Problem 68 - Chemistry The Central Science In Si Units

Question

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

Step-by-Step Solution

Answer

(a) 0.77 V (b) 1.08 V

Step 1: Write Half-Reactions

Break down the overall redox reaction into its half-reactions. For the reduction half-reaction: \( ext{Fe}^{3+} + e^- \rightarrow ext{Fe}^{2+} \). For the oxidation half-reaction: \( ext{H}_2 \rightarrow 2 ext{H}^+ + 2e^- \).

Step 2: Determine Standard Electrode Potentials

Use standard reduction potentials from a reference table: \( E^0( ext{Fe}^{3+}/ ext{Fe}^{2+}) = +0.77 ext{ V} \), \( E^0( ext{H}^+/ ext{H}_2) = 0 ext{ V} \).

Step 3: Calculate Standard EMF (E°)

Calculate the standard cell potential (E°) using the formula: \( E^0_{ ext{cell}} = E^0_{ ext{cathode}} - E^0_{ ext{anode}} = 0.77 ext{ V} - 0 ext{ V} = 0.77 ext{ V} \).

Step 4: Use the Nernst Equation for Non-Standard Conditions

Apply the Nernst equation: \( E = E^0 - \frac{RT}{nF} \ln Q \). Set conditions: \( T = 298 ext{ K} \), \( n = 2 \) moles of electrons.

Step 5: Calculate Reaction Quotient (Q)

For the reaction quotient, \( Q = \frac{[ ext{Fe}^{2+}]^2 [ ext{H}^+]^2}{[ ext{Fe}^{3+}]^2 P_{ ext{H}_2}} = \frac{(0.0010)^2 (10^{-4})^2}{(3.50)^2 (96.3 \times 10^3)} = 1.40 \times 10^{-16} \).

Step 6: Substitute into Nernst Equation

Using the Nernst equation, substitute the known values: \( E = 0.77 - \frac{8.314 \times 298}{2 \times 96485} \ln(1.40 \times 10^{-16}) \). Calculate \( E \approx 1.08 ext{ V} \).

Key Concepts

Standard Electrode Potential

The standard electrode potential, often symbolized as \( E^0 \), is a fundamental concept in electrochemistry. It refers to the potential difference between an electrode and its respective electrolytic solution under standard conditions. These standard conditions include a concentration of 1 M for ions, a pressure of 1 atm for gases, and a temperature of 25°C (298 K).

• For oxidation reactions, electrons are lost, and for reduction reactions, electrons are gained.
• Each half-cell in a voltaic cell has its own standard electrode potential.
• The standard electrode potential is crucial for determining the direction and spontaneity of a reaction.

In our exercise, the two half-reactions involve iron and hydrogen. The standard potential for the iron reaction \( ext{Fe}^{3+} + e^- \rightarrow ext{Fe}^{2+} \) is +0.77 V. Where the hydrogen half-reaction \( ext{H}_2 \rightarrow 2 ext{H}^+ + 2e^- \) is set at 0 V, which serves as a reference point. Thus, the standard cell potential \( E^0_{\text{cell}} \) is calculated by subtracting the anode potential from the cathode potential.

Nernst Equation

The Nernst Equation is a powerful tool used to calculate the cell potential at non-standard conditions. It's expressed as: \\[ E = E^0 - \frac{RT}{nF} \ln Q \] where \( E \) is the cell potential under non-standard conditions, \( R \) is the gas constant \( 8.314 \frac{J}{mol \, K} \), \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred in the redox reaction, \( F \) is the Faraday constant \( 96485 \, C/mol \), and \( Q \) is the reaction quotient.

• The Nernst Equation helps adjust the cell potential for changes in concentration, pressure, or pH levels.
• In non-standard conditions, such as when the concentrations of ions or pressure of gases differ from their standard states, the Nernst Equation comes into play.
• By accounting for these factors, it’s possible to calculate a more accurate EMF of the cell.

In our case, the Nernst equation allows us to find that the cell potential increases from the standard 0.77 V to approximately 1.08 V due to the given conditions of the solution and gases.

Reaction Quotient

The reaction quotient, symbolized as \( Q \), is a dimensionless number that provides a snapshot of a reaction at any point in time relative to its equilibrium position. It's calculated using the concentrations (or pressures) of the reactants and products involved in a reaction.

• For a given reaction \( aA + bB \leftrightarrow cC + dD \), the reaction quotient is calculated as \( Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).
• \( Q \) lets us determine which direction a reaction will shift to reach equilibrium.
• If \( Q \) is less than the equilibrium constant \( K \), the reaction proceeds forward.
• If \( Q \) is greater than \( K \), the reaction proceeds in reverse.

In our specific reaction, \( Q \) includes the concentrations of \( ext{Fe}^{2+} \), \( ext{Fe}^{3+} \), \( ext{H}^+ \), and the partial pressure of \( ext{H}_2 \). \( Q \) was calculated as \( 1.40 \times 10^{-16} \), which is used in the Nernst equation to find the cell potential under non-standard conditions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. These reactions are the cornerstone of electrochemical cells like the voltaic cell.

• Reduction is the gain of electrons, while oxidation is the loss of electrons.
• A complete redox reaction comprises two half-reactions — one for oxidation and one for reduction.
• The overall cell reaction is derived by combining these two half-reactions.

In our example, \( ext{Fe}^{3+} \) is reduced to \( ext{Fe}^{2+} \), while \( ext{H}_2 \) is oxidized to \( ext{H}^+ \). These transformations illustrate the electron flow that fuels the voltaic cell, allowing it to generate electric current.