Easy Nernst Equation Practice Questions

Concept Explanation

The Nernst Equation is a mathematical relationship used to calculate the electrical potential of an electrochemical cell at any given temperature, pressure, and concentration of reactants and products. While standard cell potentials are measured under specific conditions (1 M concentration, 1 atm pressure, and 25°C), the Nernst Equation allows chemists to determine the voltage of a cell when these variables change. This is critical for understanding redox reaction practice questions and biological processes like nerve impulse transmission. The most common form of the equation at 25°C (298.15 K) is:

E = E° - (0.0592 / n) log Q

In this equation, E represents the cell potential under non-standard conditions, E° is the standard cell potential, n is the number of moles of electrons transferred in the reaction, and Q is the reaction quotient. The reaction quotient is calculated by dividing the concentration of products by the concentration of reactants, each raised to the power of their stoichiometric coefficients. Pure solids and liquids are excluded from Q, as their activities are defined as 1. For a deeper look at the chemistry behind these potentials, you can refer to the IUPAC Gold Book definition of the Nernst Equation.

Solved Examples

The following examples demonstrate how to apply the Nernst Equation to find cell potentials under non-standard concentrations.

1. Example 1: Calculating Potential for a Zinc-Copper Cell
   Calculate the cell potential for a Zn/Cu cell where [Zn²⁺] = 0.10 M and [Cu²⁺] = 2.0 M. The standard cell potential (E°) is 1.10 V.

   1. Identify the number of electrons transferred (n). For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2.

   2. Calculate the reaction quotient (Q). Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 2.0 = 0.05.

   3. Apply the Nernst Equation: E = 1.10 - (0.0592 / 2) log(0.05).

   4. Solve: E = 1.10 - (0.0296)(-1.301) = 1.10 + 0.0385 = 1.1385 V.

2. Example 2: Determining Potential with Different Concentrations
   Find the potential of a Silver electrode in a 0.01 M AgNO₃ solution at 25°C. E° for Ag⁺/Ag is 0.80 V.

   1. The half-reaction is Ag⁺ + e⁻ → Ag, so n = 1.

   2. The reaction quotient Q = 1 / [Ag⁺] = 1 / 0.01 = 100.

   3. Apply the formula: E = 0.80 - (0.0592 / 1) log(100).

   4. Solve: E = 0.80 - (0.0592)(2) = 0.80 - 0.1184 = 0.6816 V.

3. Example 3: Concentration Cell Potential
   Calculate the E for a concentration cell with two Hydrogen electrodes. One is at pH 1.0 ([H⁺] = 0.1 M) and the other at pH 3.0 ([H⁺] = 0.001 M). E° for a concentration cell is always 0 V.

   1. The reaction is: 2H⁺ (0.1 M) → 2H⁺ (0.001 M). Here n = 2.

   2. Q = [H⁺_dilute]² / [H⁺_concentrated]² = (0.001)² / (0.1)² = 10⁻⁶ / 10⁻² = 10⁻⁴.

   3. E = 0 - (0.0592 / 2) log(10⁻⁴).

   4. E = -0.0296 * (-4) = 0.1184 V.

Practice Questions

Work through these easy Nernst Equation practice questions to build your confidence in electrochemistry calculations.

1. A galvanic cell uses the reaction: Fe(s) + Cd²⁺(aq) → Fe²⁺(aq) + Cd(s). If E° = 0.04 V, [Fe²⁺] = 0.01 M, and [Cd²⁺] = 1.0 M, what is the cell potential?

2. Calculate the reduction potential of a Hydrogen electrode at 25°C where the pressure of H₂ is 1 atm and the concentration of H⁺ is 0.5 M. (Standard reduction potential for Hydrogen is 0.00 V).

3. Determine the cell potential for the reaction: Mg(s) + Sn²⁺(aq) → Mg²⁺(aq) + Sn(s) when [Mg²⁺] = 0.05 M and [Sn²⁺] = 0.25 M. Given E° = 2.23 V.

4. Find the value of E for a Lead-Lead cell (Pb/Pb²⁺) where the anode concentration is 0.001 M and the cathode concentration is 0.1 M.

5. What is the cell potential of a Daniel cell (Zn/Cu) if [Zn²⁺] = 2.0 M and [Cu²⁺] = 0.02 M? Use E° = 1.10 V.

6. Calculate the EMF of the following cell: Ni(s) | Ni²⁺ (0.1 M) || Ag⁺ (0.01 M) | Ag(s). Given E°(Ni²⁺/Ni) = -0.25 V and E°(Ag⁺/Ag) = 0.80 V.

7. For the reaction 2Al(s) + 3Mn²⁺(aq) → 2Al³⁺(aq) + 3Mn(s), E° = 0.48 V. Calculate E when [Al³⁺] = 1.5 M and [Mn²⁺] = 0.5 M.

8. Calculate the potential of a Platinum electrode in a solution containing 0.1 M Fe²⁺ and 0.01 M Fe³⁺. E°(Fe³⁺/Fe²⁺) = 0.77 V.

