2.5.1 Electrode potentials and concentration

In the previous section, we learned the Nernst equation, which allows us to calculate the electrode potential under non-standard conditions. This section explores the physical meaning behind the equation: how and why changes in the concentration of ions affect the electrode potential.

Main Idea

The electrode potential measures how easily a species gains or loses electrons. It depends on the concentration of the ions in the redox reaction. When the concentration of ions changes, the position of equilibrium shifts according to Le Chatelier's Principle, which in turn changes the electrode potential. The Nernst equation quantitatively predicts these changes.

Core Concept

The core concept is the dynamic equilibrium of a half-cell. Consider the reduction half-reaction: Cu^{2+}_{(aq)} + 2e^- \rightleftharpoons Cu_{(s)} The standard electrode potential (E^°) is measured when the concentration of Cu^{2+} is exactly 1 mol dm^-3.

• If you increase the concentration of Cu^{2+} (by adding more copper ions), the equilibrium shifts to the right (towards the product) to counteract the change. This makes reduction easier because more ions are available to gain electrons, which increases the electrode potential.
• If you decrease the concentration of Cu^{2+} (by adding water), the equilibrium shifts to the left. This makes reduction harder because fewer ions are available to accept electrons, which decreases the electrode potential.

The same logic applies to any half-cell: increasing the concentration of the oxidized form increases the potential, while decreasing it lowers the potential.

Detailed Explanation

The textbook provides a step-by-step example to show how the Nernst equation is used to calculate the potential when the concentration is not standard.

Example: The Cu^{2+}/Cu System

Question: The standard electrode potential of the Cu^{2+}/Cu system is +0.34 V. What is the electrode potential of a solution containing 0.5 M Cu^{2+} ions?

Solution:

1. Write the Nernst equation for the half-cell: E = E^° + 0.059n log [oxidized species][reduced species] For the Cu^{2+}/Cu system, the half-reaction is: Cu^{2+} + 2e^- \rightarrow Cu So, n = 2. The oxidized species is Cu^{2+}, and the reduced species is Cu (a pure solid). The concentration of a pure solid is taken as 1. E_Cu^{2+}/Cu = E^°_Cu^{2+}/Cu + 0.0592 log [Cu^{2+}]1

• E^° = +0.34 V
• n = 2
• [Cu^{2+}] = 0.5 M

3. Calculate: E = 0.34 + 0.0295 × (-0.301) E = 0.34 - 0.0089 E = 0.331 V

Result: The electrode potential is 0.331 V. This is slightly lower than the standard potential of 0.34 V because the concentration of Cu^{2+} is less than 1 M, making reduction slightly less favorable.

Key Takeaways from the Example

• When the concentration of the oxidized species is less than 1 M, the potential is lower than the standard potential.
• When the concentration of the oxidized species is greater than 1 M, the potential is higher than the standard potential.
• The Nernst equation allows you to calculate exactly how much the potential changes.

Intermediate Level

The Nernst equation shows a logarithmic relationship between potential and concentration. This means that a ten-fold change in concentration changes the potential by a fixed amount (for a given n). For example, if you decrease the concentration of the oxidized species by a factor of 10, the log term decreases by 1, and the potential changes by 0.059n volts. The sign of the change depends on whether the concentration is above or below 1 M. The equation also shows that the effect of concentration is more pronounced for half-reactions with smaller n, since a smaller n makes the fraction 0.059n larger.

Important Points

• - Electrode potential depends on concentration.
• - Standard potential (E^°) is measured at 1 M concentration.
• - Increasing concentration of oxidized species → increases potential.
• - Decreasing concentration of oxidized species → decreases potential.
• - Nernst equation quantifies this relationship.
• - Pure solids/liquids have a concentration of 1.
• - The factor 0.059n at 25^°C shows the sensitivity of the potential to concentration.

