Problem 64 - Chemistry: The Central Science

Question

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g)-\rightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=2.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.85 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both compartments is \(5.00 ?\)

Step-by-Step Solution

Answer

(a) Under standard conditions, the emf of the cell is 0.77 V. (b) Under the given nonstandard conditions, the emf of the cell is approximately 0.89 V.

Step 1: Identify the half reactions and their standard reduction potentials

First, we need to identify the half reactions involved in the cell and their standard reduction potentials (E°). Looking at the balanced redox reaction, we see that: - Fe³⁺ is reduced to Fe²⁺: \( \mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+} \) with E° = +0.77 V - H₂ is oxidized to H⁺: \( \mathrm{H}_{2} \rightarrow 2\mathrm{H}^{+} + 2e^{-} \) with E° = 0 V (by definition)

Step 2: Calculate the standard electromotive force (emf) of the cell

We can calculate the standard emf (E°) for the cell using the following equation: \( E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \) In this case, the reduction half-reaction takes place at the cathode, and the oxidation half-reaction occurs at the anode. Therefore, \( E^\circ_\text{cell} = (+0.77\,\text{V}) - (0\,\text{V}) = +0.77\,\text{V} \) So, under standard conditions, the emf of the cell is 0.77 V.

Step 3: Apply the Nernst equation to find the emf under nonstandard conditions

We will now apply the Nernst equation when we have the given concentrations of species and pH values to find the emf for this cell. The Nernst equation is: \( E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \) Where R is the gas constant (8.314 J/mol·K), T is the temperature (assumed to be 298 K), n is the number of electrons transferred (2 in this case), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

Step 4: Calculate the reaction quotient (Q) under nonstandard conditions

To find Q for this reaction under the given conditions: \( Q = \frac{[\mathrm{Fe}^{2+}]^2[\mathrm{H}^{+}]^2}{[\mathrm{Fe}^{3+}]^2P_{\mathrm{H}_{2}}} \) Using the provided concentrations and pH values: \( Q = \frac{(0.0010\,\mathrm{M})^2 (10^{-5.00})^2}{(2.50\,\mathrm{M})^2 (0.85\,\mathrm{atm})} \) Calculating Q gives us: \( Q \approx 6.54 \times 10^{-12} \)

Step 5: Calculate the emf under nonstandard conditions

Now, we can plug the values of E°, R, T, n, F, and Q into the Nernst equation to find the emf under nonstandard conditions. \( E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \) \( E_\text{cell} = 0.77\,\text{V} - \frac{(8.314\,\text{J/mol}\cdot\text{K})(298\,\text{K})}{(2)(96,485\,\text{C/mol})} \ln (6.54 \times 10^{-12}) \) Calculating the emf gives us: \( E_\text{cell} \approx 0.89\,\text{V} \) So, under the given nonstandard conditions, the emf of the cell is approximately 0.89 V.

Key Concepts

Electromotive Force (emf)

Electromotive force, or emf, is the voltage developed by any source of electrical energy such as a battery or dynamo. In the context of a voltaic cell, emf represents the potential difference between the two electrodes. This potential difference is what drives the electron flow through the external circuit, thereby creating an electric current.
In standard conditions, the emf is calculated using the standard reduction potentials of the half reactions. To find the standard emf, you subtract the standard reduction potential of the anode (oxidation half) from that of the cathode (reduction half).
The formula for the standard emf of a cell is given by:

• \( E^ ext{°}_ ext{cell} = E^ ext{°}_ ext{cathode} - E^ ext{°}_ ext{anode} \)

This value gives us insight into the cell's ability to drive current under standard conditions, with higher emf indicating a more powerful cell.

Nernst Equation

The Nernst equation is crucial for calculating cell emf under nonstandard conditions. It accounts for variations in concentration, pressure, and temperature from the standard state. The general form of the Nernst equation for a cell can be expressed as:

• \(E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \)

Here, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged in the reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.
Using the Nernst equation helps predict the actual emf for any given concentration of reactants and products, as opposed to standard conditions. It shows how the cell potential decreases with the use of reactants or increases with the availability of products. This makes it an essential tool in electrochemistry for real-world applications.

Standard Reduction Potential

Standard reduction potential, represented as \(E^\circ\), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. The standard reduction potential is measured under standard conditions: solute concentrations of 1 M, gas pressures of 1 atm, and a temperature of 25°C (298 K).
Each half-reaction in an electrochemical cell has a standard reduction potential, and the overall cell potential can be determined by comparing these values. For example, the half-reaction with the more positive reduction potential occurs as a reduction, and the other will occur as oxidation.
By knowing the standard reduction potential, you can predict the direction of the electron flow and whether a certain redox reaction is spontaneous under standard conditions. More positive values indicate greater tendencies to gain electrons or be reduced.

Reaction Quotient (Q)

The reaction quotient, \(Q\), is a quantity that provides insight into the direction in which a chemical reaction is proceeding. It is expressed in terms of the concentrations or pressures of the reactants and products in a reaction.
For a given reaction:

• \(aA + bB \rightarrow cC + dD\)

The reaction quotient is calculated by:

• \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

Where square brackets denote concentrations, and the letters represent species coefficients from the balanced equation.
In electrochemistry, \(Q\) is used in the Nernst equation to determine how changes from standard conditions affect the emf of the cell. A shift in reaction conditions causes a change in \(Q\), which in turn alters the cell potential as calculated by the Nernst equation. Understanding \(Q\) helps in predicting the direction of the reaction and whether the reaction is at equilibrium.