Question

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?

Short answer

a. The initial cell potential is \(1.065V\). b. The cell potential after 10 hours is \(0.982V\). c. The mass of Zn after 10 hours is 193.893 g, and the mass of Cu after 10 hours is 205.889 g. d. The battery can deliver a current of 10.0 A for approximately 10.70 hours before it goes dead.

Step-by-step solution

Identify the half-reactions and the standard cell potential

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: \(E^0_{Cu^{2+}/Cu} = +0.337 V\) \(E^0_{Zn^{2+}/Zn} = -0.763 V\)

Calculate the initial cell potential using the Nernst equation

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 \(1.065V\).

Calculate the charge and change in concentrations after 10 hours

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: \(\Delta [Zn^{2+}] = \frac{Q}{nFV} = \frac{3.60 \times 10^5 C}{2 \cdot (9.6485 × 10^4 C/mol) \cdot 1L}\) \(\Delta [Zn^{2+}] = 0.09351M\) \([Zn^{2+}]_{final} = [Zn^{2+}]_{initial} + \Delta [Zn^{2+}]\) \([Zn^{2+}]_{final} = 0.10 + 0.09351 = 0.19351M\) \([Cu^{2+}]_{final} = 2.50 - 0.09351 = 2.40649M\)

Calculate the cell potential after 10 hours

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 \(0.982V\).

Calculate the mass of each electrode after 10 hours

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.

Calculate the time for the battery to go dead

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.

Key concepts

Nernst Equation

In electrochemistry, the Nernst Equation is a fundamental tool used to calculate the cell potential (voltage) for an electrochemical cell at non-standard conditions. It adjusts the standard cell potential based on the concentrations of the reactants and products. The equation is given by: \[ E_{cell} = E^0 - \frac{RT}{nF} \ln Q \] This equation takes into account:

• Standard cell potential (E^0 ext{):} The voltage of a cell under standard conditions, typically 1 M concentrations and 25°C.
• R ext{: The universal gas constant, equal to } 8.314 ext{ J/(mol K).}
• T ext{: Temperature in Kelvin.}
• n ext{: Number of electrons transferred in the redox reaction, in our example, it's } 2 ext{ for each half-reaction.}
• F ext{: Faraday's constant, } 9.6485 \times 10^4 ext{ C/mol, representing charge per mole of electrons.}
• Q ext{: The reaction quotient, calculated from the concentrations of the products divided by the reactants.}

In practice, the Nernst Equation allows us to determine how the cell potential decreases as the cell operates and concentrations change. This is crucial to understanding battery performance over time.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are integral to electrochemistry. They involve the transfer of electrons from one species to another. In a typical electrochemical cell, two half-reactions occur:

• Oxidation Reaction: Occurs at the anode. Here, zinc (Zn ext{) } loses electrons, forming } Zn^{2+} ext{ ions:} \[ Zn \rightarrow Zn^{2+} + 2e^- \]
• Reduction Reaction: Occurs at the cathode, where } Cu^{2+} ext{ ions gain electrons to form copper } (Cu) ext{ metal:} \[ Cu^{2+} + 2e^- \rightarrow Cu \]

The combination of these half-reactions forms the full redox equation, driving the flow of electrons through the circuit. Understanding these reactions helps explain why materials like zinc and copper are used in batteries, as their difference in standard reduction potential allows for the generation of electrical energy.

Electrolysis

Electrolysis involves using electrical energy to drive a non-spontaneous chemical reaction. In the context of the zinc-copper battery, it describes the changes at the electrodes when the battery is providing current for an extended period.

• Anode Reaction: Zinc atoms lose electrons (oxidation), converting into } Zn^{2+} ext{ ions, which dissolve into the solution. This reduces the mass of the zinc electrode.}
• Cathode Reaction: Copper ions in the solution gain electrons (reduction), depositing copper atoms on the electrode, increasing its mass.}

The entire process of electrolysis is guided by Faraday's laws, relating the amount of substance formed at each electrode to the amount of electricity passed through the system. It explains how the concentrations change over time and how the battery essentially "consumes" itself.

Reaction Quotient

The reaction quotient (Q ext{) is a central concept that helps track how close a system is to equilibrium. In an electrochemical cell, } Q ext{ is calculated using the ratio of the concentrations of the ionic species involved in the redox reactions.} \[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \] As the reaction proceeds, the zinc ions increase while the copper ions decrease, affecting the value of } Q ext{ and thus the cell potential.}

• Initial State: At the start, the concentrations are at their maximum differences, giving a certain } Q ext{ value that determines the initial cell potential via the Nernst Equation.}
• Final State: Over time, as the battery functions, } Q ext{ changes, reducing the voltage output until equilibrium approaches, where the cell potential drops to zero.}

Understanding } Q ext{ helps predict the battery's behavior as it "runs down," and it's a crucial factor for calculating the cell potential over time using the Nernst Equation.