Problem 99 - Chemistry Principles and Practice

Question

Another type of battery is the al 71 in which the cell reaction is $$ \mathrm{Zn}(\mathrm{s})+\mathrm{HgO}(\mathrm{s}) \rightarrow \mathrm{Hg}(\ell)+\mathrm{ZnO}(\mathrm{s}) $$ \(E^{o}=+1.35 \mathrm{~V}\) (a) What is the standard free energy change for this reaction? (b) The standard free energy change in a voltaic cell is the maximum electrical energy that the cell can produce. If the reaction in a zinc-mercury cell consumes \(1.00 \mathrm{~g}\) mercury oxide, what is the standard free energy change? (c) For how many hours could a mercury cell produce a \(10-\mathrm{mA}\) current if the limiting reactant is \(3.50 \mathrm{~g}\) mercury oxide?

Step-by-Step Solution

Answer

(a) \(-260.1 \text{ kJ/mol}\), (b) \(-1.20 \text{ kJ}\), (c) about 86.72 hours.

Step 1: Calculate Standard Free Energy Change (Part a)

The equation to calculate the standard free energy change, \( \Delta G^o \), from the standard cell potential, \( E^o \), is given by:\[ \Delta G^o = -nFE^o \]where: - \( n \) is the number of moles of electrons transferred in the reaction. For this reaction, \( n = 2 \) because two electrons are transferred. - \( F \) is Faraday's constant (approximately \( 96485 \text{ C/mol} \)). - \( E^o = 1.35 \text{ V} \).Plugging in the values:\[ \Delta G^o = -(2)(96485 \text{ C/mol})(1.35 \text{ V}) = -260109 \text{ J/mol} \]This can be expressed in kJ:\[ \Delta G^o = -260.1 \text{ kJ/mol} \]

Step 2: Calculate Standard Free Energy Change with Given Mass (Part b)

To find the standard free energy change when 1.00 g of mercury oxide is consumed, first calculate the moles of HgO.1. Molar mass of \( \text{HgO} \) is approximately \( 216.59 \text{ g/mol} \).2. Moles of \( \text{HgO} = \frac{1.00 \text{ g}}{216.59 \text{ g/mol}} = 0.00462 \text{ mol} \).The standard free energy change per mol of reaction is \(-260.1 \text{ kJ/mol}\).So, for \( 0.00462 \text{ mol}\), it would be:\[ \Delta G = -260.1 \text{ kJ/mol} \times 0.00462 \text{ mol} = -1.20 \text{ kJ} \]

Step 3: Find Cell Operating Time for Given Current (Part c)

First determine the moles of electrons available.1. Calculate moles of HgO in 3.50 g: \[ \frac{3.50 \text{ g}}{216.59 \text{ g/mol}} = 0.0162 \text{ mol} \] Each mole of HgO provides 2 moles of electrons, therefore: \( 0.0162 \text{ mol} \times 2 = 0.0324 \text{ mol} \text{ e}^- \)2. Convert moles of electrons to charge (using Faraday's constant): \[ 0.0324 \text{ mol} \times 96485 \text{ C/mol} = 3122 \text{ C} \]3. Calculate time using the current formula \( Q = It \): \( t = \frac{Q}{I} = \frac{3122 \text{ C}}{10 \times 10^{-3} \text{ A}} = 312200 \text{ s} \)4. Convert seconds to hours: \( t = \frac{312200 \text{ s}}{3600 \text{ s/h}} \approx 86.72 \text{ h} \)

Key Concepts

Standard free energy change

In electrochemistry, the standard free energy change, \( \Delta G^o \), serves as a crucial indicator of a reaction's spontaneity. In simpler terms, it tells us whether a reaction can occur on its own without needing external energy. A negative \( \Delta G^o \) value implies the reaction proceeds spontaneously, while a positive one suggests it won't happen without additional energy.
To find this value for a redox reaction like the one in the exercise involving zinc and mercury oxide, we use the equation:

• \( \Delta G^o = -nFE^o \)

Here, \( n \) represents the number of electrons transferred in the reaction, \( F \) is Faraday's constant (approximately 96485 C/mol), and \( E^o \) is the standard cell potential.
Given the equation for the zinc-mercury cell reaction has \( n = 2 \) and \( E^o = 1.35 \) V, \( \Delta G^o \) is calculated to be \(-260.1\, \text{kJ/mol}\). This confirms that the reaction can proceed spontaneously under standard conditions.

Redox reaction

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. One substance loses electrons (oxidation), while the other gains them (reduction).
In the zinc-mercury cell reaction:

• Zinc (Zn) is oxidized because it loses electrons, forming zinc oxide (ZnO).
• Mercury oxide (HgO) is reduced to mercury (Hg) as it gains those electrons.

These reactions are fundamental in electrochemistry, where the redox process creates an electric current. By understanding which substance is oxidized and which one is reduced, you can comprehend how electron flow is established.
Redox reactions are harnessed in voltaic cells to convert chemical energy into electrical energy. This makes them immensely useful in everyday applications such as batteries.

Voltaic cells

Voltaic cells, also known as galvanic cells, are devices that convert chemical energy to electrical energy using spontaneous redox reactions. In a typical voltaic cell, there are two electrodes, each in contact with a solution, connected by an external circuit and a salt bridge.
In the reaction described, the zinc-mercury cell is an example of such a cell, where zinc is the anode, and mercury oxide is the cathode. Here's how it works:

• The anode is where oxidation occurs; zinc loses electrons.
• Those electrons travel through the external circuit to the cathode.
• At the cathode, mercury oxide gains electrons, getting reduced.

The movement of electrons from the anode to the cathode through the external circuit generates an electric current, which can be harnessed to power devices.
This principle is similar across different types of batteries, making voltaic cells foundational to much of current technology. By understanding voltaic cells, you gain insight into how chemical reactions can be used effectively to produce useful electrical energy.