Problem 74 - Chemistry The Central Science In Si Units

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 \(\mathrm{MnO}_{2} ?\)

Step-by-Step Solution

Answer

(a) 5.98 g of MnO₂ is reduced. (b) 13298.3 C of charge is transferred.

Step 1: Write Balanced Half-Reactions

First, consider the balanced chemical half-reactions that occur during the discharge. For the anode (oxidation):\[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \]For the cathode (reduction):\[ \mathrm{MnO}_2 + \mathrm{H}_2\mathrm{O} + e^- \rightarrow \mathrm{MnO} + 2\mathrm{OH}^- \] Notice each reaction's stoichiometry with respect to electrons.

Step 2: Calculate Moles of Zn Consumed

Convert the mass of Zn consumed to moles using its molar mass (65.38 g/mol).\[ \text{Moles of Zn} = \frac{4.50 \, \text{g}}{65.38 \, \text{g/mol}} \approx 0.0688 \, \text{mol} \]

Step 3: Use Stoichiometry to Find Moles of MnO2 Reduced

From the balanced half-reaction, 1 mole of \( \mathrm{Zn} \) provides 2 electrons, and 1 mole of \( \mathrm{MnO}_{2} \) consumes 1 electron. Thus, 0.0688 mol Zn corresponds to 0.0688 mol electrons. \[ \text{Moles of } \mathrm{MnO}_{2} = 0.0688 \, \text{mol} \] (since each mole of \( \mathrm{MnO}_{2} \) accepts 1 mole of electrons).

Step 4: Convert Moles of MnO2 to Mass

Using the molar mass of \( \mathrm{MnO}_2 \) (86.94 g/mol), convert moles to mass.\[ \text{Mass of } \mathrm{MnO}_{2} = 0.0688 \, \text{mol} \times 86.94 \, \text{g/mol} = 5.98 \, \text{g} \]

Step 5: Calculate Charge Transferred

Each mole of Zn provides 2 moles of electrons (Faradays constant \( 96485 \text{ C/mol} \))\[ \text{Charge} = 0.0688 \, \text{mol Zn} \times 2 \, \text{mol e}^- \times 96485 \, \text{C/mol} = 13298.3 \, \text{C} \] So, the total charge transferred is 13298.3 coulombs.

Key Concepts

Alkaline Battery

An alkaline battery is one of the most common types of rechargeable batteries. It uses an alkaline electrolyte, typically potassium hydroxide, to facilitate ion transfer. These batteries are valued for their ability to produce a stable voltage and have a longer shelf life compared to other dry-cell batteries. Inside an alkaline battery, chemical reactions occur between the zinc at the anode and manganese dioxide at the cathode.

• At the anode, zinc undergoes an oxidation reaction.
• At the cathode, manganese dioxide is reduced.

This combination makes the alkaline battery a popular choice for household electronics.

Half-Reactions

In electrochemistry, a half-reaction breaks down the oxidation-reduction process into two simpler parts: oxidation at the anode and reduction at the cathode. This division helps us understand and calculate the movement of electrons in electrochemical cells.

• Oxidation involves the loss of electrons. For the reaction in the exercise, zinc gives away electrons: \[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \]
• Reduction involves the gain of electrons. This happens at the cathode, where manganese dioxide accepts electrons:\[ \mathrm{MnO}_2 + \mathrm{H}_2\mathrm{O} + e^- \rightarrow \mathrm{MnO} + 2\mathrm{OH}^- \]

Writing balanced half-reactions is crucial for understanding the stoichiometry and calculating how much material is consumed or produced.

Oxidation-Reduction

The terms oxidation and reduction describe a pair of processes that always occur together, hence the term 'redox'. In the context of the alkaline battery, these reactions are key to its operation.

• Oxidation is the process where a substance loses electrons. For zinc in the exercise, it transforms from a neutral atom to a positive ion by losing two electrons.
• Reduction is the gain of electrons. In the case of manganese dioxide, it accepts electrons, allowing it to undergo a chemical transformation.

This electron transfer drives the generation of electrical energy which you use in devices powered by these batteries.

Coulombs of Electrical Charge

The quantity of electrical charge that moves through the circuit is measured in coulombs. It’s essential to find the coulombs transferred to understand how much energy a battery can provide before being depleted.
The number of coulombs is calculated by considering the amount of charge passed by each mole of electrons. Using Faraday's constant, which is approximately 96485 C/mol, we can calculate the total charge. In our exercise:

• Each mole of zinc gives away 2 moles of electrons.
• Using the relation: \[ \text{Charge} = 0.0688 \times 2 \times 96485 = 13298.3 \text{ C} \]

This indicates the total amount of electrical charge that the reaction can transport, providing insights into the battery's capacity.