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Calculating cell potential (voltaic cells)

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Intros
Lessons
  1. Will a redox reaction occur?
  2. Reduction potential
  3. Will a redox reaction happen spontaneously?
  4. Will a redox reaction happen spontaneously? Worked example 1
  5. Will a redox reaction happen spontaneously? Worked example 2
  6. Cell potential, free energy and equilibrium.
  7. The Nernst equation (non-standard cell potential).
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Examples
Lessons
  1. Use a standard potential table1 to predict whether a spontaneous redox reaction will occur.
    1. Will a reaction between Cu (I) and nickel metal occur spontaneously in a redox process?
    2. Will a reaction between Al (III) and scandium metal occur spontaneously in a redox process?
    3. Will a reaction between vanadium metal and Scandium (III) occur spontaneously in a redox process?
  2. Calculate the equilibrium constant K using the cell potential and calculate the cell potential at nonstandard conditions.
    1. The oxidation of zinc metal (Zn) by copper ions (Cu2+) to form Zn2+ ions and Cu (s) happens spontaneously in a redox reaction at standard conditions at 298K. Calculate the equilibrium constant K for this process using the following standard potential data:

      Cu2+ + 2e-  ⇌  \, \rightleftharpoons \, Cu where E0red = +0.34 V
      Zn2+ + 2e-  ⇌  \, \rightleftharpoons \, Zn where E0red = -0.76 V
    2. During the reaction, the reaction quotient Q is measured to be 1.8*108. Use the Nernst equation to find the nonstandard cell potential at these conditions.
      Is the cell potential increasing, decreasing or unchanged during the reaction?
Topic Notes
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Introduction to Calculating Cell Potential in Voltaic Cells

Understanding how to calculate cell potential in voltaic cells is crucial for predicting spontaneous redox reactions. Our introductory video provides a comprehensive overview of this essential concept in electrochemistry. Cell potential, measured in volts, represents the driving force behind electron flow in a voltaic cell. To determine cell potential, we rely on standard electrode reduction potentials, which are tabulated values for various half-reactions. These potentials allow us to predict the direction of electron flow and the overall spontaneity of redox reactions. By comparing the reduction potentials of different half-cells, we can calculate the cell potential and determine whether a reaction will occur spontaneously. This knowledge is fundamental for designing efficient batteries, understanding corrosion processes, and developing new electrochemical technologies. As you delve deeper into this topic, you'll discover how cell potential calculations play a vital role in various applications of electrochemistry.

Understanding Standard Reduction Potentials

Standard reduction potentials play a crucial role in predicting and understanding redox reactions in chemistry. These potentials provide a quantitative measure of the tendency of chemical species to be reduced, which is essential for determining the direction and spontaneity of electron transfer reactions. The concept of standard reduction potentials is fundamental to electrochemistry and has wide-ranging applications in fields such as battery technology, corrosion science, and environmental chemistry.

At the heart of this concept is the table of standard reduction potentials, a powerful tool that chemists use to compare the relative strengths of oxidizing and reducing agents. This table is structured as a list of half-reactions, each representing the reduction of a chemical species. The reactions are arranged in order of decreasing reduction potential, with the most positive potentials at the top and the most negative at the bottom.

The structure of the table is designed to provide immediate insights into the behavior of different substances in redox reactions. Each half-reaction is written as a reduction, showing the species gaining electrons. Alongside each reaction is its corresponding standard reduction potential, measured in volts (V). These potentials are determined under standard conditions: 25°C, 1 atm pressure, and 1 M concentration for aqueous solutions.

Interpreting the table of standard reduction potentials requires understanding that the values represent the potential difference between the half-reaction and the standard hydrogen electrode (SHE), which is arbitrarily assigned a potential of 0.00 V. Positive values indicate half-reactions that are more easily reduced than the hydrogen electrode, while negative values indicate those that are less easily reduced.

The ordering of substances in the table is significant and provides valuable information about their behavior as oxidizing or reducing agents. Species near the top of the table, with more positive reduction potentials, are stronger oxidizing agents. They have a greater tendency to accept electrons and be reduced. Conversely, species near the bottom of the table, with more negative reduction potentials, are stronger reducing agents. These substances have a greater tendency to donate electrons and be oxidized.

For example, fluorine (F2) is found near the top of the table with a standard reduction potential of +2.87 V, making it one of the strongest oxidizing agents. This means that fluorine has a strong tendency to accept electrons and be reduced to fluoride ions (F-). On the other hand, lithium (Li) is found near the bottom of the table with a standard reduction potential of -3.04 V, indicating that it is a powerful reducing agent. Lithium readily gives up electrons to form Li+ ions.

