Question

Using data in Appendix \(\mathrm{E}\), calculate the standard emf for each of the following reactions: (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{~F}^{-}(a q)\) (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) \(\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \mathrm{Cu}^{+}(a q) \longrightarrow 2 \mathrm{Hg}(l)+2 \mathrm{Cu}^{2+}(a q)\)

Short answer

The standard emf for the given reactions are: (a) 2.87 V (b) 3.21 V (c) 1.21 V (d) -0.40 V

Step-by-step solution

a) Identify half-reactions for reaction (a)

The balanced redox reaction is given as: \[ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{~F}^{-}(a q) \] The half-reactions in this case can be described as: 1. Oxidation: \(\mathrm{H_2 \longrightarrow 2 H^+ + 2e^-}\) (hydrogen loses 2 electrons) 2. Reduction: \(\mathrm{F_{2} + 2e^{-} \longrightarrow 2 F^{-}}\) (fluorine gains 2 electrons)

b) Use Appendix E to find E° for each half-reaction in reaction (a)

Using the reduction potential table from Appendix E, look for the E° values for both half-reactions: 1. E°(oxidation) for \(\mathrm{H_2 \longrightarrow 2 H^+ + 2e^-}\) = 0 V (The standard hydrogen electrode is used as a reference, so its potential is always zero) 2. E°(reduction) for \(\mathrm{F_{2} + 2e^{-} \longrightarrow 2 F^{-}}\) = 2.87 Volts

c) Calculate the standard emf for reaction (a)

The standard emf (Ecell°) of the cell is the difference between the reduction potential of the reduction half-reaction and the oxidation half-reaction. Ecell° = E°(reduction) - E°(oxidation) = 2.87 V - 0 V = 2.87 V So, the standard emf for reaction (a) is 2.87 V. Similarly, repeat the process for reactions (b), (c), and (d) by identifying the half-cell reactions, looking up their standard reduction potentials using Appendix E, and calculating the difference.

d) Repeat steps a, b, and c for reactions (b), (c), and (d)

For reaction (b): 1. Oxidation: \(\mathrm{Ca \longrightarrow Ca^{2+} + 2e^{-}}\) 2. Reduction: \(\mathrm{Cu^{2+} + 2e^{-} \longrightarrow Cu}\) E°(oxidation) = -2.87 V and E°(reduction) = 0.34 V Ecell° = E°(reduction) - E°(oxidation) = 0.34 V - (-2.87 V) = 3.21 V For reaction (c): 1. Oxidation: \(\mathrm{3Fe^{2+} \longrightarrow Fe + 2Fe^{3+} + e^{-}}\) 2. Reduction: \(\mathrm{Fe^{2+} + e^{-} \longrightarrow Fe}\) E°(oxidation) = -0.44 V and E°(reduction) = 0.77 V Ecell° = E°(reduction) - E°(oxidation) = 0.77 V - (-0.44 V) = 1.21 V For reaction (d): 1. Oxidation: \(\mathrm{Hg_{2}^{2+} + 2e^{-} \longrightarrow 2Hg}\) 2. Reduction: \(\mathrm{2Cu^{+} \longrightarrow Cu^{2+} + Cu}\) E°(oxidation) = 0.92 V and E°(reduction) = 0.52 V Ecell° = E°(reduction) - E°(oxidation) = 0.52 V - 0.92 V = -0.40 V So, the standard emf for the given reactions are as follows: (a) 2.87 V (b) 3.21 V (c) 1.21 V (d) -0.40 V

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between species. This transfer gives rise to changes in the oxidation states of the atoms involved. This is crucial in the field of electrochemistry, which we'll explore later in this article. In these reactions, one species loses electrons and is oxidized, while another gains electrons and is reduced. An everyday example is the rusting of iron where iron is oxidized and oxygen is reduced.

An understanding of redox reactions is pivotal when calculating the standard electromotive force (emf) of a reaction. This is because the standard emf is a measure of the driving force behind the flow of electrons from the oxidizing agent to the reducing agent. Thus, the concept of redox reactions forms the foundation for many calculations in electrochemistry, including battery design, corrosion, and even the function of biological cells.

Half-Reactions

When dealing with redox reactions, it's beneficial to split the overall reaction into two separate equations known as half-reactions. These half-reactions show electrons explicitly and help us understand the individual processes of reduction and oxidation. Each half-reaction has its own standard reduction potential which is reflective of the reactants' affinity for electrons. The more positive the reduction potential, the greater the species' tendency to gain electrons and be reduced.

In studying half-reactions, it's important to balance the number of electrons lost in oxidation with those gained in reduction to ensure that the complete redox reaction is electrically neutral. When calculating standard emf, carefully identifying and analyzing these half-reactions is step one – it allows you to determine oxidation and reduction potentials separately before combining them to find the overall cell potential.

Standard Reduction Potential

The standard reduction potential, denoted as E°, is crucial to understanding electrochemical cells. It is measured in volts and indicates the tendency of a chemical species to gain electrons and be reduced, under standard conditions which are 25°C, 1 M concentration, and 1 atm pressure. The greater the value of E°, the more likely the species will be reduced. The standard hydrogen electrode (SHE) has an assigned potential of 0 Volts, which serves as a reference for measuring all other half-reactions.

For the purpose of calculating standard emf, you'll use the standard reduction potentials of half-reactions from tables, such as Appendix E in the example given. E° values are used to compare the relative strengths of various oxidizing and reducing agents and also to predict whether a reaction is feasible under standard conditions. For instance, when a reduction potential is compared with that of the SHE, you can tell if a substance is a better reducing agent than hydrogen.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationships between electricity and chemical reactions. It encompasses the study of both electrical energy as a product of spontaneous chemical reactions, reflected in galvanic cells, and the use of electrical energy to drive non-spontaneous reactions, seen in electrolytic cells. Understanding electrochemistry is vital for numerous applications including batteries, fuel cells, and electroplating.

The calculation of standard emf falls within this domain, serving as the quantitative measure of a cell's ability to generate an electric current. Therefore, learning how to calculate emf is not only foundational in mastering electrochemistry but also in applying this knowledge to real-world scenarios where control over chemical reactions using electrical energy, or vice versa, is required.