9. A concentration cell consists of two Copper electrodes. The solutions are 0.05 M Cu²⁺ and 0.5 M Cu²⁺. Calculate the cell potential at 25°C.

10. Determine the pH of a solution if the potential of a hydrogen electrode is -0.177 V at 25°C (P = 1 atm).

Answers & Explanations

Review the step-by-step solutions below to check your work on these cell potential calculations practice questions.

1. Answer: 0.099 V.
   n = 2. Q = [Fe²⁺]/[Cd²⁺] = 0.01/1.0 = 0.01.
   E = 0.04 - (0.0592/2) log(0.01) = 0.04 - (0.0296)(-2) = 0.04 + 0.0592 = 0.0992 V.

2. Answer: -0.018 V.
   Reaction: 2H⁺ + 2e⁻ → H₂. n = 2. Q = P(H₂)/[H⁺]² = 1 / (0.5)² = 4.
   E = 0.00 - (0.0592/2) log(4) = -0.0296 * 0.602 = -0.0178 V.

3. Answer: 2.25 V.
   n = 2. Q = 0.05 / 0.25 = 0.2.
   E = 2.23 - (0.0296) log(0.2) = 2.23 - (0.0296)(-0.699) = 2.23 + 0.0207 = 2.2507 V.

4. Answer: 0.059 V.
   Concentration cell, E° = 0. n = 2. Q = [dilute]/[conc] = 0.001/0.1 = 0.01.
   E = 0 - (0.0296) log(0.01) = -0.0296 * (-2) = 0.0592 V.

5. Answer: 1.041 V.
   n = 2. Q = 2.0 / 0.02 = 100.
   E = 1.10 - (0.0296) log(100) = 1.10 - (0.0296 * 2) = 1.10 - 0.0592 = 1.0408 V.

6. Answer: 0.961 V.
   E°_cell = 0.80 - (-0.25) = 1.05 V. Reaction: Ni + 2Ag⁺ → Ni²⁺ + 2Ag. n = 2.
   Q = [Ni²⁺]/[Ag⁺]² = 0.1 / (0.01)² = 0.1 / 0.0001 = 1000.
   E = 1.05 - (0.0296) log(1000) = 1.05 - (0.0296 * 3) = 1.05 - 0.0888 = 0.9612 V.

7. Answer: 0.467 V.
   n = 6. Q = [Al³⁺]² / [Mn²⁺]³ = (1.5)² / (0.5)³ = 2.25 / 0.125 = 18.
   E = 0.48 - (0.0592/6) log(18) = 0.48 - 0.00987 * 1.255 = 0.48 - 0.0124 = 0.4676 V.

8. Answer: 0.711 V.
   Reaction: Fe³⁺ + e⁻ → Fe²⁺. n = 1. Q = [Fe²⁺]/[Fe³⁺] = 0.1 / 0.01 = 10.
   E = 0.77 - (0.0592/1) log(10) = 0.77 - 0.0592 = 0.7108 V.

9. Answer: 0.0296 V.
   E° = 0. n = 2. Q = 0.05 / 0.5 = 0.1.
   E = 0 - (0.0296) log(0.1) = -0.0296 * (-1) = 0.0296 V.

10. Answer: pH 3.0.
    E = -0.0592 * pH. -0.177 = -0.0592 * pH.
    pH = -0.177 / -0.0592 = 2.99 ≈ 3.0.

Quick Quiz

Frequently Asked Questions

What is the Nernst Equation used for?

The Nernst Equation is used to calculate the cell potential of an electrochemical cell under non-standard conditions, such as varying concentrations or temperatures. It bridges the gap between theoretical standard potentials and real-world chemical environments.

Why is 0.0592 used in the Nernst Equation?

The value 0.0592 is a constant derived from (RT/F) multiplied by the natural log to base-10 conversion factor (2.303) at 25°C. It simplifies calculations by combining the gas constant, temperature, and Faraday's constant into one number.

How do you find 'n' for the Nernst Equation?

You find 'n' by looking at the balanced half-reactions of the redox process and identifying the total number of electrons exchanged. For more practice on this, see our guide on balancing redox practice questions.

Can the Nernst Equation be used for gases?

Yes, the Nernst Equation can be used for gases by substituting the partial pressure of the gas into the reaction quotient (Q) in place of concentration. This is common in hydrogen electrode calculations where pressure is measured in atmospheres.

What does a negative cell potential indicate?

A negative cell potential calculated via the Nernst Equation indicates that the reaction is non-spontaneous in the forward direction under those specific conditions. This means the reverse reaction is actually the spontaneous one.

How does temperature affect the Nernst Equation?

Temperature is directly proportional to the slope of the potential change in the Nernst Equation; as temperature increases, the effect of concentration deviations from standard states becomes more pronounced. For most classroom problems, 298K is assumed unless stated otherwise.

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