Problem 36 - Chemistry: The Central Science

Question

Using data in Appendix \(\mathrm{E}\), calculate the standard emf for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathbf{F}^{-}(a q)\) (b) \(\mathrm{Cu}^{2+}(a q)+\mathrm{Ca}(\mathrm{s}) \longrightarrow \mathrm{Cu}(s)+\mathrm{Ca}^{2+}(a q)\) (c) \(3 \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q)\) (d) \(\mathrm{Hg}_{2}{\underline{\phantom{xx}}}^{2+}(a q)+2 \mathrm{Cu}^{+}(a q) \longrightarrow 2 \mathrm{Hg}(l)+2 \mathrm{Cu}^{2+}(a q)\)

Step-by-Step Solution

Answer

The standard emf for each of the given reactions are as follows: (a) 2.87 V (b) 3.207 V (c) -1.211 V (d) 0.490 V

Step 1: Write the half-reactions

For the given reaction, we can identify the oxidation and reduction half-reactions as follows: Oxidation: \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(a q) + 2 e^{-}\) Reduction: \(\mathrm{F}_{2}(g) + 2 e^{-} \rightarrow 2 \mathrm{F}^{-}(a q)\)

Step 2: Find standard reduction potentials

Using Appendix E, we find the standard reduction potentials for the half-reactions: \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(a q) + 2 e^{-}\) : E° = 0.000 V (as H2 is the standard reference) \(\mathrm{F}_{2}(g) + 2 e^{-} \rightarrow 2 \mathrm{F}^{-}(a q)\) : E° = 2.87 V

Step 3: Calculate standard emf

Subtract the oxidation half-reaction potential from the reduction half-reaction potential to find the standard emf of the reaction: E°cell = E°red - E°ox E°cell = 2.87 V - 0.000 V E°cell = 2.87 V So, the standard emf for reaction (a) is 2.87 V. #b)# Cu2+(aq) + Ca(s) -> Cu(s) + Ca2+(aq) Similarly, we need to write the half-reactions.

Step 1: Write the half-reactions

The oxidation and reduction half-reactions are: Oxidation: \(\mathrm{Ca}(s) \rightarrow \mathrm{Ca}^{2+}(a q) + 2 e^{-}\) Reduction: \(\mathrm{Cu}^{2+}(a q) + 2 e^{-} \rightarrow \mathrm{Cu}(s)\)

Step 2: Find standard reduction potentials

Using Appendix E, we find: \(\mathrm{Ca}(s) \rightarrow \mathrm{Ca}^{2+}(a q) + 2 e^{-}\) : E° = -2.87 V \(\mathrm{Cu}^{2+}(a q) + 2 e^{-} \rightarrow \mathrm{Cu}(s)\) : E° = 0.337 V

Step 3: Calculate standard emf

E°cell = E°red - E°ox E°cell = 0.337V - (-2.87 V) E°cell = 3.207 V The standard emf for reaction (b) is 3.207 V. #c)# 3 Fe2+(aq) -> Fe(s) + 2 Fe3+(aq) Again, write the half-reactions.

Step 1: Write the half-reactions

Oxidation: \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q) + e^{-}\) Reduction: \(\mathrm{Fe}^{2+}(a q) + 2 e^{-} \rightarrow \mathrm{Fe}(s)\) Note that we need to balance the electrons in the oxidation half-reaction by multiplying it by 2: Oxidation: \(2\mathrm{Fe}^{2+}(a q) \rightarrow 2\mathrm{Fe}^{3+}(a q) + 2 e^{-}\)

Step 2: Find standard reduction potentials

Using Appendix E, we find: \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q) + e^{-}\) : E° = 0.771 V (multiply by 2 to account for the balanced half-reaction) \(\mathrm{Fe}^{2+}(a q) + 2 e^{-} \rightarrow \mathrm{Fe}(s)\) : E° = -0.440 V

Step 3: Calculate standard emf

E°cell = E°red - E°ox E°cell = (-0.440 V) - (0.771 V) E°cell = -1.211 V The standard emf for reaction (c) is -1.211 V. #d)# Hg2 2+(aq) + 2 Cu+(aq) -> 2 Hg(l) + 2 Cu2+(aq) Finally, write the half-reactions for the last reaction.