Remember

• - Nernst equation: E = E^° + 0.059n log [Ox][Red] (at 25^°C).
• - Higher [Ox] → higher potential.
• - Lower [Ox] → lower potential.
• - **Solids/liquids = 1**.
• - n = number of electrons.
• - A 10-fold change in concentration changes the potential by 0.059n V.

Common Mistakes

• - Forgetting to include the concentration of the reduced species. If the reduced species is an ion in solution (e.g., Fe^{3+} + e^- \rightarrow Fe^{2+}), both concentrations must be included in the ratio.
• - Using the wrong n value. Always check the balanced half-reaction for the number of electrons transferred.
• - Including pure solids or liquids in the concentration ratio. They are assigned a value of 1 and do not appear in the ratio.
• - Misinterpreting the sign. If the log term is negative, the potential is lower than E^°; if positive, it is higher.

Quick Revision

The electrode potential of a half-cell depends on the concentration of the ions involved. According to Le Chatelier's Principle, increasing the concentration of the oxidized species makes reduction easier and increases the potential, while decreasing it makes reduction harder and lowers the potential. The Nernst equation quantitatively predicts this: E = E^° + 0.059n log [Ox][Red]. Pure solids and liquids are assigned a concentration of 1. The textbook example shows that for the Cu^{2+}/Cu half-cell, decreasing the concentration from 1 M to 0.5 M lowers the potential from 0.34 V to 0.331 V.

60-Second Revision

Electrode potential depends on ion concentration. The Nernst equation at 25^°C is E = E^° + 0.059n log [Ox][Red]. Higher concentration of the oxidized species increases the potential; lower concentration decreases it. Pure solids and liquids have a concentration of 1. The effect is logarithmic: a 10-fold change in concentration changes the potential by 0.059n V.

Frequently Asked Questions (FAQs)

Q1. How does the concentration of ions affect electrode potential? Answer: The electrode potential depends on the concentration of the ions in the half-cell. According to Le Chatelier's Principle, if you increase the concentration of the oxidized species (the reactant in reduction), the equilibrium shifts to the right, making reduction easier and increasing the electrode potential. Conversely, decreasing the concentration shifts the equilibrium to the left, making reduction harder and decreasing the potential.

• E = actual electrode potential
• E^° = standard electrode potential (+0.34 V)
• 0.059 = constant at 25°C
• 2 = number of electrons transferred (n)
• [Cu^{2+}] = concentration of the oxidized species (since the reduced species, Cu, is a solid and its concentration is 1)

• Q3. Calculate the electrode potential of a Cu^{2+}/Cu half-cell with a concentration of 0.5 M Cu^{2+}.
• Answer: Using the Nernst equation:
• E = 0.34 + 0.0592 log(0.5)
• E = 0.34 + 0.0295 × (-0.301)
• E = 0.34 - 0.0089 = 0.331 V

Q4. Why is the concentration of a pure solid or liquid taken as 1 in the Nernst equation? Answer: The concentration (or more precisely, the activity) of a pure solid or pure liquid is constant and does not depend on the amount of the substance. Therefore, its effective concentration is taken as 1, and it does not appear in the reaction quotient.

Q5. What is the effect of a ten-fold increase in the concentration of the oxidized species on the electrode potential? Answer: A ten-fold increase in the concentration of the oxidized species changes the log term by +1. The potential increases by 0.059n volts. For example, if n=2, the potential increases by approximately 0.0295 V.

Q6. How does the electrode potential change if the concentration of the reduced species increases? Answer: If the concentration of the reduced species (when it is an ion in solution) increases, the ratio [Ox][Red] decreases. The log term becomes more negative, and the electrode potential decreases.

Q7. What is a concentration cell? Answer: A concentration cell is a galvanic cell in which both electrodes are made of the same material and are immersed in solutions of the same ions, but at different concentrations. The cell potential arises solely from the difference in concentration, as described by the Nernst equation. The half-cell with the lower concentration acts as the anode (oxidation), and the one with the higher concentration acts as the cathode (reduction).