The relative positions of substances in the table allow chemists to predict the direction of electron flow in redox reactions. When two species are involved in a redox reaction, electrons will flow from the species with the more negative reduction potential to the one with the more positive potential. This principle is fundamental in determining whether a reaction will occur spontaneously.

For instance, if we consider a reaction between copper ions (Cu2+) and zinc metal (Zn), we can use the table to predict the outcome. The standard reduction potential for Cu2+ to Cu is +0.34 V, while for Zn2+ to Zn it is -0.76 V. Since zinc has a more negative reduction potential, it will act as the reducing agent, donating electrons to copper ions. The reaction Cu2+ + Zn Cu + Zn2+ will occur spontaneously.

Understanding standard reduction potentials and how to use the table is essential for predicting the feasibility of redox reactions, designing electrochemical cells, and selecting appropriate materials for various applications. In battery design, for example, materials with a large difference in reduction potentials are chosen to maximize the cell voltage. In corrosion prevention, understanding the relative positions of metals in the table helps in selecting appropriate sacrificial anodes to protect more noble metals.

In conclusion, the concept of standard reduction potentials and the associated table are powerful tools in chemistry. They provide a systematic way to compare the oxidizing and reducing strengths of different species, predict the direction of electron flow in redox reactions, and understand the behavior of substances in electrochemical processes. Mastering this concept opens up a wealth of applications in both theoretical and practical chemistry, making it an indispensable part of a chemist's toolkit.

Calculating Cell Potential

Understanding how to calculate cell potential is crucial in electrochemistry. The cell potential equation, also known as the cell potential formula, is a fundamental tool used to determine the voltage of an electrochemical cell. This process involves identifying the reduction and oxidation half-reactions and applying the standard reduction potential values to calculate the overall cell potential.

To begin calculating cell potential, we first need to identify the reduction and oxidation half-reactions. The standard reduction potential table is an essential resource for this step. This table lists various half-reactions and their corresponding standard reduction potentials. The half-reaction with the more positive reduction potential will be the reduction half-reaction, while the other will be the oxidation half-reaction.

Once we have identified the half-reactions, we can proceed with the cell potential calculation. The cell potential equation is as follows:

E°cell = E°reduction - E°oxidation

Where E°cell is the standard cell potential, E°reduction is the standard reduction potential of the reduction half-reaction, and E°oxidation is the standard reduction potential of the oxidation half-reaction.

Here's a step-by-step guide on how to calculate cell potential:

1. Identify the reduction half-reaction: Look for the half-reaction with the more positive standard reduction potential in the table.

2. Identify the oxidation half-reaction: The remaining half-reaction will be the oxidation half-reaction.

3. Write down the standard reduction potentials for both half-reactions.

4. Apply the cell potential equation: Subtract the standard reduction potential of the oxidation half-reaction from the standard reduction potential of the reduction half-reaction.

5. Calculate the result: Perform the subtraction to obtain the standard cell potential.

It's important to note that the sign of the calculated cell potential is crucial in determining the spontaneity of the reaction. A positive cell potential indicates that the reaction is spontaneous under standard conditions, while a negative cell potential suggests that the reaction is non-spontaneous.

For example, let's consider a zinc-copper cell. The relevant half-reactions and their standard reduction potentials are:

Cu² + 2e Cu E° = +0.34 V (reduction)

Zn² + 2e Zn E° = -0.76 V (oxidation)

Applying the cell potential equation:

E°cell = E°reduction - E°oxidation

E°cell = (+0.34 V) - (-0.76 V)

E°cell = +1.10 V

The positive cell potential indicates that this reaction is spontaneous under standard conditions.

Understanding how to calculate cell potential is essential for various applications in electrochemistry, including the design of batteries, fuel cells, and electroplating processes. By mastering the cell potential equation and the process of identifying reduction and oxidation half-reactions, you can predict the behavior of electrochemical systems and determine the feasibility of specific reactions.

In conclusion, calculating cell potential involves identifying the reduction and oxidation half-reactions, using the standard reduction potential table, and applying the cell potential formula. The resulting value not only provides information about the voltage of the electrochemical cell but also indicates whether the reaction will occur spontaneously. This knowledge is fundamental in understanding and predicting the behavior of electrochemical systems in various scientific and industrial applications.

Predicting Spontaneous Reactions

Understanding how to use calculated cell potentials to predict whether a redox reaction will occur spontaneously is a crucial skill in electrochemistry. This knowledge allows chemists to determine the feasibility of various chemical reactions and design more efficient electrochemical systems. The key to this prediction lies in the relationship between positive cell potentials and spontaneous reactions.

