Nernst Equation: Videos & Practice Problems

The Nernst equation connects cell potential (E_cell) to concentrations under non-standard conditions, where concentrations differ from 1 molar. It is expressed as E_cell = E_cell^° - (RT/nF) ln(Q). Here, R is the gas constant, T is temperature, n is the number of electrons transferred, and F is Faraday's constant. As the reaction progresses, the reaction quotient (Q) changes, affecting E_cell until equilibrium is reached, where E_cell equals zero. This relationship also ties to Gibbs free energy (ΔG) and the equilibrium constant (K).

Nernst Equation

The Nernst equation establishes a quantitative relationship between the concentrations of reactants and products in a redox reaction and the cell potential, particularly under non-standard conditions. Standard conditions are defined as having a concentration of 1 molar, a temperature of 25 degrees Celsius, a pH of 7, and a pressure of 1 atmosphere. The equation is expressed as:

$$E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q$$

Here, \(E_{cell}\) represents the cell potential under non-standard conditions, while \(E^{\circ}_{cell}\) is the standard cell potential. The variables in the equation are defined as follows: \(R\) is the gas constant, equal to 8.314 J/(mol·K), \(T\) is the temperature in Kelvin (298.15 K at standard conditions), \(n\) is the number of electrons transferred in the reaction, and \(F\) is Faraday's constant, approximately 96485 C/mol. The term \(Q\) is the reaction quotient, which is the ratio of the concentrations of products to reactants.

When the concentrations of the reactants and products deviate from standard conditions, the Nernst equation allows for the calculation of the cell potential. As the electrochemical cell operates, the concentrations of the electrolyte change, leading to an increase in \(Q\) and a corresponding decrease in \(E_{cell}\) until it reaches zero, indicating that the battery is dead and equilibrium has been achieved.

At equilibrium, the Nernst equation can be modified to relate the standard cell potential to the equilibrium constant \(K\):

$$E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln K$$

When \(E_{cell}\) equals zero, the equation simplifies to:

$$0 = E^{\circ}_{cell} - \frac{RT}{nF} \ln K$$

From this, we can derive that:

$$K = e^{\frac{nE^{\circ}_{cell}}{RT}}$$

Alternatively, using logarithmic functions, the relationship can be expressed as:

$$K = 10^{\frac{nE^{\circ}_{cell}}{0.05916}}$$

Additionally, the connection between Gibbs free energy (\(ΔG\)) and the equilibrium constant can be established. The equation for non-standard conditions is:

$$ΔG = ΔG^{\circ} - RT \ln K$$

At equilibrium, \(ΔG\) equals zero, leading to:

$$0 = ΔG^{\circ} - RT \ln K$$

From which we can derive:

$$K = e^{\frac{ΔG^{\circ}}{RT}}$$

These equations illustrate the intricate relationships between cell potential, equilibrium constants, and Gibbs free energy, emphasizing the importance of the Nernst equation in electrochemistry. Understanding these connections is crucial for analyzing electrochemical systems, particularly when concentrations are not at standard levels.

Nernst Equation Calculations 1

In electrochemistry, understanding concentration cells is crucial for determining cell potentials. A concentration cell consists of two half-cells with different concentrations of the same ion, and it operates based on the principles of oxidation and reduction. The cell notation typically includes the anode, a physical break, and the cathode. In this case, we analyze a silver and copper concentration cell.

At the anode, silver solid (Ag) oxidizes to silver ions (Ag⁺), represented by the half-reaction:

Ag (s) → Ag⁺ (aq) + e^-

Here, the charge changes from 0 to +1, necessitating the addition of one electron to balance the reaction.

At the cathode, copper ions (Cu²⁺) are reduced to solid copper (Cu), represented by the half-reaction:

Cu²⁺ (aq) + 2e^- → Cu (s)

In this case, the charge changes from +2 to 0, requiring the addition of two electrons to balance the reaction. To ensure the number of electrons transferred is equal, the silver half-reaction is multiplied by 2:

2Ag (s) → 2Ag⁺ (aq) + 2e^-

Next, we apply the Nernst equation to calculate the cell potentials under non-standard conditions. The Nernst equation is given by:

E = E_standard - (0.05916/n) log(Q)

For the anode (silver), the standard potential is 0.79 V, and since we have 2 electrons transferred, we calculate:

E_anode = 0.79 V - (0.05916/2) log(Ag⁺²)

Given that the concentration of Ag⁺ is 0.0010 M, we find:

Q = [Ag⁺]² = (0.0010)² = 1.0 x 10^-6

Thus, log(Q) = log(1.0 x 10^-6) = -6, leading to:

E_anode = 0.79 V - (0.05916/2)(-6) = 0.976 V

For the cathode (copper), the standard potential is 0.339 V. The calculation is as follows:

E_cathode = 0.339 V - (0.05916/2) log(1/[Cu²⁺])

With [Cu²⁺] = 1 M, we have:

Q = 1/1 = 1, thus log(1) = 0, leading to:

E_cathode = 0.339 V

To find the overall cell potential, we subtract the anode potential from the cathode potential:

E_cell = E_cathode - E_anode = 0.339 V - 0.976 V = -0.637 V

This negative cell potential indicates that the reaction is non-spontaneous. In practical terms, this means that silver cannot be oxidized by copper ions without an external energy source, such as a battery, which would drive the electrons from the silver to the copper electrode. This scenario exemplifies an electrolytic cell, where external energy is required to facilitate the reaction.

To further your understanding, practice with additional examples by identifying the anode and cathode, setting up the Nernst equation for each half-cell, and determining the overall cell potential to assess spontaneity.

Nernst Equation Calculations 1

To determine the spontaneity and cell potential of an electrochemical cell, we analyze the cell notation and the half-reactions involved. In this case, we have a platinum electrode serving as an inert conductor, with potassium bromide (KBr) in the anode compartment. The bromide ions (Br^-) undergo oxidation to form bromine (Br₂) in the liquid state. The half-reaction for this process is:

Oxidation Half-Reaction:
2 Br^-(aq) → Br₂(l) + 2 e^-

In the cathode compartment, iron(II) ions (Fe²⁺) are reduced to solid iron (Fe). The reduction half-reaction is:

Reduction Half-Reaction:
Fe²⁺ + 2 e^- → Fe(s)

Next, we apply the Nernst equation to calculate the cell potentials. The Nernst equation is given by:

Nernst Equation:
E = E_° - (0.05916 V/n) log(Q)

For the oxidation half-reaction, the standard reduction potential (E_°) is 1.078 V. The reaction involves the transfer of 2 electrons (n = 2), and the reaction quotient (Q) is calculated as:

Q = 1 / [Br^-]² = 1 / (0.01)² = 10000

Substituting these values into the Nernst equation gives:

E_oxidation = 1.078 V - (0.05916 V/2) log(10000) = 1.078 V - 0.05916 V/2 × 4 = 0.960 V

For the reduction half-reaction, the standard reduction potential (E_°) is -0.440 V. The reaction quotient (Q) is:

Q = 1 / [Fe²⁺] = 1 / 1 = 1

Substituting these values gives:

E_reduction = -0.440 V - (0.05916 V/2) log(1) = -0.440 V - 0 = -0.440 V

Now, we can find the overall cell potential (E_cell) by subtracting the oxidation potential from the reduction potential:

E_cell = E_reduction - E_oxidation = -0.440 V - 0.960 V = -1.400 V

Since the overall cell potential is negative, this indicates that the process is non-spontaneous, characterizing it as an electrolytic cell. For a process to be spontaneous, the overall cell potential must be greater than 0, which would indicate a galvanic or voltaic cell.

Nernst Equation Calculations 2

In a standard voltaic cell, the reaction involving 2 moles of hydrogen ions (H⁺) and 1 mole of solid tin (Sn) produces 1 mole of tin(II) ions (Sn²⁺) and hydrogen gas (H₂). The electromotive force (EMF) or cell potential can be influenced by various factors, particularly when operating under non-standard conditions. The Nernst equation is essential for understanding these changes, expressed as:

E = E_° - \(\frac{0.05916}{n}\) log(Q)

Here, E_° is the standard cell potential, n is the number of moles of electrons transferred in the reaction, and Q is the reaction quotient, defined as the ratio of the concentrations of products to reactants. For this reaction, Q can be represented as:

Q = \(\frac{[Sn^{2+}]}{(p_{H_2})/([H^+]^2)}\)

In this expression, solids are omitted, and the pressure of hydrogen gas is considered. Adjusting the pH directly affects the concentration of H⁺ ions, which in turn influences Q and the cell potential. An increase in pH indicates a decrease in H⁺ concentration, leading to an increase in Q and a subsequent decrease in cell potential. Conversely, lowering the pH increases H⁺ concentration, decreasing Q and increasing the cell potential.

