Nernst Equation Practice Questions with Answers

The Nernst equation is a mathematical relationship that connects the reduction potential of an electrochemical reaction to the standard electrode potential, temperature, and activities of the chemical species involved. Mastering this equation is essential for students in general chemistry and biochemistry, as it allows for the calculation of cell voltages under non-standard conditions. Whether you are preparing for a midterm or looking to study for exams to get straight A’s, understanding the nuances of electrochemical potentials is a critical step in your academic journey.

Concept Explanation

The Nernst equation is a fundamental formula in electrochemistry used to calculate the electrical potential of an electrochemical cell at any given temperature, pressure, and reactant concentration. In standard conditions, we use the standard cell potential (E°); however, most real-world applications, such as the biological signals in your nervous system or the discharge of a lithium-ion battery, occur under non-standard conditions. The equation accounts for these variations by incorporating the reaction quotient (Q).

The general form of the Nernst equation at any temperature is:

E = E° - (RT / nF) ln Q

Where:

• E is the cell potential under non-standard conditions (V).

• E° is the standard cell potential (V).

• R is the universal gas constant (8.314 J/(mol·K)).

• T is the absolute temperature (K).

• n is the number of moles of electrons transferred in the balanced redox reaction.

• F is Faraday's constant (96,485 C/mol).

• Q is the reaction quotient, calculated as [Products] / [Reactants].

At the standard biological temperature of 25°C (298.15 K), the equation is often simplified using base-10 logarithms to make calculations faster:

E = E° - (0.0592 / n) log Q

This simplified version is what most students use for chemistry practice problems involving aqueous solutions at room temperature. The relationship demonstrates that as the concentration of reactants increases, the cell potential becomes more positive, whereas an increase in product concentration drives the potential down. This principle is vital for understanding how equilibrium is reached in chemical systems, which you can further explore in our guide on reaction order practice questions.

Solved Examples

To help you grasp the application of these formulas, here are three detailed examples ranging from simple concentration cells to complex redox reactions.

Example 1: Calculating Potential for a Zinc Half-Cell

Calculate the electrode potential (E) for a zinc electrode immersed in a 0.01 M Zn²⁺ solution at 25°C. The standard reduction potential (E°) for Zn²⁺/Zn is -0.76 V.

1. Identify the half-reaction: Zn²⁺(aq) + 2e⁻ → Zn(s). Here, n = 2.

2. Identify the reaction quotient: Q = 1 / [Zn²⁺] because the activity of a solid metal is 1. Q = 1 / 0.01 = 100.

3. Apply the simplified Nernst equation: E = E° - (0.0592 / n) log Q.

4. Substitute values: E = -0.76 - (0.0592 / 2) log(100).

5. Solve: log(100) = 2. E = -0.76 - (0.0296 * 2) = -0.76 - 0.0592 = -0.8192 V.

Example 2: Spontaneous Cell Potential

A galvanic cell consists of a Cu²⁺/Cu electrode ([Cu²⁺] = 2.0 M) and an Ag⁺/Ag electrode ([Ag⁺] = 0.005 M). Calculate the cell potential at 25°C. (E° Cu²⁺/Cu = +0.34 V, E° Ag⁺/Ag = +0.80 V).

1. Determine the standard cell potential: E°cell = E°cathode - E°anode. Silver has the higher reduction potential, so it is the cathode. E°cell = 0.80 - 0.34 = 0.46 V.

2. Write the balanced equation: 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq). Here, n = 2.

3. Calculate Q: Q = [Cu²⁺] / [Ag⁺]² = 2.0 / (0.005)² = 2.0 / 0.000025 = 80,000.

4. Apply Nernst: E = 0.46 - (0.0592 / 2) log(80,000).

5. Solve: log(80,000) ≈ 4.903. E = 0.46 - (0.0296 * 4.903) = 0.46 - 0.145 = 0.315 V.

Example 3: Finding Unknown Concentration

A hydrogen electrode (E° = 0.00 V) is placed in a solution of unknown pH at 25°C. If the measured potential against a standard hydrogen electrode (SHE) is 0.236 V, what is the pH? (Assume P_H2 = 1 atm).

1. Reaction: 2H⁺ + 2e⁻ → H₂(g). n = 2.

2. The SHE is 0.00 V. The measured E is 0.236 V.

3. E = E° - (0.0592 / 2) log(P_H2 / [H⁺]²).

4. 0.236 = 0 - 0.0296 * log(1 / [H⁺]²). Using log properties: log(1/[H⁺]²) = -2 log[H⁺].

5. 0.236 = -0.0296 * (-2 log[H⁺]) = 0.0592 * log[H⁺].

6. Note: pH = -log[H⁺]. Therefore, 0.236 = 0.0592 * (-pH).

7. pH = -0.236 / 0.0592 ≈ 4.0.

Practice Questions

Test your knowledge with these Nernst equation practice questions. They range from basic calculations to conceptual challenges involving temperature changes.

1. Calculate the reduction potential of a Platinum electrode in a solution where [Fe³⁺] = 1.0 M and [Fe²⁺] = 0.1 M at 25°C. (E° Fe³⁺/Fe²⁺ = 0.771 V).

2. A concentration cell is made of two copper electrodes. One is in 0.001 M Cu²⁺ and the other in 1.0 M Cu²⁺. Calculate the cell potential (E) at 298 K.

3. Determine the value of n (number of electrons) for a reaction where E° = 0.20 V and E = 0.14 V at 25°C when Q = 100.

4. Calculate the potential of a Daniell cell (Zn-Cu) at 25°C when [Zn²⁺] = 0.5 M and [Cu²⁺] = 0.02 M. (E°cell = 1.10 V).

5. If the temperature of a cell is increased from 298 K to 350 K, while all concentrations remain 1.0 M, how does the cell potential (E) change? (Explain conceptually).