Cell potentials, also known as electromotive force (EMF), are calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. The resulting value provides valuable information about the spontaneity of a redox reaction. A positive cell potential indicates that the reaction will occur spontaneously, while a negative cell potential suggests that the reaction is non-spontaneous under standard conditions.

The relationship between positive cell potentials and spontaneous reactions is rooted in thermodynamics. A positive cell potential corresponds to a negative change in Gibbs free energy (ΔG), which is the criterion for a spontaneous process. This connection allows us to predict the direction of electron flow and the overall feasibility of a redox reaction.

To determine which reaction will occur spontaneously given standard electrode reduction potentials, we need to compare the reduction potentials of the half-reactions involved. The species with the more positive reduction potential will be reduced, while the species with the more negative reduction potential will be oxidized. This comparison helps us identify the direction of electron flow and the overall spontaneity of the reaction.

Let's consider an example to illustrate this concept. Suppose we have two half-reactions:

1. Cu² + 2e Cu (E° = +0.34 V)
2. Zn² + 2e Zn (E° = -0.76 V)

To determine which reaction will occur spontaneously, we compare the reduction potentials. Since copper has a more positive reduction potential (+0.34 V) than zinc (-0.76 V), copper will be reduced, and zinc will be oxidized. The overall reaction can be written as:

Cu² + Zn Cu + Zn²

The cell potential for this reaction can be calculated by subtracting the reduction potential of the anode (Zn) from the reduction potential of the cathode (Cu):

E°cell = E°cathode - E°anode = 0.34 V - (-0.76 V) = 1.10 V

The positive cell potential of 1.10 V indicates that this reaction will occur spontaneously under standard conditions.

It's important to note that the spontaneity of a reaction can change under non-standard conditions. Factors such as concentration, temperature, and pressure can affect the cell potential and, consequently, the spontaneity of the reaction. In such cases, the Nernst equation can be used to calculate the cell potential under specific conditions.

Another example to consider is the reaction between silver and iron:

1. Ag + e Ag (E° = +0.80 V)
2. Fe² + 2e Fe (E° = -0.44 V)

Comparing the reduction potentials, we can see that silver has a more positive potential and will be reduced, while iron will be oxidized. The spontaneous reaction can be written as:

2Ag + Fe 2Ag + Fe²

The cell potential for this reaction is:

E°cell = E°cathode - E°anode = 0.80 V - (-0.44 V) = 1.24 V

Again, the positive cell potential confirms that this reaction will occur spontaneously under standard conditions.

In conclusion, calculated cell potentials serve as a powerful tool for predicting the spontaneity of redox reactions. Positive

Relationship Between Cell Potential and Gibbs Free Energy

The connection between cell potential and Gibbs free energy is a fundamental concept in electrochemistry, providing crucial insights into the spontaneity and feasibility of chemical reactions. This relationship is essential for understanding the behavior of electrochemical cells and predicting the direction of chemical processes.

Cell potential, denoted as E°cell, is a measure of the electrical potential difference between two half-cells in an electrochemical cell. It represents the driving force for electron transfer in a redox reaction. On the other hand, Gibbs free energy (ΔG) is a thermodynamic quantity that indicates the amount of useful work obtainable from a system at constant temperature and pressure.

The sign of the cell potential is directly related to the sign of ΔG, and this relationship provides valuable information about reaction spontaneity. When the cell potential is positive (E°cell > 0), the corresponding Gibbs free energy change is negative (ΔG < 0). This scenario indicates that the reaction is spontaneous and will proceed forward without any external input of energy. Conversely, when the cell potential is negative (E°cell < 0), the Gibbs free energy change is positive (ΔG > 0), signifying a non-spontaneous reaction that requires an input of energy to proceed.

The mathematical relationship between cell potential and Gibbs free energy is expressed by the equation:

ΔG = -nFE°cell

Where:

  • ΔG is the change in Gibbs free energy
  • n is the number of electrons transferred in the reaction
  • F is Faraday's constant (96,485 C/mol)
  • E°cell is the standard cell potential

This equation is of paramount importance in electrochemistry as it quantitatively links the electrical and thermodynamic properties of a system. It allows scientists and engineers to calculate the Gibbs free energy change from measured cell potentials and vice versa, providing a powerful tool for predicting and analyzing chemical reactions.

The significance of this relationship extends beyond theoretical calculations. In practical applications, it helps in determining the maximum electrical work that can be obtained from a galvanic cell or the minimum electrical energy required to drive a non-spontaneous reaction in an electrolytic cell. This knowledge is crucial in the design and optimization of batteries, fuel cells, and electrochemical processes in industry.

