Problem 63 - Chemistry: The Central Science

Question

A voltaic cell utilizes the following reaction: \(4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)-\cdots 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=0.010 \mathrm{M}, \mathrm{P}_{\mathrm{O}_{2}}=0.50 \mathrm{~atm}\), and the \(\mathrm{pH}\) of the so- lution in the cathode compartment is \(3.50 ?\)

Step-by-Step Solution

Answer

The emf of the voltaic cell under standard conditions is \(2.000 \mathrm{V}\). Under the given conditions with specific concentrations, pressure, and pH, the emf of the cell is approximately \(1.554 \mathrm{V}\).

Step 1: Identify the half-cell reactions

The original cell reaction can be broken down into two half-reactions: Oxidation: \(4 \mathrm{Fe}^{2+}(a q) \rightarrow 4 \mathrm{Fe}^{3+}(a q) + 4 e^-\) Reduction: \(\mathrm{O}_2(g) + 4 \mathrm{H}^{+}(a q) + 4 e^- \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

Step 2: Calculate the standard EMF for the cell (Part A)

We'll use standard reduction potential values to find the EMF under standard conditions. The standard reduction potential for the half-reactions are: \(E_{red\mathrm{(O_2)}}^o = +1.229 \mathrm{V}\) \(E_{red\mathrm{(Fe^{3+}/Fe^{2+})}}^o = +0.771 \mathrm{V}\) As the reduction of \(\mathrm{Fe}^{3+}\) is the reverse of the given oxidation reaction, the standard potential for oxidation will be equal to the negative of the reduction potential: \(E_{ox\mathrm{(Fe^{2+}/Fe^{3+})}}^o = -E_{red\mathrm{(Fe^{3+}/Fe^{2+})}}^o = -0.771 \mathrm{V}\) Now we can calculate the standard cell potential: \(E_{cell}^o = E_{red}^o - E_{ox}^o = 1.229 - (-0.771) = 2.000 \mathrm{V}\) So under standard conditions, the emf of this cell is \(2.000 \mathrm{V}\).

Step 3: Modify the Nernst equation for the given concentrations and pressure (Part B)

The Nernst equation relates the cell EMF to concentrations of the species involved: \(E_{cell} = E_{cell}^o - \frac{RT}{nF} \ln Q\) where \(R\) is the gas constant, \(T\) is the temperature (in Kelvin), \(n\) is the number of electrons transferred, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient. For this reaction, we have: \(Q = \frac{[\mathrm{Fe}^{3+}]^4[\mathrm{H_2O}]^2}{[\mathrm{Fe}^{2+}]^4[\mathrm{O_2}] [\mathrm{H^+}]^4}\) Now we need to incorporate the given concentrations and pressure of the species and the pH value to determine the concentration of \(\mathrm{H}^{+}\) ions. \([\mathrm{Fe}^{2+}] = 1.3 \mathrm{M}\) \([\mathrm{Fe}^{3+}] = 0.010 \mathrm{M}\) \(P_{\mathrm{O}_2} = 0.50 \mathrm{~atm}\) \(pH = 3.50\) As \(pH = -\log_{10} [\mathrm{H}^+]\), we have: \([\mathrm{H}^+] = 10^{-3.50} = 3.16 \times 10^{-4} \mathrm{M}\) Also, for the partial pressure of \(\mathrm{O}_2\), we can find the concentration using the ideal gas law and assuming the cell's volume is \(1 \mathrm{L}\): \([\mathrm{O_2}] = \frac{P_{\mathrm{O}_2}}{RT} = \frac{0.50 \mathrm{~atm}}{0.0821 \frac{L \cdot atm}{K \cdot mol} \times 298K} = 0.0203 \mathrm{M}\)

Step 4: Calculate the emf of the cell under given conditions (Part B)

Now we can plug in the concentration values into the modified Nernst equation: \(E_{cell} = 2.000 - \frac{8.314 \mathrm{J/(mol \cdot K)} \times 298 \mathrm{K}}{4 \times 96485 \mathrm{C/mol}} \ln \frac{(0.010)^4 \times (1)^2}{(1.3)^4 \times 0.0203 \times (3.16 \times 10^{-4})^4}\) \(E_{cell} \approx 2.000 - \frac{0.0257}{4} \ln \frac{(0.010)^4}{(1.3)^4 \times 0.0203 \times (3.16 \times 10^{-4})^4} \approx 1.554 \mathrm{V}\) So the emf of this cell when using given concentrations and pressure under the given pH is approximately \(1.554 \mathrm{V}\).

