Problem 74 - Chemistry: The Central Science

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) The mass of \(\mathrm{MnO}_{2}\) reduced at the cathode is approximately \(2.95\) g. (b) The charge transferred between \(\mathrm{Zn}\) and \(\mathrm{MnO}_{2}\) is approximately \(261000\) C.

Step 1: Calculate moles of Zinc

To find the moles of Zinc, we will use its molar mass, which is approximately \(65.38 \mathrm{~g/mol}\). We have: Moles of \(\mathrm{Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}}\) Moles of \(\mathrm{Zn} = \frac{4.50 \mathrm{~g}}{65.38 \mathrm{~g/mol}}\)

Step 2: Moles of \(\mathrm{MnO}_{2}\) reduced

During the discharge of the battery, \(\mathrm{Zn}\) and \(\mathrm{MnO}_{2}\) react according to the following balanced equation: $$\mathrm{Zn} + 2 \mathrm{MnO}_{2} \rightarrow \mathrm{ZnO} + \mathrm{Mn}_{2}\mathrm{O}_{3}$$ From the stoichiometry of the reaction, we see that one mole of \(\mathrm{Zn}\) reacts with two moles of \(\mathrm{MnO}_{2}\). So, we can find the moles of \(\mathrm{MnO}_{2}\) reduced using: Moles of \(\mathrm{MnO}_{2} = 2 \times \text{moles of Zn}\)

Step 3: Calculate mass of \(\mathrm{MnO}_{2}\)

To find the mass of \(\mathrm{MnO}_{2}\) reduced, we will use its molar mass, which is approximately \(86.94 \mathrm{~g/mol}\). We have: Mass of \(\mathrm{MnO}_{2} = \text{moles of MnO}_2 \times \text{molar mass of MnO}_2\) (a) The mass of \(\mathrm{MnO}_{2}\) reduced at the cathode can be calculated by plugging the values obtained in Step 1 and Step 2 into the above equation.

Step 4: Calculate the charge transferred

The charge transferred between the two half-reactions can be calculated using Faraday's law: $$Q = n \times F$$ Where \(Q\) is the charge in coulombs, \(n\) is the total moles of electrons transferred, and \(F\) is Faraday's constant (\(96485 \mathrm{~C/mol}\)). For each mole of \(\mathrm{Zn}\) consumed, two moles of electrons are transferred, as shown by the balanced equation. Hence: \(n = 2 \times \text{moles of Zn}\) (b) The charge transferred can be calculated by plugging the values obtained in Step 1 and the value of \(F\) into the above equation. So, follow the steps to determine: (a) The mass of \(\mathrm{MnO}_{2}\) reduced at the cathode and (b) The charge transferred between \(\mathrm{Zn}\) and \(\mathrm{MnO}_{2}\).

Key Concepts

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It relies heavily on the balanced chemical equation, which provides the necessary ratios needed to perform these calculations.
In the exercise we're discussing, we see stoichiometry at work in several steps. First, it is used to calculate the moles of zinc (\( \mathrm{Zn} \)). Knowing the mass of zinc, and with its molar mass known as well, we can find how many moles of zinc are involved. The formula used is:

• \( \text{Moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} \)

Next, stoichiometry helps to determine how much manganese dioxide (\( \mathrm{MnO}_2 \)) is reduced. The balanced chemical equation tells us that every mole of zinc reacts with two moles of \( \mathrm{MnO}_2 \). Therefore:

• \( \text{Moles of MnO}_2 = 2 \times \text{moles of Zn} \)

Here, stoichiometry provides a clear path from reactants to products, ensuring that the calculations reflect the real chemical process happening in the battery.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are essential to electrochemistry as they involve the transfer of electrons between substances. In our alkaline battery example, zinc undergoes oxidation, while manganese dioxide undergoes reduction.
Oxidation is the loss of electrons. For zinc, this can be represented by the reaction:

• \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^- \)

On the other hand, manganese dioxide (\( \mathrm{MnO}_2 \)) undergoes reduction, gaining electrons in the process. Its half-reaction is generally written as:

• \( \mathrm{MnO}_2 + \mathrm{H}^+ + \mathrm{e}^- \rightarrow \mathrm{MnO(OH)} \)

These two half-reactions together drive the electric current in the battery by moving electrons from zinc, which is oxidized at the anode, to manganese dioxide, which is reduced at the cathode.
Being able to understand and balance these redox reactions is crucial in predicting the behavior of electrochemical cells.

Faraday's Law

Faraday's Law of Electrolysis is a key principle that allows chemists to link the amount of electric charge to the amount of substance converted at an electrode in an electrochemical cell. This law states that the amount of substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the circuit.
To put it simply, for zinc (\( \mathrm{Zn} \)) oxidized in our example, Faraday's law allows us to calculate the total charge in coulombs by using:

• \( Q = n \times F \)

Here, \( Q \) is the charge, \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant (\( 96485 \mathrm{~C/mol} \)).
Given that for every mole of zinc consumed, two moles of electrons are transferred, the mole value \( n = 2 \times \text{moles of Zn} \) is used. This relationship is the backbone of calculations regarding charge transfer in electrochemistry, empowering us to quantify the electrical output or requirements of a chemical process.