Q8. How does the Nernst equation relate to Le Chatelier's Principle? Answer: The Nernst equation is a quantitative expression of Le Chatelier's Principle in electrochemistry. A change in the concentration of a species shifts the equilibrium position of the half-reaction, which changes the electrode potential. The Nernst equation quantifies this change.

Multiple Choice Questions (MCQs)

• Q1. How does the electrode potential change if the concentration of the oxidized species increases?
• A. It decreases.
• B. It increases.
• C. It remains the same.
• D. It becomes zero.
• Correct Answer: B
• Explanation: Increasing the concentration of the oxidized species makes reduction easier, which increases the electrode potential.

• Q2. In the Nernst equation, what is the concentration of a pure solid taken as?
• A. 0
• B. 1
• C. Its molarity
• D. Its density
• Correct Answer: B
• Explanation: The concentration of a pure solid is constant and is assigned a value of 1 in the Nernst equation.

• Q3. For the Cu^{2+}/Cu half-cell, what is the value of n in the Nernst equation?
• A. 1
• B. 2
• C. 3
• D. 4
• Correct Answer: B
• Explanation: The balanced half-reaction is Cu^{2+} + 2e^- \rightarrow Cu, so n = 2.

• Q4. If the concentration of Cu^{2+} is decreased from 1 M to 0.5 M, what happens to the electrode potential?
• A. It increases.
• B. It decreases.
• C. It remains the same.
• D. It becomes zero.
• Correct Answer: B
• Explanation: Decreasing the concentration of the oxidized species makes reduction harder, so the potential decreases.

Q5. What is the Nernst equation for the Zn^{2+}/Zn half-cell? A. E = E^° + 0.0592 log [Zn^{2+}] B. E = E^° - 0.0592 log [Zn^{2+}] C. E = E^° + 0.0591 log [Zn^{2+}] D. E = E^° + 0.0592 ln [Zn^{2+}] Correct Answer: A Explanation: For the reduction Zn^{2+} + 2e^- \rightarrow Zn, n=2, and the concentration of the reduced species (Zn) is 1, so the equation is E = E^° + 0.0592 log [Zn^{2+}].

• Q6. What happens to the electrode potential if the concentration of the oxidized species is less than 1 M?
• A. It is higher than the standard potential.
• B. It is lower than the standard potential.
• C. It is equal to the standard potential.
• D. It becomes zero.
• Correct Answer: B
• Explanation: If the concentration is less than 1 M, the log term is negative, so the potential is lower than the standard potential.

• Q7. In the Nernst equation, the term [reduced species] for the half-reaction Fe^{3+} + e^- \rightarrow Fe^{2+} is the concentration of:
• A. Fe^{3+}
• B. Fe^{2+}
• C. Fe
• D. Electrons
• Correct Answer: B
• Explanation: The reduced species is the product of the reduction half-reaction, which is Fe^{2+}.

Q8. What is the relationship between the Nernst equation and Le Chatelier's Principle? A. The Nernst equation is a mathematical expression of Le Chatelier's Principle for electrochemical cells. B. They are unrelated. C. The Nernst equation contradicts Le Chatelier's Principle. D. The Nernst equation is only valid at equilibrium. Correct Answer: A Explanation: The Nernst equation shows how changes in concentration shift the equilibrium and thus change the potential, which is a direct application of Le Chatelier's Principle.

Q9. If the standard electrode potential of a half-cell is +0.34 V, and the concentration of the oxidized species is 0.1 M (with n=2), what is the new potential? (Use 0.059 for the constant). A. 0.31 V B. 0.34 V C. 0.37 V D. 0.40 V Correct Answer: A Explanation: E = 0.34 + 0.0592 log(0.1) = 0.34 + 0.0295(-1) = 0.34 - 0.0295 = 0.3105 V, which is approximately 0.31 V.

• Q10. A ten-fold increase in the concentration of the oxidized species will change the potential by:
• A. 0.059n volts
• B. -0.059n volts
• C. 0.059 volts
• D. n × 0.059 volts
• Correct Answer: A
• Explanation: A ten-fold increase makes the log term +1, so the potential increases by 0.059n volts.

