Question

The cathode reaction in the Leclanché cell is given by: $$ 2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s) $$ If a Leclanché cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.The cathode reaction in the Leclanché cell is given by: $$2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s)$$ If a Leclanché cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.

Short answer

The current will last approximately 246.57 hours.

Step-by-step solution

Determine the molar mass of MnO2

Calculate the molar mass of \( \mathrm{MnO}_2 \). The molar mass of manganese (\( \mathrm{Mn} \)) is approximately 54.94 g/mol, and that of oxygen (\( \mathrm{O} \)) is about 16.00 g/mol. Therefore, the molar mass of \( \mathrm{MnO}_2 \) is: \( 54.94 + 2 \times 16.00 = 86.94 \text{ g/mol} \).

Calculate moles of MnO2

Using the initial mass of \( \mathrm{MnO}_2 \) and its molar mass, calculate the number of moles present. \[ \text{Moles of MnO}_2 = \frac{4.0 \text{ g}}{86.94 \text{ g/mol}} \approx 0.046 \text{ mol} \]

Determine the number of moles of electrons exchanged

According to the balanced cathode reaction, 2 moles of electrons are consumed for every 2 moles of \( \mathrm{MnO}_2 \). Thus, the moles of electrons are equal to the moles of \( \mathrm{MnO}_2 \): \[ 0.046 \text{ mol \( \mathrm{MnO}_2 \)} \rightarrow 0.046 \text{ mol electrons} \]

Calculate total charge exchanged

Calculate the total charge (\( Q \)) using the formula \( Q = n \cdot F \), where \( n \) is the number of moles of electrons and \( F = 96485 \text{ C/mol} \) is the Faraday constant.\[ Q = 0.046 \text{ mol} \times 96485 \text{ C/mol} = 4438.31 \text{ C} \]

Calculate time in seconds for the current

The time in seconds can be found using the formula \( t = \frac{Q}{I} \), where \( I \) is the current.\[ t = \frac{4438.31 \text{ C}}{0.0050 \text{ A}} = 887662 \text{ seconds} \]

Convert seconds to hours

Convert the time from seconds to hours by dividing by the number of seconds per hour (3600 seconds per hour).\[ t = \frac{887662}{3600} \approx 246.57 \text{ hours} \]

Key concepts

Electrochemistry

Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical reactions. It is essential in understanding how batteries, like the Leclanché cell, generate electrical energy.

In a typical electrochemical cell, a chemical reaction occurs at the electrodes, leading to the flow of electrons through an external circuit. This flow of electrons is what we perceive as electric current. In the Leclanché cell, the chemical reactions take place between manganese dioxide (\(\text{MnO}_2\)) and zinc ions in an acidic or electrolytic medium.

Each component in this process plays a crucial role; the electrolyte facilitates the movement of ions, while the electrodes are the sites where oxidation and reduction reactions occur. This movement of electrons from the anode to the cathode through an external circuit is vital for the functioning of the cell.

Cathode Reaction

In the context of a Leclanché cell, the cathode reaction is a key chemical process where reduction occurs. Reduction involves the gain of electrons, and in this specific case, \(\text{MnO}_2\) is reduced as it accepts electrons.

The reaction at the cathode can be written as:

• \(2 \text{MnO}_2(s) + \text{Zn}^{2+}(aq) + 2 e^- \rightarrow \text{ZnMn}_2 \text{O}_4(s)\)

During this reaction, the manganese dioxide undergoes a transformation as it gains electrons, converting into zinc manganate (ZnMn₂O₄).

The cathode reaction is crucial because it directly influences the amount of current the cell can deliver. The more effectively this reaction progresses, the better the cell's ability to generate electricity. Additionally, it determines the lifetime of the battery by consuming the reactants over time.

Faraday's Constant

Faraday's constant, denoted by \(F\), is a fundamental value in electrochemistry. It represents the charge of one mole of electrons, approximately \(96485 \text{ C/mol}\), which is integral to calculations involving electrochemical processes.

In the exercise at hand, Faraday's constant helps us convert moles of electrons involved in the reaction into an electrical charge. This conversion is critical for understanding how long a cell can supply a certain current.

Using Faraday's constant, we can determine the total charge passed in the cell with the equation:

• \(Q = n \times F\)

Here, \(Q\) is the total charge, \(n\) is the number of moles of electrons involved, and \(F\) is Faraday's constant. With this, one can calculate the time it takes for the cell to run out, given a specific current output. It's a fundamental concept in understanding and calculating the efficiencies and capabilities of electrochemical cells.