Moreover, the relationship between cell potential and Gibbs free energy allows for the prediction of reaction equilibria. At equilibrium, the cell potential becomes zero, corresponding to a Gibbs free energy change of zero. This concept is utilized in determining equilibrium constants and understanding the factors that influence the extent of chemical reactions.

In the context of corrosion science, this relationship helps in understanding the thermodynamic driving forces behind metal degradation processes. By analyzing the cell potentials of different metal-electrolyte systems, corrosion engineers can predict the likelihood and severity of corrosion reactions, aiding in the development of effective corrosion prevention strategies.

The interplay between cell potential and Gibbs free energy also has implications in biological systems. Many biochemical processes, such as cellular respiration and photosynthesis, involve electron transfer reactions. Understanding the energetics of these processes through the lens of electrochemistry provides insights into the efficiency and directionality of vital biological functions.

In conclusion, the relationship between cell potential and Gibbs free energy is a cornerstone concept in electrochemistry, bridging the gap between electrical phenomena and thermodynamic principles. It provides a quantitative framework for understanding reaction spontaneity, predicting chemical behavior, and designing efficient electrochemical systems. As research in renewable energy, advanced materials, and biotechnology continues to advance, this fundamental relationship will undoubtedly play a crucial role in shaping future innovations and technologies.

Non-Standard Conditions and the Nernst Equation

In the realm of electrochemistry, understanding non-standard conditions is crucial for accurately predicting and analyzing the behavior of electrochemical cells. While standard conditions provide a baseline for comparison, real-world applications often involve scenarios that deviate from these idealized states. This is where the concept of non-standard conditions comes into play, and the Nernst equation becomes an invaluable tool for calculating cell potential under these varied circumstances.

Non-standard conditions in electrochemical cells refer to any situation where the concentrations of reactants and products differ from the standard state (typically 1 M for solutions, 1 atm for gases, and pure solids or liquids for other species), or when the temperature deviates from the standard 25°C (298.15 K). These conditions are common in practical applications and can significantly affect the cell potential and overall performance of an electrochemical system.

The Nernst equation, named after the German chemist Walther Nernst, provides a mathematical relationship between the standard cell potential (E°) and the cell potential (E) under non-standard conditions. This equation is fundamental in electrochemistry and is expressed as:

E = E° - (RT/nF) ln Q

Where:

  • E is the cell potential under non-standard conditions
  • E° is the standard cell potential
  • R is the universal gas constant (8.314 J/mol·K)
  • T is the temperature in Kelvin
  • n is the number of electrons transferred in the cell reaction
  • F is the Faraday constant (96,485 C/mol)
  • Q is the reaction quotient

The Nernst equation allows us to calculate how the cell potential changes as the concentrations of reactants and products vary from their standard states. This is particularly useful in real-world applications where maintaining standard conditions is often impractical or impossible.

To use the Nernst equation effectively, one must first determine the standard cell potential (E°) from standard reduction potentials of the half-reactions involved. Next, calculate the reaction quotient (Q) based on the actual concentrations or pressures of the species in the cell. Finally, plug these values, along with the temperature and number of electrons transferred, into the equation to find the cell potential under the given non-standard conditions.

For example, consider a zinc-copper cell operating at 25°C with [Zn2+] = 0.1 M and [Cu2+] = 0.01 M. The standard cell potential (E°) for this cell is 1.10 V. Using the Nernst equation, we can calculate the actual cell potential:

E = 1.10 V - (0.0592 V / 2) log(0.1 / 0.01) = 1.13 V

This example demonstrates how the cell potential can differ from the standard value due to non-standard concentrations of reactants and products.

The practical applications of the Nernst equation are vast and diverse. In analytical chemistry, it forms the basis for potentiometric measurements, allowing for the determination of ion concentrations in solution. This principle is utilized in pH meters, ion-selective electrodes, and various other electrochemical sensors.

In the field of corrosion science, the Nernst equation helps predict the conditions under which metals are likely to corrode, enabling the development of more effective corrosion prevention strategies. The equation is also crucial in the design and optimization of batteries and fuel cells, where understanding how cell potential varies with changes in reactant and product concentrations is essential for maximizing performance and efficiency.

Furthermore, the Nernst equation plays a significant role in biological systems, particularly in understanding membrane potentials and the function of ion channels in cells. It helps explain how neurons generate and transmit electrical signals, a fundamental process in the nervous system.