Multiple Choice Questions (MCQs)

• Q1. What does the Nernst equation allow us to calculate?
• A. The standard potential of a cell.
• B. The potential of a cell under non-standard conditions.
• C. The concentration of a solution.
• D. The number of electrons transferred.
• Correct Answer: B
• Explanation: The Nernst equation relates the standard potential to the actual potential under non-standard conditions, accounting for concentration changes.

• Q2. In the Nernst equation for a half-cell, what is the concentration of a pure solid electrode taken as?
• A. 0
• B. 1
• C. Its molar mass
• D. Its density
• Correct Answer: B
• Explanation: The concentration (activity) of a pure solid is constant and is assigned a value of 1.

• Q3. If the concentration of the oxidized species in a half-cell increases, what happens to the electrode potential?
• A. It decreases.
• B. It increases.
• C. It remains unchanged.
• D. It becomes zero.
• Correct Answer: B
• Explanation: Increasing the concentration of the oxidized species makes reduction easier, increasing the potential.

• Q4. What is the value of n for the half-reaction Fe^{3+} + e^- \rightarrow Fe^{2+}?
• A. 1
• B. 2
• C. 3
• D. 4
• Correct Answer: A
• Explanation: One electron is transferred in this half-reaction, so n = 1.

• Q5. In the Nernst equation, the term 0.059n is used at what temperature?
• A. 0°C
• B. 25°C
• C. 100°C
• D. 273°C
• Correct Answer: B
• Explanation: The constant 0.059 is derived at 25°C (298 K).

• Q6. If the concentration of the oxidized species is less than 1 M, what is the sign of the log term in the Nernst equation?
• A. Positive
• B. Negative
• C. Zero
• D. It depends on the reduced species
• Correct Answer: B
• Explanation: The logarithm of a number less than 1 is negative, so the term is negative, lowering the potential.

Q7. What is the Nernst equation for the Ag^+/Ag half-cell? A. E = E^° + 0.0591 log [Ag^+] B. E = E^° - 0.0591 log [Ag^+] C. E = E^° + 0.0592 log [Ag^+] D. E = E^° + 0.059 ln [Ag^+] Correct Answer: A Explanation: For the half-reaction Ag^+ + e^- \rightarrow Ag, n=1, and the concentration of the reduced species (Ag) is 1, so the equation is E = E^° + 0.0591 log [Ag^+].

Q8. A concentration cell is a cell in which: A. Both electrodes are different and have different concentrations. B. Both electrodes are the same but immersed in solutions of different concentrations. C. The concentrations are always 1 M. D. The cell potential is always zero. Correct Answer: B Explanation: A concentration cell uses identical electrodes in solutions of the same ion at different concentrations, and the potential arises from the concentration difference.

Q9. If the standard potential of a half-cell is +0.80 V, and the concentration of the oxidized species is 0.01 M (n=1), what is the new potential? (Use 0.059). A. 0.80 V B. 0.74 V C. 0.86 V D. 0.68 V Correct Answer: B Explanation: E = 0.80 + 0.0591 log(0.01) = 0.80 + 0.059(-2) = 0.80 - 0.118 = 0.682 V, which is approximately 0.68 V. Wait, let's recalculate: 0.80 - 0.118 = 0.682 V. The closest option is 0.68 V. However, the textbook uses 0.059, so 0.059 × (-2) = -0.118. 0.80 - 0.118 = 0.682 V. The option B is 0.74 V, which is incorrect. Let's correct the MCQ: The answer is 0.68 V, but the option D is 0.68 V. So the correct answer is D.

• Q10. What is the effect of a ten-fold decrease in the concentration of the oxidized species on the electrode potential?
• A. It increases by 0.059n V.
• B. It decreases by 0.059n V.
• C. It increases by 2 × 0.059n V.
• D. It remains the same.
• Correct Answer: B
• Explanation: A ten-fold decrease makes the log term -1, so the potential decreases by 0.059n V.