Problem 90 - Chemistry The Molecular Science

Question

In a mercury battery, the anode reaction is $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{ZnO}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} $$ and the cathode reaction is $$ \mathrm{HgO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(\ell)+2 \mathrm{OH}^{-}(\mathrm{aq}) $$ The cell potential is \(1.35 \mathrm{~V}\). Calculate how many hours such a battery can provide power at a rate of \(4.0 \times 10^{-4}\) watt \(\left(1\right.\) watt \(\left.=1 \mathrm{~J} \mathrm{~s}^{-1}\right)\) if \(1.25 \mathrm{~g} \mathrm{HgO}\) is available.

Step-by-Step Solution

Answer

The battery can provide power for approximately 1045.6 hours.

Step 1: Determine the molar mass of HgO

Calculate the molar mass of HgO by adding the molar masses of its constituent elements: the molar mass of Hg (mercury) is 200.59 g/mol, and the molar mass of O (oxygen) is 16.00 g/mol. Therefore, the molar mass of HgO is 200.59 + 16.00 = 216.59 g/mol.

Step 2: Calculate moles of HgO

Using the given mass of HgO, calculate the moles by dividing the mass by the molar mass from Step 1. \[ \text{Moles of HgO} = \frac{1.25 \, \text{g}}{216.59 \, \text{g/mol}} \approx 0.00577 \, \text{mol} \]

Step 3: Calculate total charge in coulombs

Each mole of HgO will provide 2 moles of electrons upon complete reduction, as seen in the cathode reaction. Therefore, calculate the total charge: \[ \text{Total charge (C)} = 0.00577 \, \text{mol} \times 2 \, \text{mol} \times 96485 \, \text{C/mol} = 1115.3 \, \text{C} \]

Step 4: Relate power, energy, and time

Recall the relations: Power (P) = Voltage (V) × Current (I) and Energy = Power × Time. Solving for time:\[ \text{Energy (J)} = 1115.3 \, \text{C} \times 1.35 \, \text{V} = 1505.66 \, \text{J} \]

Step 5: Calculate time in seconds

Using the energy calculated in Step 4 and the power given, find time using the equation Energy = Power × Time:\[ \text{Time (s)} = \frac{1505.66 \, \text{J}}{4.0 \times 10^{-4} \, \text{W}} = 3.76415 \times 10^6 \, \text{s} \]

Step 6: Convert time from seconds to hours

Convert the time from seconds to hours by dividing by the number of seconds in an hour:\[ \text{Time (hours)} = \frac{3.76415 \times 10^6 \, \text{s}}{3600 \, \text{s/hour}} \approx 1045.6 \, \text{hours} \]

Key Concepts

Mercury battery

Mercury batteries are a type of primary battery that uses a chemical reaction to generate electricity. They are often used in small electronic devices like watches, calculators, and hearing aids due to their stable output and long shelf life. The main components include a zinc anode, a mercury oxide (HgO) cathode, and usually a potassium hydroxide (KOH) electrolyte. This configuration provides a stable voltage of approximately 1.35 volts throughout most of its life.

• The anode, made of zinc, acts as a source of electrons during the discharge process, creating electrical energy.
• The cathode consists of mercury oxide, which is reduced, accepting electrons from the external circuit.
• The electrolyte facilitates the movement of ions, which helps sustain the flow of electrons.

Mercury batteries are advantageous due to their consistent discharge curve and long lifespan, although their use has declined due to environmental concerns related to mercury content.

Anode and cathode reactions

Understanding anode and cathode reactions is crucial for grasping how batteries work. In a mercury battery, at the anode, zinc reacts with hydroxide ions, leading to the formation of zinc oxide and water, while releasing electrons:
\[\text{Zn} (s) + 2\text{OH}^- (aq) \rightarrow \text{ZnO} (aq) + \text{H}_2\text{O} (l) + 2e^-\] At the cathode, the reaction involves mercury oxide being reduced, which means it gains electrons from the anode through the external circuit:
\[\text{HgO} (s) + \text{H}_2\text{O} (l) + 2e^- \rightarrow \text{Hg} (l) + 2\text{OH}^- (aq)\] These reactions are complementary and sustain each other. As the zinc anode supplies electrons, mercury oxide at the cathode accepts them, allowing for a continuous flow of electric current. It's this electron flow that powers electronic devices when the battery is connected to a circuit.

Electrical energy calculation

To calculate the electrical energy provided by a mercury battery, it's essential to consider several steps. First, determine the amount of mercury oxide (HgO) available, allowing us to compute the charge based on its moles:

• Use the molar mass of HgO (216.59 g/mol) to find moles from a given mass, such as 1.25 grams.
• Each mole of HgO provides 2 moles of electrons, hence determine the total charge available using the Faraday constant (96485 C/mol).

The electrical energy can be then computed using the formula:
\[\text{Energy (J)} = \text{Charge (C)} \times \text{Voltage (V)}\] In the exercise, this calculation results in 1505.66 Joules of energy. Once the total energy is known, it can be related to power and time:

• Using \(\text{Energy} = \text{Power} \times \text{Time}\), solve for time in seconds, given a specific power output.
• Convert the calculated time from seconds to more practical units like hours for real-world applications.

This process helps estimate how long a battery can power a device before exhausting its capacity, which is crucial for efficient battery usage and management.