In conclusion, the concept of non-standard conditions and the

Conclusion

In this article, we've explored the crucial concept of cell potential calculations and their role in predicting spontaneous redox reactions. Understanding these principles is fundamental to mastering electrochemistry. The introduction video provided a solid foundation for grasping these complex ideas, making them more accessible to learners. By comprehending cell potential, students can accurately determine whether a redox reaction will occur spontaneously or require external energy input. This knowledge is invaluable for solving practical electrochemistry problems in various scientific and industrial applications. We encourage readers to apply these concepts to real-world scenarios, reinforcing their understanding through hands-on practice. For those seeking to deepen their knowledge, exploring additional resources on electrochemistry is highly recommended. Remember, mastering cell potential calculations is a key step in unlocking the fascinating world of redox reactions and their wide-ranging applications in chemistry and beyond.

Will a Redox Reaction Occur? Understanding Reduction Potential

In this guide, we will explore the steps to determine whether a redox reaction will occur by understanding and using reduction potentials. This process involves using standard reduction potential tables to predict the spontaneity of redox reactions and calculate cell potential.

Step 1: Introduction to Redox Reactions and Spontaneity

Redox reactions involve the transfer of electrons between two substances. One substance gets reduced (gains electrons), and the other gets oxidized (loses electrons). The spontaneity of a redox reaction refers to whether the reaction will occur on its own without external energy input. To predict this, we use standard reduction potentials.

Step 2: Understanding Standard Reduction Potentials

Standard reduction potentials are values that indicate the tendency of a chemical species to be reduced, measured in volts. These values are listed in a table known as the standard reduction potential table. Each entry in the table represents a half-reaction, showing the reduction process of a species along with its associated potential.

Step 3: Using the Standard Reduction Potential Table

The standard reduction potential table is organized with the highest reduction potentials at the top and the lowest at the bottom. Substances at the top are excellent oxidizing agents because they are easily reduced. Conversely, substances at the bottom are good reducing agents because they are easily oxidized.

Step 4: Identifying the Half-Reactions

To determine if a redox reaction will occur, identify the two half-reactions involved: one for reduction and one for oxidation. The reduction half-reaction will be found in the table as it is, while the oxidation half-reaction will be the reverse of a reduction half-reaction listed in the table.

Step 5: Calculating the Cell Potential

Calculate the cell potential (Ecell) by combining the standard reduction potentials of the two half-reactions. The formula is:

Ecell = Ereduction + Eoxidation

Note that the oxidation potential is the negative of the reduction potential for the same species. Therefore, if the reduction potential of a species is -0.76V, its oxidation potential will be +0.76V.

Step 6: Determining Spontaneity

If the calculated cell potential (Ecell) is positive, the redox reaction is spontaneous and will occur on its own. If Ecell is negative, the reaction is non-spontaneous and will not occur without external energy input.

Step 7: Example Calculation

Consider a reaction between zinc and copper ions. The standard reduction potentials are:

  • Zn2+ + 2e- Zn (Ereduction = -0.76V)
  • Cu2+ + 2e- Cu (Ereduction = +0.34V)

For the oxidation half-reaction of zinc, we reverse the reduction reaction:

  • Zn Zn2+ + 2e- (Eoxidation = +0.76V)

Now, calculate the cell potential:

Ecell = Ereduction (Cu2+/Cu) + Eoxidation (Zn/Zn2+)

Ecell = +0.34V + (+0.76V) = +1.10V

Since Ecell is positive, the reaction between zinc and copper ions is spontaneous.

Step 8: Key Points to Remember

When using the standard reduction potential table, remember:

  • Reduction potentials are listed for reduction reactions.
  • Oxidation potentials are the negative of the reduction potentials.
  • A positive cell potential indicates a spontaneous reaction.
  • Hydrogen ions (H+) are defined as 0 volts for reference.
  • Substances with multiple oxidation states will appear multiple times in the table.

FAQs

Here are some frequently asked questions about calculating cell potential in voltaic cells:

1. What is the cell potential formula?

The cell potential formula is E°cell = E°reduction - E°oxidation, where E°cell is the standard cell potential, E°reduction is the standard reduction potential of the reduction half-reaction, and E°oxidation is the standard reduction potential of the oxidation half-reaction.

2. How is the cell potential calculated?

To calculate cell potential, follow these steps:

  1. Identify the reduction and oxidation half-reactions.
  2. Look up the standard reduction potentials for both half-reactions.
  3. Apply the cell potential formula: E°cell = E°reduction - E°oxidation.
  4. Perform the calculation to obtain the standard cell potential.

3. How do you calculate cell potential in AP Chem?

In AP Chemistry, calculate cell potential using the same method described above. Additionally, you may need to use the Nernst equation for non-standard conditions: E = E° - (RT/nF) ln Q, where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is temperature in Kelvin, n is the number of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.