6. Calculate the equilibrium constant (K) for a reaction at 25°C if the standard cell potential (E°) is 0.45 V and n = 2. (Hint: At equilibrium, E = 0 and Q = K).

7. A lead-acid battery cell has E° = 2.04 V. If the concentration of sulfuric acid drops such that Q = 1.5 × 10³, calculate the new cell potential at 25°C (n = 2).

8. What is the potential of a silver electrode (Ag⁺/Ag, E° = 0.80 V) in a saturated solution of AgCl (Ksp = 1.8 × 10⁻¹⁰) at 25°C?

9. Calculate the E for the reaction: Mg(s) + Sn²⁺(aq, 0.01 M) → Mg²⁺(aq, 1.0 M) + Sn(s). (E° Mg²⁺/Mg = -2.37 V, E° Sn²⁺/Sn = -0.14 V).

10. At what ratio of [Zn²⁺]/[Cu²⁺] will the potential of a Daniell cell be exactly 1.00 V at 25°C? (E°cell = 1.10 V).

Answers & Explanations

Review the solutions below to verify your work and understand the logic behind each calculation.

1. Answer: 0.830 V
Reaction: Fe³⁺ + e⁻ → Fe²⁺ (n=1). E = 0.771 - (0.0592/1) log(0.1 / 1.0). log(0.1) = -1. E = 0.771 - (0.0592 * -1) = 0.771 + 0.0592 = 0.8302 V.

2. Answer: 0.0888 V
For concentration cells, E° = 0. The reaction is Cu²⁺(1.0M) → Cu²⁺(0.001M). n = 2. E = 0 - (0.0592/2) log(0.001 / 1.0). log(10⁻³) = -3. E = -0.0296 * (-3) = 0.0888 V.

3. Answer: n = 2
0.14 = 0.20 - (0.0592 / n) log(100). -0.06 = - (0.0592 / n) * 2. -0.06 = -0.1184 / n. n = 0.1184 / 0.06 ≈ 1.97, which rounds to 2.

4. Answer: 1.059 V
E = 1.10 - (0.0592 / 2) log(0.5 / 0.02). log(25) ≈ 1.398. E = 1.10 - (0.0296 * 1.398) = 1.10 - 0.041 = 1.059 V.

5. Answer: It stays the same (E = E°)
If all concentrations are 1.0 M, then Q = 1. Since log(1) = 0, the term (RT/nF)lnQ becomes zero regardless of the temperature. Thus, E = E°.

6. Answer: 1.6 × 10¹⁵
At equilibrium, E = 0, so E° = (0.0592 / n) log K. 0.45 = (0.0592 / 2) log K. 0.45 = 0.0296 log K. log K = 15.20. K = 10^(15.20) ≈ 1.58 × 10¹⁵.

7. Answer: 1.946 V
E = 2.04 - (0.0592 / 2) log(1.5 × 10³). log(1500) ≈ 3.176. E = 2.04 - (0.0296 * 3.176) = 2.04 - 0.094 = 1.946 V.

8. Answer: 0.215 V
First, find [Ag⁺] from Ksp: [Ag⁺] = sqrt(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M. E = 0.80 - 0.0592 log(1 / 1.34 × 10⁻⁵). log(74626) ≈ 4.87. E = 0.80 - (0.0592 * 4.87) = 0.80 - 0.288 = 0.512 V. (Note: Calculation depends on Q definition; here we assume reduction Ag⁺ + e⁻ → Ag).

9. Answer: 2.17 V
E°cell = -0.14 - (-2.37) = 2.23 V. n = 2. E = 2.23 - (0.0296) log(1.0 / 0.01). log(100) = 2. E = 2.23 - (0.0296 * 2) = 2.23 - 0.0592 = 2.17 V.

10. Answer: 2387:1
1.00 = 1.10 - (0.0296) log Q. -0.10 = -0.0296 log Q. log Q = 3.378. Q = 10^(3.378) ≈ 2387. The ratio [Zn²⁺]/[Cu²⁺] must be 2387.

Quick Quiz

Frequently Asked Questions

What is the physical significance of the Nernst equation?

The Nernst equation quantifies the relationship between chemical energy and electrical energy under non-standard conditions. It essentially describes how the "driving force" of a chemical reaction changes as reactants are consumed and products accumulate.

Why is the number 0.0592 used in the simplified equation?

The value 0.0592 is the result of combining the constants R, T (at 298.15 K), and the conversion factor from natural log (ln) to base-10 log (2.303), all divided by Faraday's constant. It simplifies calculations for problems set at room temperature.

Can the Nernst equation be used for gases?

Yes, for reactions involving gases, the partial pressure of the gas (in atmospheres) is used in the reaction quotient Q instead of molar concentration. This is similar to how you would handle partial pressures in Dalton’s Law practice questions.

What happens to the Nernst equation at equilibrium?

At chemical equilibrium, the cell potential (E) becomes zero because there is no longer a net drive for electrons to flow. In this state, the reaction quotient (Q) is equal to the equilibrium constant (K), allowing the equation to be used to calculate K from E°.

Does temperature affect the cell potential?

Yes, temperature is a direct variable in the Nernst equation (T). An increase in temperature will amplify the effect of the concentration gradient (the log Q term) on the total potential, though the specific direction of change depends on whether Q is greater or less than 1.

How do you determine 'n' in the Nernst equation?

The value of 'n' is determined by balancing the redox half-reactions and finding the total number of electrons exchanged. For example, in a reaction between Zinc (Zn → Zn²⁺ + 2e⁻) and Silver (Ag⁺ + e⁻ → Ag), the silver half-reaction must be multiplied by two, making n = 2.

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