Problem 40 - Chemistry: An Atoms-Focused Approach

Question

Thomas Edison, the inventor of the incandescent lightbulb, developed a voltaic cell that delivers \(1.4 \mathrm{V}\) of cell potential in an alkaline electrolyte based on the following cell reaction: \(3 \mathrm{Fe}(s)+8 \mathrm{NiO}(\mathrm{OH})(s)+4 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 8 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Fe}_{3} \mathrm{O}_{4}(s) $$ a. Assign oxidation numbers to each element in each of the nickel and iron compounds. b. How many electrons are transferred in the overall reaction? c. What is the value of \(\Delta G_{\text {cell }} ?\)

Step-by-Step Solution

Answer

Answer: The change in Gibbs free energy, \(\Delta G_{\text{cell}}\) for the given redox reaction in the voltaic cell is -1,084,632 J/mol.

Step 1: Assign oxidation numbers

Assign oxidation numbers to each element in each of the given nickel and iron compounds. Using established rules for assigning oxidation numbers, we can determine the oxidation numbers for each element. For the reactants: -\(\mathrm{Fe}(s)\) has an oxidation number of 0 (element in its elemental form). -\(\mathrm{NiO(OH)}(s)\): O has an oxidation number of -2, H has +1, and Ni has +2 (Balancing the overall charge to 0) For the products: -\(\mathrm{Ni(OH)_2}(s)\): O has an oxidation number of -2, H has +1, and Ni has +2 (Balancing the overall charge to 0) -\(\mathrm{Fe_3O_4}(s)\): O has an oxidation number of -2, and \(\mathrm{Fe}\) has +8/3 to balance the overall charge to 0 (One-third of the Fe atoms have an oxidation number of +2, while the other two-thirds have +3)

Step 2: Calculate electrons transferred

To find the number of electrons transferred in the overall reaction: 1. Write the half-reactions for oxidation and reduction processes. 2. Balance the electrons in each half-reaction. 3. Add the half-reactions, and compute the total number of electrons transferred. Oxidation half-reaction: \(3\mathrm{Fe} \rightarrow \mathrm{Fe_3O_4} + 8\mathrm{e^-}\) Reduction half-reaction: \(8\mathrm{NiO(OH)} \rightarrow 8\mathrm{Ni(OH)_2} + 8\mathrm{e^-}\) The total number of electrons transferred in the overall reaction is 8.

Step 3: Calculate \(\Delta G_{\text {cell }}\)

To determine the change in Gibbs free energy, \(\Delta G_{\text {cell}}\), we can use the following equation: \(\Delta G_{\text {cell }}=-nFE_{\text {cell }}\) Here, \(\Delta G_{\text {cell }}\) is the change in Gibbs free energy, n is the number of electrons transferred in the overall reaction, F is the Faraday constant (96,485 coulombs/mol), and \(E_{\text {cell }}\) is the cell potential. Since we know that the voltaic cell delivers 1.4 V of cell potential, and the number of electrons transferred is 8, we can compute the change in Gibbs free energy: \(\Delta G_{\text {cell }} = -8 \times 96,485\,\text{C/mol} \times 1.4\,\text{V}\) \(\Delta G_{\text {cell }} = -1,084,632\,\text{J/mol}\) So, the value of \(\Delta G_{\text {cell}}\) is -1,084,632 J/mol.

Key Concepts

Oxidation Numbers

In redox reactions, oxidation numbers help track the transfer of electrons between elements. Think of oxidation numbers as a bookkeeping system for electrons during chemical changes. They indicate whether an element is losing or gaining electrons during a reaction.
A few basic rules to assign oxidation numbers include:

• Elements in their natural state, like \(\mathrm{Fe}(s)\), have an oxidation number of 0.
• Oxygen is usually -2, and hydrogen is typically +1 in compounds.
• The total oxidation numbers in a neutral compound should sum to zero.

In the cell reaction developed by Thomas Edison, iron (\(\mathrm{Fe}(s)\)) starts with an oxidation number of 0, and in \(\mathrm{Fe_3O_4}(s)\), the oxidation numbers change with two-thirds of the iron atoms having +3 and one-third having +2.
Nickel in \(\mathrm{NiO(OH)}(s)\) and \(\mathrm{Ni(OH)_2}(s)\) maintains an oxidation number of +2 despite the change in compounds due to balancing with other elements' charges, reinforcing the role of maintaining electrical neutrality.

Gibbs Free Energy

The concept of Gibbs Free Energy (\(\Delta G\)) connects chemical reactions to energy changes. It measures the maximum usable work obtained from a thermodynamic system at constant temperature and pressure.
The equation used in electrochemistry is:\[ \Delta G_{\text {cell }} = -nFE_{\text {cell }} \]Where:

• \(\Delta G_{\text {cell }}\) is the change in Gibbs free energy for the cell reaction.
• \(n\) is the number of moles of electrons transferred, which is 8 in this specific example.
• \(F\) is the Faraday constant, approximately 96,485 C/mol.
• \(E_{\text{cell}}\) is the cell potential, 1.4 V in Edison's cell.

By computing, we find \(\Delta G_{\text {cell }}\) equals -1,084,632 J/mol, meaning the reaction delivers that much energy. A negative \(\Delta G\) indicates a spontaneous reaction under the given conditions.

Electrochemistry

Electrochemistry bridges chemical reactions and electricity. It involves studying chemical processes that can produce electric currents or be driven by a current. A voltaic cell, like Edison's, uses spontaneous redox reactions to convert chemical energy into electrical energy.
Key components in electrochemical cells include:

• Anodes and cathodes: The anode is where oxidation happens, and the cathode is where reduction occurs.
• Electrolytes: These are ionically conductive solutions that allow the flow of ions.
• Cell potential (\(E_{\text{cell}}\)): This potential difference between the anode and cathode, measured in volts, indicates the cell's ability to conduct electricity.

Edison's cell achieved a practical potential of 1.4 V, a result of a carefully orchestrated redox reaction involving nickel and iron compounds. This potential provides the driving force for electron flow within the cell, demonstrating electrochemical principles in action.