4. How do you measure cell potential?

Cell potential is measured using a voltmeter connected to the electrodes of the electrochemical cell. The voltmeter reads the potential difference between the cathode and anode, giving the cell potential in volts.

5. What is the relationship between cell potential and Gibbs free energy?

The relationship between cell potential and Gibbs free energy is expressed by the equation ΔG = -nFE°cell, where ΔG is the change in Gibbs free energy, n is the number of electrons transferred, F is Faraday's constant, and E°cell is the standard cell potential. A positive cell potential indicates a negative ΔG, meaning the reaction is spontaneous.

Prerequisite Topics

Understanding the concept of calculating cell potential in voltaic cells is a crucial aspect of electrochemistry. While there are no specific prerequisite topics listed for this subject, it's important to recognize that a strong foundation in basic chemistry principles is essential for grasping this more advanced concept.

To fully comprehend the intricacies of calculating cell potential in voltaic cells, students should have a solid understanding of fundamental chemical concepts. These include atomic structure, chemical bonding, and the principles of oxidation and reduction reactions. Familiarity with these topics will provide the necessary groundwork for exploring the more complex aspects of electrochemistry.

Additionally, a good grasp of basic mathematical skills is crucial when dealing with cell potential calculations. Students should be comfortable with algebraic manipulations and unit conversions, as these will be frequently used in determining cell potentials and related quantities.

Understanding the concept of electrochemical cells and their components is also vital. This includes knowledge of electrodes, electrolytes, and the movement of ions within a cell. Familiarity with these elements will help students visualize the processes occurring in voltaic cells and better comprehend the factors influencing cell potential.

Moreover, an introduction to redox reactions and their significance in electrochemistry is essential. Students should be able to identify oxidation and reduction half-reactions, as well as understand the concept of electron transfer between species. This knowledge forms the basis for understanding how voltaic cells generate electrical energy.

Lastly, familiarity with the electrochemical series and standard reduction potentials is crucial for calculating cell potentials. Students should be comfortable using these tools to predict the direction of electron flow and determine the overall cell potential of a voltaic cell.

While there may not be specific prerequisite topics listed, it's clear that a strong foundation in general chemistry, mathematics, and basic electrochemistry concepts is necessary for successfully tackling the subject of calculating cell potential in voltaic cells. By ensuring a solid understanding of these fundamental areas, students will be better equipped to explore the intricacies of electrochemical calculations and their practical applications in various fields of science and technology.

In this lesson, we will learn:

  • To understand the use of standard potentials in redox reactions.
  • How to predict the spontaneity of a redox reaction using a standard potential table.
  • How to calculate cell potential to prove the spontaneity of a redox reaction.
  • How the cell potential relates to Gibbs free energy.
  • To calculate the cell potential at non-standard conditions with the Nernst equation.

Notes:

  • We learned in Introduction to electrochemistry that electrochemical cells use redox reactions to convert chemical potential energy into electrical energy. The amount of energy released depends on what is reacting.
    Because of the energy involved when reduction or oxidation takes place, some pairs of substances will react in a redox process spontaneously!
    To find out if a redox reaction will be spontaneous or not, we need to use the table of Standard Reduction Potentials (or just ‘standard potentials’). Here are some key points about it:
    • The table is a list of substances with the amount of energy (in volts) they release in a reduction half-equation. This means a few things:
      • The substances near the top of the table are excellent oxidizing agents, so they are likely to get reduced in a redox reaction.
      • The substances near the bottom of the table are poor oxidizing agents. On the flip side, these substances are good reducing agents, which means they release lots of energy when they get oxidized themselves. The table always shows reduction reading left to right though, so for oxidation reactions, read the equation right to left.
    • Hydrogen ions being reduced to H2 is defined at zero volts. This is why some of the equations have negative values; it just means the substances are less oxidizing than H + ions.
    • Because some substances (e.g. transition metals) have multiple oxidation states, they will appear multiple times in the table. These are showing different reactions (e.g. an oxidation or reduction from two different states).

  • Because redox reactions involve one reduction and one oxidation (NO EXCEPTIONS), a redox reaction will be built from two half-equations with one from the table being read left to right (reduction) and one going from right to left (oxidation). These two half-cells would be joined together, the half-equation higher on the table will be reduced and the lower half-equation will be oxidized.

  • Whether a redox reaction is spontaneous can also be found mathematically using the values in the table. Using the E0red values, the reaction is spontaneous if E0cell is a positive value. E0cell is worked out by:

    E0cell = E0red (reduction) + E0ox (oxidation)

    Remember, the table ONLY gives E0red values! Find E0red and reverse the sign to get E0ox.