Other factors that affect the EMF include:

• Increasing the concentration of Sn²⁺ at the anode raises the numerator in the Q expression, increasing Q and decreasing the cell potential.
• Increasing the pressure of hydrogen gas at the cathode also raises Q, resulting in a decrease in cell potential.

Thus, any action that alters the concentrations of reactants or products, such as changing pH or adjusting gas pressures, will affect the EMF of the cell. Understanding these relationships is crucial for predicting the behavior of electrochemical cells under varying conditions.

Nernst Equation Calculations 2

In electrochemistry, the half-cell potential is a crucial concept that helps us understand the behavior of electrochemical reactions. When considering a half-cell reaction at a standard temperature of 25 degrees Celsius, it is important to recognize how changes to the reaction can affect the half-cell potential, denoted as E.

When a half-cell reaction is reversed, the sign of the half-cell potential also reverses. For instance, if the original half-cell potential is positive, reversing the reaction will yield a negative potential, and vice versa. In this case, the half-cell potential was initially given, and upon reversing the reaction, it became +0.260 volts. This change is significant because it indicates the direction of electron flow in the electrochemical cell.

Additionally, multiplying or dividing the half-cell reaction by any integer does not affect the half-cell potential. This means that while the stoichiometry of the reaction may change, the overall potential remains constant. Therefore, when analyzing half-cell reactions, it is essential to focus on the direction of the reaction and the corresponding sign of the potential, while recognizing that scaling the reaction does not alter the potential value.

In summary, reversing a half-cell reaction changes the sign of the half-cell potential, while multiplying or dividing the reaction by a number does not affect the potential. Understanding these principles is vital for predicting the behavior of electrochemical cells and their potentials.

Here’s what students ask on this topic:

What is the Nernst equation and how is it used in electrochemistry?

The Nernst equation is used to calculate the cell potential (E_cell) of an electrochemical cell under non-standard conditions. It is expressed as:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(Q\right)$

Here, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of electrons transferred, and F is Faraday's constant (96485 C/mol). Q is the reaction quotient. The equation helps in understanding how the cell potential changes with varying concentrations of reactants and products.

How does the Nernst equation relate to Gibbs free energy?

The Nernst equation is closely related to Gibbs free energy (ΔG). The relationship is given by:

$\mathrm{\Delta G}=-\mathrm{nF}{E}_{\mathrm{cell}}$

At equilibrium, ΔG = 0, and the Nernst equation can be rearranged to connect the equilibrium constant (K) with the standard cell potential (E_cell°):

$E_{\mathrm{cell}}=\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(K\right)$

This shows how the cell potential under standard conditions is related to the equilibrium constant and Gibbs free energy.

What are the standard conditions in the Nernst equation?

Standard conditions in the Nernst equation refer to a set of specific conditions under which the cell potential is measured. These include:

• Concentration of 1 M for all reactants and products
• Temperature of 25°C (298.15 K)
• Pressure of 1 atm
• pH of 7

Under these conditions, the cell potential is denoted as E_cell°. When conditions deviate from these standards, the Nernst equation is used to calculate the cell potential.

How do you convert the Nernst equation from natural logarithm (ln) to common logarithm (log)?

To convert the Nernst equation from natural logarithm (ln) to common logarithm (log), you multiply the ln term by 2.303. The Nernst equation in terms of ln is:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{\mathrm{RT}}{\mathrm{nF}}ln\left(Q\right)$

Multiplying by 2.303, it becomes:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{0.05916}{n}log\left(Q\right)$

This conversion is useful for calculations involving common logarithms.

What is the significance of the reaction quotient (Q) in the Nernst equation?

The reaction quotient (Q) in the Nernst equation represents the ratio of the concentrations of products to reactants at any point in time. It is given by:

$Q=\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}$

Q affects the cell potential (E_cell) as the reaction progresses. When Q = 1, the cell potential is at its standard value (E_cell°). As Q changes, it indicates the shift in the reaction towards equilibrium, affecting E_cell until it reaches zero at equilibrium.