Step 1: Write the half-reactions

Oxidation: \(2\mathrm{Cu}^{+}(a q) \rightarrow 2\mathrm{Cu}^{2+}(a q) + 2 e^{-}\) Reduction: \(\mathrm{Hg}_{2}^{2+}(a q) + 2 e^{-} \rightarrow 2\mathrm{Hg}(l)\)

Step 2: Find standard reduction potentials

Using Appendix E, we find: \(2\mathrm{Cu}^{+}(a q) \rightarrow 2\mathrm{Cu}^{2+}(a q) + 2 e^{-}\) : E° = 2 x 0.153 V (multiply by 2 to account for balanced half-reaction) \(\mathrm{Hg}_{2}^{2+}(a q) + 2 e^{-} \rightarrow 2\mathrm{Hg}(l)\) : E° = 0.796 V

Step 3: Calculate standard emf

E°cell = E°red - E°ox E°cell = 0.796 V - (2 x 0.153 V) E°cell = 0.796 V - 0.306 V E°cell = 0.490 V The standard emf for reaction (d) is 0.490 V.

Key Concepts

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. This fascinating field explores how chemical energy is converted to electrical energy and vice versa, a process that is central to the operation of batteries, fuel cells, and electrolysis. In the context of electrochemistry, an electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy.

For students grappling with the fundamentals of electrochemistry, it's crucial to understand that the key reactions occurring within these cells are oxidation (the loss of electrons) and reduction (the gain of electrons). Every electrochemical process involves an electron transfer between two substances, occurring through two separate half-reactions taking place at different electrodes in a cell. Mastering the concept of electron transfer lays the groundwork for understanding more complex principles such as standard reduction potentials, half-reactions, and the functioning of Galvanic cells.

Standard Reduction Potentials

Standard reduction potentials are values that measure the tendency of a chemical species to acquire electrons and thereby be reduced. These potentials are measured in volts and are determined under standard conditions, which typically include a solute concentration of 1 M, a pressure of 1 atm for gases, and a temperature of 25°C (298 K). The standard hydrogen electrode (SHE) is used as the reference point and is assigned a potential of 0.000 V.

When solving electrochemistry problems like the textbook example, it is essential to reference a table of standard reduction potentials to ascertain the voltage associated with each half-reaction. By comparing the potentials of the two half-reactions, you can predict which species will be oxidized and which will be reduced in a cell. Students should take care to note the sign of the standard reduction potential; a positive value indicates a greater tendency to be reduced, while a negative value implies a preference for oxidation when considering the reverse reaction.

Half-Reactions

Half-reactions are a way of breaking down the overall reaction in an electrochemical cell into its two separate parts—the oxidation part and the reduction part. By doing so, it becomes easier to analyze and balance the electron transfers occurring during the reaction. An essential step in solving electrochemistry exercises is writing out these half-reactions correctly.

In each half-reaction, electrons are either produced or consumed. In an oxidation half-reaction, electrons are shown as products, indicating their release by the species being oxidized. Conversely, in a reduction half-reaction, electrons are reactants, highlighting their consumption by the species being reduced. Balancing half-reactions involves ensuring that the number of electrons lost in the oxidation process is equal to the number gained in the reduction process. This balance is fundamental to the stoichiometry of the reactants and products and helps in calculating the cell's standard electromotive force (emf).

Galvanic Cells

Galvanic cells, also known as voltaic cells, are a type of electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. They consist of two different metals connected by a salt bridge or permeable membrane and immersed in solutions of their respective ions.

In a Galvanic cell, the anode is the electrode where oxidation occurs, and it is connected to the negative terminal of the battery. The cathode, on the other hand, is the electrode where reduction takes place and is connected to the positive terminal. Electrons flow from the anode to the cathode through an external circuit, generating an electrical current. The movement of ions within the cell occurs through the salt bridge, maintaining electrical neutrality. Understanding the functions and components of a Galvanic cell is pivotal for students as it deepens their understanding of how electricity is harvested from chemical reactions and provides practical knowledge for dealing with real-world applications.