Key Concepts

Standard EMF Calculation

The electromotive force (EMF) of a voltaic, or galvanic, cell under standard conditions is a measure of the potential difference between two electrodes. It is an important parameter because it indicates the cell's ability to do work, specifically electrical work. To calculate the standard EMF, we use standard reduction potentials, which are found in tables and represent the voltage associated with a reduction reaction compared to the standard hydrogen electrode.

Here's how we calculate standard EMF for a cell:

• Identify the two half-reactions occurring in the cell: the oxidation reaction and the reduction reaction.
• Look up the standard reduction potential (\(E_red^o\)) for each half-reaction.
• Remember, if a half-reaction is being reversed as it's an oxidation instead of a reduction, change the sign of its standard potential.

You then use the formula:\[E_{cell}^o = E_{red}^o - E_{ox}^o\]where \(E_{red}^o\) is the standard potential of the reduction reaction and \(E_{ox}^o\) is that of the oxidation reaction (usually a reversed reduction reaction).
The overall cell potential under standard conditions helps gauge whether the cell reaction is spontaneous. A positive \(E_{cell}^o\) denotes a spontaneous cell reaction, as seen in the exercise where the EMF of 2.000 V indicates a strong drive for the reaction to proceed.

Nernst Equation

The Nernst equation extends the concept of the EMF to non-standard conditions. It considers concentration effects, temperature, and partial pressures to find the cell's actual EMF during operation. This adaptability is vital since real cells rarely operate under standard conditions.

The Nernst equation is given by:\[E_{cell} = E_{cell}^o - \frac{RT}{nF} \ln Q\]where:

• \(E_{cell}\) is the EMF of the cell under specific conditions.
• \(E_{cell}^o\) is the standard EMF.
• \(R\) is the universal gas constant \(8.314 \, \text{J/(mol K)}\).
• \(T\) is the temperature in Kelvin. Standard is 298 K (25°C), but it can vary.
• \(n\) is the number of moles of electrons transferred in the cell reaction.
• \(F\) is Faraday's constant \(96485 \, \text{C/mol e}^-\).
• \(Q\) is the reaction quotient, reflecting the hydronium concentration, partial pressures, and molar concentrations. It's calculated based on the products over reactants involved in the half-reactions.

Apply concentration and partial pressure values within the Nernst equation helps understand how cell potential changes, as demonstrated in the exercise. Given conditions cause the EMF to drop from 2.000 V (standard) to approximately 1.554 V. This adjustment displays the importance of real-world variables and the potential decreases as the reactant concentration increases and product concentration decreases.

Half-Cell Reactions

In electrochemistry and when dealing with voltaic cells, understanding half-cell reactions is crucial. Each half-cell reaction involves either oxidation or reduction. These reactions occur in separate compartments of a voltaic cell and ultimately dictate the overall cell reaction and potential.

In our initial example exercise:

• The oxidation half-reaction involves the conversion of ferrous ion \(\mathrm{Fe}^{2+}\) to ferric ion \(\mathrm{Fe}^{3+}\), releasing electrons. It is represented as: \[\mathrm{4 \, Fe^{2+}(aq) \rightarrow 4 \, Fe^{3+}(aq) + 4e^-}\]
• The reduction half-reaction sees oxygen gas \(\mathrm{O}_2\) gaining electrons in the presence of protons to form water, a simple but highly stable compound: \[\mathrm{O_2(g) + 4 \, H^{+}(aq) + 4 \,e^- \rightarrow 2 \, H_2O(l) }\]

Remember, half-cell reactions are crucial since:

• They help identify where each part of the full cell reaction happens: oxidation at the anode and reduction at the cathode.
• They lay the groundwork for calculating potential using standard reduction potentials.

By dividing the whole cell reaction into half-reactions, we understand the detailed electron flow and energy changes, crucial when considering the EMF and the cell's energetic efficiency.