(∆m_{Zn} = 3.732 mol \times 65.38 g/mol\) \(∆m_{Zn} ≈ 243.73 g\)

For the Cu electrode: \(∆m_{Cu} = n \times M_{Cu}\) \(∆m_{Cu} = 3.732 mol \times 63.55 g/mol\) \(∆m_{Cu} ≈ 237.08 g\) Step 4: Calculate how long the battery can deliver a current of 10.0A before it goes dead Since we already determined that the battery goes dead after 10 hours of operation at 10A, the answer is simply 10 hours.