Question

During the discharge of an alkaline battery, 4.50 g of 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 Zn to \(\mathrm{MnO}_{2} ?\)

Short answer

(a) Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of Zn consumed × 86.94 g/mol = \( \frac{4.50}{65.38} \) moles × 86.94 g/mol ≈ 5.92 g (b) Coulombs transferred = moles of electrons transferred × 96,485 C/mol = \( ( \frac{4.50}{65.38} \) moles × 2) × 96,485 C/mol ≈ 13294 C

Step-by-step solution

Convert the mass of Zn consumed to moles

To convert the mass of Zn consumed (4.50 g) to moles, we'll use the formula: Moles = mass / molar mass where the molar mass of Zn is 65.38 g/mol. Moles of Zn consumed = \( \frac{4.50}{65.38} \) moles

Find the number of moles of \(\mathrm{MnO}_{2}\) reduced

In the alkaline battery, the balanced half-reactions are as follows: Anode: \(Zn(s) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2e^-\) Cathode: \(MnO_{2}(s) + H_{2}O(l) + 2e^- \rightarrow MnOOH(s) + 2OH^-(aq)\) By comparing the anode and cathode half-reactions, we can see that the ratio of Zn to \(\mathrm{MnO}_{2}\) is 1:1. This means that the moles of \(\mathrm{MnO}_{2}\) reduced will be equal to the moles of Zn consumed. Moles of \(\mathrm{MnO}_{2}\) reduced = moles of Zn consumed

Convert moles of \(\mathrm{MnO}_{2}\) reduced to mass

To convert the moles of \(\mathrm{MnO}_{2}\) reduced to mass, we'll use the formula: Mass = moles × molar mass where the molar mass of \(\mathrm{MnO}_{2}\) is 86.94 g/mol. Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of \(\mathrm{MnO}_{2}\) reduced × 86.94 g/mol

Calculate the number of coulombs transferred

Now that we know the number of moles of Zn consumed, we can calculate the number of electrons transferred during the discharge. From the anode half-reaction, we know that for each mole of Zn consumed, 2 moles of electrons are transferred. Therefore: Moles of electrons transferred = moles of Zn consumed × 2 To find the number of coulombs transferred, we need to convert the moles of electrons to coulombs using Faraday's constant: Coulombs = moles of electrons transferred × Faraday's constant where Faraday's constant is 96,485 C/mol. Coulombs transferred = moles of electrons transferred × 96,485 C/mol

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes in which the oxidation state of atoms changes. In alkaline batteries, these reactions facilitate the flow of electrons. They consist of two half-reactions: oxidation and reduction.

In the problem, zinc (Zn) undergoes oxidation at the anode, losing electrons to become zinc ions. This is expressed as:

• Anode Reaction: \(Zn(s) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2e^-\)

At the cathode, manganese dioxide (\(MnO_2\)) is reduced, gaining electrons to form \(MnOOH\):

• Cathode Reaction: \(MnO_{2}(s) + H_{2}O(l) + 2e^- \rightarrow MnOOH(s) + 2OH^-(aq)\)

Understanding these reactions helps conceptualize how the battery generates electricity through the movement of electrons.

Molar Mass Calculations

Molar mass calculations are essential for converting a substance's mass to moles, which plays a crucial role in stoichiometry.

To find the number of moles of a substance, you use the formula: \[\text{Moles} = \frac{\text{mass}}{\text{molar mass}}\] In the given exercise, zinc has a molar mass of 65.38 g/mol. With 4.50 g of zinc consumed, the moles of zinc are:

• \(\frac{4.50}{65.38}\) moles of \(Zn\)

Similarly, for \(MnO_2\), knowing its molar mass is 86.94 g/mol allows us to convert the moles to mass. This step ensures accuracy in determining how much material is involved in the reaction.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. Within these cells, there are two electrodes: an anode and a cathode.

In an alkaline battery:

• The **anode** is where oxidation occurs. For the battery in question, zinc serves as the anode, releasing electrons.
• The **cathode** is where reduction occurs. Here, \(MnO_2\) gains electrons.

This setup allows for the consistent flow of electrons through the external circuit, powering devices. Understanding how these components interact is fundamental to grasping battery operation and effectiveness.

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis connects the amount of substance that undergoes a reaction to the electrical charge applied. It states that the charge in coulombs is proportional to the amount of substance converted at the electrode.

Using Faraday's constant (96,485 C/mol), the number of moles of electrons transferred can be converted into electrical charge:

• For every mole of \(Zn\) consumed, two moles of electrons are produced.
• The relationship is expressed as: \[ \text{Coulombs} = \text{moles of electrons} \times 96,485 \text{C/mol} \]

This allows us to determine the overall charge transferred during the battery's discharge, linking the chemical nature of the reaction to its electrical output.