Question

Discuss the separate half-reactions and net cell reaction that occur in a flashlight battery. If you have a current of \(.01\) amp, how long can the current flow if the battery contains \(25.0 \mathrm{~g}\) of a paste of \(\mathrm{MnO}_{2}, \mathrm{NH}_{4} \mathrm{Cl}\), and \(\mathrm{ZnCl}_{2}\) of which \(20 \%\) is \(\mathrm{MnO}_{2} ?\)

Short answer

The half-reactions and net cell reaction for a flashlight battery consisting of \(\mathrm{MnO}_{2}\), \(\mathrm{NH}_{4} \mathrm{Cl}\), and \(\mathrm{ZnCl}_{2}\) are as follows: Reduction: \(\mathrm{MnO}_{2} + \mathrm{NH}_{4}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{MnO}(s) + \mathrm{NH}_{3}(g)\) Oxidation: \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^{-}\) Net cell reaction: \(\mathrm{MnO}_{2} + 2\mathrm{NH}_{4}^{+} + \mathrm{Zn}(s) \rightarrow \mathrm{MnO}(s) + 2\mathrm{NH}_{3}(g) + \mathrm{Zn}^{2+}\) Given a battery paste (25 g) consisting of \(20 \%\) \(\mathrm{MnO}_{2}\), the moles of \(\mathrm{MnO}_{2}\) are \(0.0575 \, \textrm{mol}\). The total charge is thus \(0.0575 \, \textrm{F}\). With a current of \(0.01 \, A\), the battery can maintain a continuous flow for \(5.75 \, \mathrm{h}\).

Step-by-step solution

Identify the half-reactions and net cell reaction

In a flashlight battery, two half-reactions occur: reduction and oxidation. In this specific battery with a paste of \(\mathrm{MnO}_{2}, \mathrm{NH}_{4} \mathrm{Cl}\), and \(\mathrm{ZnCl}_{2}\), the reactions are as follows: Reduction (in the cathode) \(\mathrm{MnO}_{2} + \mathrm{NH}_{4}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{MnO}(s) + \mathrm{NH}_{3}(g) \) Oxidation (in the anode) \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^{-} \) Now, we can combine these half-reactions to find the net cell reaction. Net cell reaction: \(\mathrm{MnO}_{2} + 2\mathrm{NH}_{4}^{+} + \mathrm{Zn}(s) \rightarrow \mathrm{MnO}(s) + 2\mathrm{NH}_{3}(g) + \mathrm{Zn}^{2+} \)

Calculate the number of moles of MnO2

Given quantity of the battery paste is \(25.0 \, g\), and \(20 \%\) is \(\mathrm{MnO}_{2}\). We need to calculate the number of moles of \(\mathrm{MnO}_{2}\). First, find the mass of MnO2: \(25.0 \mathrm{~g} * 0.20 = 5.0 \mathrm{~g} \) Now we can find the moles of MnO2 by dividing the mass by its molar mass: \(\textrm{Moles of MnO}_{2} = \frac{5.0 \, \textrm{g}}{86.94 \, \textrm{g/mol}} = 0.0575 \, \textrm{mol}\)

Calculate the total charge of MnO2

In the reduction half-reaction, one mole of \(\mathrm{MnO}_{2}\) consumes one mole of electron, so total charge can be calculated as: \(\textrm{Total charge} = 0.0575 \, \textrm{mol}\ e^{-} * \textrm{1 Faraday/mol} \) \(\textrm{Total charge} = 0.0575 \, \textrm{F}\)

Calculate the time of continuous current flow

Now that we have the total charge, we can find the flow time of the battery given the current of \(0.01 \, A\). We will use the following equation: \(\textrm{Time} = \frac{\textrm{Total charge}}{\textrm{Current}} \) \(\textrm{Time} = \frac{0.0575 \, \textrm{F}}{0.01 \, \mathrm{A}} = 5.75 \, \mathrm{h}\) So, the battery can provide a continuous current of \(0.01 \, A\) for \(5.75 \, \mathrm{h}\).

Key concepts

Half-Reactions in Battery Chemistry

Understanding half-reactions is crucial to grasp the chemistry behind batteries, including the common flashlight battery. In any electrochemical cell, reactions occur in two separate parts: the anode and the cathode. These parts host the oxidation and reduction processes, respectively.

The reduction half-reaction typically occurs at the cathode, where the chemical species gain electrons. In contrast, the oxidation half-reaction happens at the anode, involving the loss of electrons. For flashlight batteries, the specific half-reactions incorporating the substances inside, such as \(\mathrm{MnO}_{2}\), \(\mathrm{NH}_{4}\mathrm{Cl}\), and \(\mathrm{ZnCl}_{2}\), highlight how these materials contribute to the overall electrical energy production.

Within the educational context, explaining half-reactions allows students to deconstruct the complexity of the net cell reaction, thereby gaining a fundamental comprehension of the processes that enable a battery to generate current.

Net Cell Reaction of a Flashlight Battery

The net cell reaction in a battery sums up the overall chemical changes that occur during the discharge cycle. It is found by combining the oxidation and reduction half-reactions while ensuring that the electrons lost in the oxidation process match the electrons gained in the reduction process.

In a flashlight battery, the net cell reaction can be understood as the chemical equation that represents the substance changes from reactants to products. By studying the net cell reaction, students can see the big picture of how a simple flashlight operates on a chemical level. This reaction is central to battery functionality because it describes the entire sequence of chemical events that leads to current flow within the device.

Understanding the Connection

When we connect the half-reactions stated in the step-by-step solution, we're able to comprehend how the solid \(\mathrm{MnO}_{2}\) interacts with \(\mathrm{Zn}\), through \(\mathrm{NH}_{4}^{+}\) ions, to produce \(\mathrm{MnO}(s)\), \(\mathrm{NH}_{3}(g)\), and \(\mathrm{Zn}^{2+}\), which is the essence of the flashlight's chemistry.

Current and Charge Calculation for Battery Operation

Current and charge calculations are essential topics in the study of electrochemistry and battery operations. A demonstrative example is when figuring out how long a flashlight battery can power a device.

To find the duration a battery can supply a current, we must calculate the total charge available in the battery. Then, we relate this charge to the current provided using the equation \(\textrm{Time} = \frac{\textrm{Total charge}}{\textrm{Current}}\).

Practical Application

The step-by-step solution demonstrates the practical application of these calculations. By quantifying the amount of \(\mathrm{MnO}_{2}\) and its corresponding charge capacity, students can link the theoretical concepts of charge and current to real-life battery usage.

This meaningful connection between theory and practice ensures that students not only learn how to perform the calculations but also understand the significance of these calculations in the context of everyday electronic devices.