    WORKED EXAMPLE 1:1

    Will a redox reaction between Bi3+ and In occur spontaneously?
    First, use the table to find an equation with these compounds and their E0red. Below are all the half-equations equations containing the substances in the question, taken from a standard potential table1:

    Bi3+ + 3e- ⇌\enspace \rightleftharpoons \enspace BiE0red = +0.20 V
    In+ + e- ⇌\enspace \rightleftharpoons \enspace InE0red = -0.14 V
    In3+ + 3e- ⇌\enspace \rightleftharpoons \enspace InE0red = -0.34 V

    The second step is to see if one substance can be reduced (left side of the reaction arrow) and if the other can be oxidized (right side of the reaction arrow). For a redox to occur we need both!
    In reading the data, we see:
    • Bi3+ can be reduced into Bi (Bi3+ is on the left of an equation).
    • In+ can be reduced into In, which means In can be oxidized into In+ because In is on the right side of the equation.
    • The reduction of In3+ into In, which means In can be oxidized into In3+.
    We can also see, reading the table from top to bottom (highest E0cell to lowest) that Bi3+ is higher on the table than In. Therefore, the reduction of Bi3+ with the oxidation of In is a spontaneous reaction.

    To find this mathematically, we use the electrode potential expression:

    E0cell = E0red (reduction) + E0ox (oxidation)

    The E0cell must be a positive value for the redox reaction to be spontaneous. We have two possible reactions:
    • Reaction 1: The reduction of Bi3+ to Bi (E0red = +0.20 V) with the oxidation of In to In+ (E0red = -0.14 V)
    • Reaction 2: The reduction of Bi3+ to Bi (E0red +0.20 V) with the oxidation of In to In3+ (E0red = -0.34 V)

    For the oxidation of indium we need to reverse the sign to get E0ox instead of E0red. The calculations are:
    • Reaction 1: E0cell = +0.20 + +0.14 V = +0.34 V Therefore the reaction with Bi3+ and In is spontaneous.
    • Reaction 2: E0cell = +0.20 V + +0.34 V = +0.54V Therefore the reaction with Bi3+ and In is spontaneous.


    WORKED EXAMPLE 2:1

    Will the oxidation of copper metal (Cu) by zinc ions (Zn2+) happen spontaneously in a redox reaction?

    Use the table1 to find equations with these substances in them:
    Cu+ + e- ⇌\enspace \rightleftharpoons \enspace Cu E0red = +0.52 V
    Cu2+ + 2e- ⇌\enspace \rightleftharpoons \enspace Cu E0red = +0.34 V
    Zn2+ + 2e- ⇌\enspace \rightleftharpoons \enspace Zn E0red = -0.76 V

    Reading the data we see a possibility for two reactions:
    • Reaction 1: The Cu / Cu+ equation is higher up the table than Zn/Zn2+. This means the reduction potential is higher for copper than it is for zinc. Therefore, oxidation of zinc (with the reduction of copper ions) would occur spontaneously but oxidation of copper metal would not occur spontaneously.
    • Reaction 2: The Cu / Cu2+ equation is higher on the table than Zn/Zn2+. Therefore like reaction 1 above the oxidation of copper would not occur spontaneously.

    Mathematically we can calculate the cell potential, E0cell:

    E0cell = E0red (reduction) + E0ox (oxidation)

    For “the oxidation of copper metal (Cu) by zinc ions (Zn2+)” in the question, zinc will be reduced (use the E0red value above) and copper will be oxidized (reverse the sign in the E0red above to get E0ox) so the calculations would be:

    E0cell = E0red (reduction) + E0ox (oxidation)

    Reaction 1: E0cell = -0.76 V + +0.52 V = -0.24 V Therefore the reaction with Cu and Zn2+ is not spontaneous.

    Reaction 2: E0cell = -0.76 V + +0.34 V = -0.42 V Therefore the reaction with Cu and Zn2+ is not spontaneous.

  • We learned in Entropy and Gibbs free energy that ΔG\Delta G is a ‘driving force’ for a reaction. It is a finite source of potential energy that drives a reaction towards equilibrium when ΔG\Delta G = 0. In other words, ΔG\Delta G pushes a reaction until there is an equal force for the forward and the reverse reaction.
    We saw the equation that relates standard Gibbs free energy change to a reaction’s equilibrium constant, K:

    • ΔG=−RTIn (K)\Delta G = - RTIn \, (K)

    Where:
    • R = universal gas constant (8.314 J K-1 mol-1)
    • T = temperature (K)
    • K = the equilibrium constant

    There is a similar equation for cell potential and its relation to the standard Gibbs free energy change:

    ΔG0=−nFE0\Delta G^{0} = - nFE^{0}

    Where:
    • n = the number of moles of electrons being transferred in the reaction.
    • F = the Faraday constant, which is 96485 C mol-1 (Coulombs per mole). This is the amount of charge a mole of electrons carries.
    • E0 = the standard cell potential. This is in volts or J C-1 so the units will cancel to joules.

