Question

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2}\) ?

Short answer

In conclusion, (a) the mass of MnO2 reduced at the cathode during the discharge is approximately 120 g, and (b) approximately 13,300 coulombs of electrical charge are transferred from Zn to MnO2.

Step-by-step solution

Write the balanced reaction equation

The balanced equation for the given reaction is: \(Zn + 2MnO_{2} + 2H_{2}O \longrightarrow Zn(OH)_{4}^{2-} + 2MnOOH \) Note that 1 mole of Zn reacts with 2 moles of MnO2.

Calculate the number of moles of Zn consumed

To find the number of moles of Zn consumed, we can use the provided mass and the molar mass of Zn. \(moles~of~Zn = \frac{mass~of~Zn}{molar~mass~of~Zn} \) The molar mass of Zn is 65.38 g/mol. Given that 4.50 g of Zn is consumed: \(moles~of~Zn = \frac{4.50~g}{65.38~g/mol} = 0.06884~mol \)

Calculate the number of moles of MnO2 reduced

Using the stoichiometry of the balanced equation (1 mole of Zn reacts with 2 moles of MnO2): \(moles~of~MnO_{2} = 2 \times moles~of~Zn \) \(moles~of~MnO_{2} = 2 \times 0.06884~mol = 0.1377~mol \)

Calculate the mass of MnO2 reduced

Using the number of moles and the molar mass of MnO2 (Mn: 54.94 g/mol, O: 16.00 g/mol): \(mass~of~MnO_{2} = moles~of~MnO_{2} \times molar~mass~of~MnO_{2} \) \(mass~of~MnO_{2} = 0.1377~mol \times (54.94~g/mol + 2 \times 16.00~g/mol)\) \(mass~of~MnO_{2} = 0.1377~mol \times 86.94~g/mol = 120~g \) So, the mass of MnO2 reduced is approximately 120 g.

Calculate the number of electrons transferred per mole of Zn

From the balanced equation: 1 mole of Zn loses 2 electrons (as it forms Zn(OH)4 with a charge of 2-), and 2 moles of MnO2 gain 2 electrons to form 2 moles of MnOOH. So, the number of electrons transferred per mole of Zn is 2.

Calculate the total charge transferred

Using Faraday's law: Charge (Coulombs) = moles of Zn × number of electrons transferred per mole of Zn × Faraday constant (96,485 C/mol) Charge (C) = 0.06884 mol × 2 × 96,485 C/mol Charge (C) = 13,300 C Around 13,300 coulombs of electrical charge are transferred from Zn to MnO2. In conclusion, (a) The mass of MnO2 reduced at the cathode during the discharge is approximately 120 g. (b) Approximately 13,300 coulombs of electrical charge are transferred from Zn to MnO2.

Key concepts

Zinc Anode Reaction

In an alkaline battery, the zinc anode plays a crucial role in facilitating the discharge process. The zinc metal (Zn) undergoes an oxidation reaction when the battery is discharging. Oxidation means that the zinc gives away electrons. Specifically, zinc at the anode loses two electrons to form zincate ions:
\[ \mathrm{Zn} + 2\mathrm{OH}^- \rightarrow \mathrm{Zn(OH)_4}^{2-} + 2\mathrm{e}^- \]

• The zinc atoms turning into zincate ions is the loss part of the oxidation process. It's like zinc is losing this specific charge in the process.
• The generation of free electrons through this process is key because they are used in generating electricity.

To calculate the amount of zinc used, one needs to know the mass of zinc available, and then convert it into moles using its molar mass. For example, if you start with 4.50 grams of zinc, using its molar mass (65.38 g/mol), you can find out how many moles are involved in the reaction.
Thus:
\[ \text{moles of Zn} = \frac{4.50~\text{g}}{65.38~\text{g/mol}} \approx 0.06884~\text{mol} \]This calculation helps determine the extent of the reaction and how much manganese dioxide can be reduced at the other electrode.

MnO2 Reduction

The cathode of an alkaline battery involves the reduction of manganese dioxide (MnO_2). Reduction is the opposite of oxidation. It involves gaining electrons. For each zinc atom oxidized at the anode, two MnO2 molecules at the cathode accept a total of two electrons, forming Mn(OH)_2:
\[ 2\mathrm{MnO}_2 + 2\mathrm{e}^- + 2\mathrm{H}_2O \rightarrow 2\mathrm{MnO(OH)} + 2\mathrm{OH}^- \]

• For the reduction reaction, manganese dioxide receives electrons from the zinc oxidation at the other side of the battery.
• It's important to know that two MnO2 molecules reduce for every one Zn atom oxidized.

So, to find how much MnO2 is reduced, we multiply the number of moles of Zn by two because two MnO2 react per one Zn:
\[ \text{moles of MnO}_2 = 2 \times 0.06884 \approx 0.1377 \text{ mol} \]Using MnO2's molar mass (86.94 g/mol), the mass reduction is:
\[ \text{mass of MnO}_2 = 0.1377\text{ mol} \times 86.94\text{ g/mol} \approx 11.97 \text{ g} \]This indicates how much MnO2 gets used up at the cathode when an equal amount of Zn is consumed at the anode.

Faraday's Law

Faraday's Law of Electrolysis explains the relationship between the quantity of substance altered at an electrode and the total electricity used in the process. It states that the amount of substance either oxidized or reduced is directly proportional to the quantity of electric charge passed through the substance. This principle is important for understanding how batteries operate because it connects chemical changes to electric current.

• The number of moles of electrons transferred can be linked to the amount of charge through Faraday's constant (96,485 C/mol).
• It helps in calculating how much charge is transferred, using the formula:
  \[ \text{Charge (C)} = \text{moles of } \text{Zn} \times \text{number of electrons transferred per mole of Zn} \times \text{Faraday's constant} \]

In our example, 0.06884 moles of Zn contribute to the transfer of electrons. Since each mole transfers two electrons:
\[ \text{Charge (C)} = 0.06884 \text{ mol} \times 2 \times 96,485 \text{ C/mol} = 13,300 \text{ C} \]This calculation explains the link between the discharged mass and the total electricity output of the battery.

Electrochemical Reactions

Electrochemical reactions are chemical reactions that involve the transfer of electrons. They are a fundamental component of how batteries function, transforming chemical energy into electrical energy. It's like a dance of oxidation and reduction happening simultaneously.

• At the anode, oxidation occurs, creating electrons while the metal loses electrons.
• At the cathode, the reduction reaction happens, using those free electrons.

Together these create a flow of electrical current through a connected circuit.
The key insight into electrochemical reactions is their reliance on redox chemistry. Think of two partners in a dance: the anode loses electrons (makes them free), and the cathode accepts them (takes in the free ones):

• This transfer results in an electric current because electrons move from negative to positive points within the battery.
• Electrochemical reactions maintain this movement as long as there is reactant available.

By controlling materials and reactions, engineers design various batteries for different power needs—demonstrating how chemistry can power technology.