Problem 67 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) The emf of the cell under standard conditions is \(0.46\,\text{V}\). (b) The emf of the cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3\,\mathrm{M}, \left[\mathrm{Fe}^{3+}\right]=0.010\,\mathrm{M}, P_{\mathrm{o}_{2}}=0.50\,\mathrm{~atm}\), and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50\) is approximately \(0.32\,\text{V}\).

Step 1: Write the half-reactions and determine the standard reduction potentials

First, we need to write the half-reactions: Oxidation half-reaction: \(4 \mathrm{Fe}^{2+}(aq) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+4 \mathrm{e}^-\) Reduction half-reaction: \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+ 4 \mathrm{e}^-\longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) The standard reduction potential values are: \(E^0_{Fe^{3+}/Fe^{2+}} = +0.77\,\mathrm{V}\) and \(E^0_{O_2/H_2O} = +1.23\,\mathrm{V}\)

Step 2: Calculate the standard cell potential (emf) under standard conditions

To calculate the standard emf of the voltaic cell, we can use the equation: \(E_\text{cell}^0 = E_\text{cathode}^0 - E_\text{anode}^0\) In our case, the cathode reaction is the reduction half-reaction and the anode reaction is the oxidation half-reaction, so our equation is: \(E_\text{cell}^0 = E_\text{O_2/H_2O}^0 - E_\text{Fe^{3+}/Fe^{2+}}^0\) Substitute the values of the standard reduction potentials: \(E_\text{cell}^0 = (+1.23\,\mathrm{V}) - (+0.77\,\mathrm{V})\) \(E_\text{cell}^0 = +0.46\,\text{V}\) The emf under standard conditions is \(0.46\,\text{V}\).

Step 3: Determine the cell potential (emf) under nonstandard conditions using the Nernst equation

To find the emf under nonstandard conditions, we can use the Nernst equation: \(E_\text{cell} = E_\text{cell}^0 - \cfrac{RT}{nF} \ln{Q}\) Where: - \(R\) is the universal gas constant, \(8.31\,\text{J}/\text{K·mol}\), - \(T\) is the temperature in Kelvin (assume \(298\,\text{K}\)), - \(n\) is the number of moles of electrons transferred (in our case, \(n=4\)), - \(F\) is the Faraday constant, \(96485\,\text{C}/\text{mol}\), - \(Q\) is the reaction quotient, which is defined as: \(Q = \cfrac{\left[\mathrm{Fe}^{3+}\right]^4 \cdot P_{O_2}}{\left[\mathrm{Fe}^{2+}\right]^4 \cdot \left[\mathrm{H}^+\right]^4}\) Substitute the given values into the reaction quotient equation: \(Q = \cfrac{(0.010\,\text{M})^4 \cdot (0.50\,\text{atm})}{(1.3\,\text{M})^4 \cdot (10^{-3.5})^4}\) Now, substitute the values into the Nernst equation: \(E_\text{cell} = 0.46\,\text{V} - \cfrac{8.31\,\text{J}/\text{K·mol} \cdot 298\,\text{K}}{4 \cdot 96485\,\text{C}/\text{mol}} \ln{Q}\) Solving it, we get: \(E_\text{cell} \approx 0.32\,\text{V}\) The emf of the cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3\,\mathrm{M}, \left[\mathrm{Fe}^{3+}\right]=0.010\,\mathrm{M}, P_{\mathrm{o}_{2}}=0.50\,\mathrm{~atm}\), and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50\) is approximately \(0.32\,\text{V}\).

Key Concepts

Nernst equation

The Nernst equation is a crucial tool in electrochemistry. It allows us to calculate the cell potential under non-standard conditions. When conditions deviate from standard states (1 M concentration, 1 atm pressure, and 298 K temperature), the cell potential changes. The Nernst equation helps account for these variations: \[E_{\text{cell}} = E_{\text{cell}}^0 - \frac{RT}{nF} \ln{Q}\] Where:

• E_\text{cell} is the cell potential under the given conditions.
• E_\text{cell}⁰ is the standard cell potential.
• R is the universal gas constant (8.31 J/K·mol).
• T is the temperature in Kelvin.
• n is the number of moles of electrons transferred.
• F is the Faraday constant (96485 C/mol).
• Q is the reaction quotient, reflecting the concentrations and pressures of products and reactants.

The reaction quotient is calculated based on the equilibrium state of the reaction, considering all the reactant and product concentrations and partial pressures. This equation allows chemists to predict how changes in conditions will affect the voltage generated by the cell.

standard cell potential

In electrochemistry, the standard cell potential (emf) is the voltage across a voltaic cell when all reactants and products are in their standard states. For galvanic cells (spontaneous reactions), the potential is calculated as:\[E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0\] Here’s a step-by-step view:

• The standard potential for each half-reaction is measured under standard conditions: 1 M concentration for aqueous species, 1 atm for gases, and usually 25°C (298 K).
• The more positive the standard reduction potential, the stronger the oxidizing agent.
• For example, in the cell reaction provided, the reduction half-reaction involves \( \mathrm{O}_2(g) \) and is assigned \( E^0 = +1.23\, \mathrm{V} \) while the oxidation half-reaction is \( \mathrm{Fe}^{2+}/\mathrm{Fe}^{3+}\) with \( E^0 = +0.77\, \mathrm{V} \).

Subtracting the anode potential from the cathode potential gives the standard cell potential. The more positive the result, the more favorable the cell reaction under standard conditions.

half-reactions

In any electrochemical cell, the overall redox reaction is split into two half-reactions: oxidation and reduction. Breaking a reaction into these components helps in understanding how electrons are transferred.

• The oxidation half-reaction focuses on the species losing electrons. For the given reaction, \( 4 \mathrm{Fe}^{2+} \rightarrow 4 \mathrm{Fe}^{3+} + 4 \mathrm{e}^- \) represents that \( \mathrm{Fe}^{2+} \) ions are oxidized, losing electrons.
• The reduction half-reaction focuses on the species gaining electrons. Here, the reaction \( \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^- \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \) shows that \( \mathrm{O}_2 \) is reduced, gaining electrons to form water.
• Each half-reaction is associated with its own potential, determined by the tendency of a species to be reduced or oxidized.
• Balancing half-reactions is crucial and involves ensuring the number of electrons lost equals those gained.

Understanding these half-reactions is critical in calculating cell potentials and predicting the direction of electron flow.