    Do you notice the similarities? Both are expressions of ΔG\Delta G for mixtures at standard conditions showing which direction the reaction favours, in order to reach equilibrium. The only real difference is that the first is for for gases and the second is for aqueous species in a redox cell.
    • A negative E0 yields a positive ΔG0\Delta G^{0}, so the mixture will form more reactants in order to reach equilibrium, which has a greater quantity of reactants than products.
    • A positive E0 yields a negative ΔG0\Delta G^{0}, so the mixture will form more products in order to reach equilibrium, which has a greater quantity of products than reactants.

    You can therefore treat E0 like ΔG\Delta G: a driving force of a reaction towards equilibrium. To avoid having to use the constants, you can equate the two expressions above:

    −RTln (K)=−nFE0-RTln \, (K) = -nFE^{0}

    Let’s solve for E0cell using this:

    RTnF\large \frac{RT} {nF}In (K)=E0In \, (K) = E^{0}


    If you are at room temperature of 298 K, this expression for RT/F cancels out as:

    (8.314JK mol) (298 K)n (96485Cmol)\large \frac{(8.314 \frac{J} {K \, mol} ) \, (298 \, K )} {n \, (96485 \frac{C} {mol}) } In (K) In \, (K) \, →  \, (8.314 J) (298)n (96485 C)\large \frac{(8.314 \, J) \, (298) } {n \, (96485 \, C) } In (K)=In \, (K) = 0.0257JCn\large \frac{0.0257 \frac{J} {C} } {n} In (K)=E0In \, (K) = E^{0}

    Recall that joules per coulomb is a volt, so:

    E0=E^{0} = 0.0257 Vn\large \frac{0.0257 \, V } {n }In (K)In \, (K)

    This can be used to find KK if you are given the standard cell potential.

  • Like with gas equilibria, we are normally at nonstandard conditions throughout the redox reaction!
    • In redox cells, standard conditions means there is an equal quantity of products and reactants, usually 1M concentration of both. These are the conditions for which we calculate E0E^{0}.
      Redox reactions will only be in this state briefly, probably at the beginning of the reaction when the solutions are prepared.

    Concentrations are changing over time as the system approaches equilibrium, so ΔG\Delta G and the cell potential are also both changing. We can use the following equation when not at standard conditions to show how cell potential changes as we approach equilibrium:

    ΔG=−nFE\Delta G = - nFE

    Consider the expression we used in Entropy and Gibbs free energy:

    ΔG=ΔG0+RTIn (Q)\Delta G = \Delta G^{0} + RTIn \, (Q)

    Because ΔG0=−nFE0\Delta G^{0}= -nFE^{0} and ΔG=−nFE \Delta G = -nFE, we can create the expression:

    −nFE=−nFE0+RTln (Q)-nFE = -nFE^{0} + RTln \, (Q)

    Dividing both sides of this expression by −nF-nF to yield:

    E=E0−RTnFln (Q)E = E^{0} - \frac{RT}{nF} ln \, (Q)

    This is the Nernst equation used to predict cell potential at nonstandard conditions.
    As shown above, if you are at 298K you can simplify it:

    E=E0+E = E^{0} + (8.314) (298)−n (96485)\large \frac{(8.314) \, (298) } {-n \, (96485) } In (Q) In \, (Q) \, → E=E0− \, E = E^{0} - 0.0257n\large \frac{0.0257} {n} In (Q)In \, (Q)

    This is often derived further using the change of base rule (log10Q= (log_{10} Q = ln(Q)ln(10))\large \frac{ln(Q)} {ln(10)}) to convert ln(Q)ln(Q) into log(Q)log(Q):

    log10Q=log_{10} Q = ln(Q)ln(10)\large \frac{ln(Q)} {ln(10)}  \, → E=E0− \, E = E^{0} - 0.0257 ∗ In (10)n \large \frac{0.0257 \, * \, In \, (10)} {n} log (Q) log \, (Q) \, → E=E0− \, E = E^{0} - 0.0592n\large \frac{0.0592} {n} log (Q)log \, (Q)

    This is the most common form of the Nernst equation.



1 Source for standard potential data: ATKINS, P. W., & DE PAULA, J. (2006).?Atkins' Physical chemistry. Oxford, Oxford University Press.