Question

This oxidation-reduction reaction in acidic solution is spontaneous: \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow\) \(5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

Short answer

The anode is where Fe(II) is being oxidized, and the cathode is where MnO4- is being reduced. Electrons flow from anode to cathode through the external wire. Cations flow through the salt bridge from the anode solution to the cathode solution, and anions flow from the cathode solution to the anode solution. At the anode's surface, Fe2+ ions are oxidized to Fe3+ ions and lose an electron, which is transferred to the platinum foil and the wire. The emf of the cell under standard conditions is 2.28 V, and the emf of the cell with given concentrations is 2.47 V.

Step-by-step solution

1. Identify the anode and cathode

From the balanced redox reaction \[5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\] We can identify the half-reactions: Oxidation: \(\ce{Fe^2+ -> Fe^3+ + e-}\) Reduction: \(\ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O}\) The anode is where the oxidation reaction occurs, and the cathode is where the reduction reaction occurs. Therefore, Fe(II) is being oxidized at the anode, and MnO4- is being reduced at the cathode.

2. Sketch the cell and indicate the directions

Left beaker contents: Solution of KMnO4 and H2SO4. Anode: Platinum foil in contact with the FeSO4 solution. Right beaker contents: Solution of FeSO4. Cathode: Platinum foil in contact with the KMnO4 + H2SO4 solution. Electrons flow from anode to cathode through the external wire, which includes a voltmeter. Positive ions (cations) flow through the salt bridge from the anode solution to the cathode solution. Negative ions (anions) flow through the salt bridge from the cathode solution to the anode solution.

3. Sketch the process at the atomic level at the anode's surface

At the anode's surface, Fe2+ ions are being oxidized to Fe3+ ions. They lose an electron, which is then transferred to the platinum foil and subsequently to the wire that connects both foils. As the Fe2+ ions become Fe3+ ions, they leave the surface of the platinum foil and enter the solution.

4. Calculate the emf under standard conditions

Under standard conditions, the emf can be calculated using the cell potential difference: \(E_{cell} = E_{cathode} - E_{anode} = E^0_{red} - E^0_{ox}\) We can find the standard reduction potential for both half-reactions from a standard reduction potential table: \(E^0_{red}\) (MnO4- to Mn2+) = +1.51 V \(E^0_{ox}\) (Fe2+ to Fe3+) = -0.77 V Then, \(E_{cell}^0 = 1.51 - (-0.77) = 2.28 \,\text{V}\)

5. Calculate the emf with given concentrations

To calculate the emf of the cell at the given concentrations, we will use the Nernst equation: \[E = E^0 - \frac{0.05916}{n} \log{Q}\] where n is the number of electrons transferred (5, as seen from the balanced reaction), Q is the reaction quotient, and E is the emf at nonstandard conditions. The reaction quotient is calculated as follows: \[Q = \frac{[\mathrm{Fe}^{3+}]^5[\mathrm{Mn}^{2+}]}{[\mathrm{Fe}^{2+}]^5[\mathrm{MnO_4^-}][\mathrm{H^+}]^8}\] Plugging in the given concentrations, we have: \[Q = \frac{(2.5 \times 10^{-4})^5 (0.001)}{(0.10)^5(1.50)(1)^8} = 3.125 \times 10^{-17}\] Now, we can plug in the values into the Nernst equation: \[E = 2.28 - \frac{0.05916}{5} \log{(3.125 \times 10^{-17})} = 2.28 + 0.189 = \boxed{2.47\,\text{V}}\]

Key concepts

Electrochemical Cells

Electrochemical cells are the foundation of harnessing chemical reactions to produce electrical energy. They consist essentially of two half-cells where oxidation and reduction reactions occur separately. These two processes jointly facilitate the flow of electrons from the anode to the cathode through an external circuit, which is how electrical current is generated.

In each half-cell, a metal electrode is submerged in a solution containing ions of the same metal. An example would be a strip of zinc in a solution of zinc sulfate. The electrode where oxidation happens, loses electrons and is known as the anode, whereas the cathode is where reduction occurs, gaining electrons. It is essential to note that, for the electron flow to persist, the electric circuit has to be closed, which is also complemented by an ionic circuit often closed by a salt bridge or porous membrane to allow the ionic flow, balancing the charge and completing the circuit.

In addition to the basic anode and cathode setup, real-life applications of electrochemical cells include batteries and various types of sensors and fuel cells, each designed with specific materials and configurations to optimize their efficiency for the intended applications.

Galvanic Cells

Galvanic cells, also known as voltaic cells, are a type of electrochemical cell that spontaneously converts chemical energy into electrical energy through redox reactions. This is the cell type described in the initial exercise. The primary feature distinguishing a galvanic cell from other types of electrochemical cells is its spontaneous nature; the reactions happen without the need for an external power source.

The setup often involves two separate half-cells connected by a salt bridge and an external wire. In the provided exercise, a platinum foil in the FeSO₄ solution acts as the anode while another platinum foil in the KMnO₄ and H₂SO₄ solution serves as the cathode. Electrons flow from the anode to the cathode through the wire, which is measured by a voltmeter, while the salt bridge maintains charge balance by allowing ion migration.

Galvanic cells have real-world implications in battery technology, where a series of such cells can be combined to provide a higher voltage, or adjusted for higher capacity, according to the requirements of the device they power.

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a chemical species to gain electrons, thus to be reduced. It is a half-cell potential when all reactants and products are in their standard states at a pressure of 1 bar and concentration of 1 M, typically measured at the temperature of 298 K (25°C).

In the step-by-step solution, two standard reduction potentials are given: +1.51 V for the reduction of MnO₄^- to Mn²⁺ and -0.77 V for the oxidation of Fe²⁺ to Fe³⁺. The more positive the reduction potential, the greater the species' ability to gain electrons. The standard reduction potential is crucial in predicting the direction of electron flow and cell potential in an electrochemical cell. By comparing the standard reduction potentials of the half-reactions, one can predict that the substance with the higher reduction potential will be reduced at the cathode, and the substance with the lower potential will be oxidized at the anode in a spontaneous redox reaction.

Nernst Equation

The Nernst equation allows the determination of the cell potential under non-standard conditions, accounting for the temperature, pressure, and reactant/product concentrations. It is mathematically expressed as:
\[E = E^0 - \frac{0.05916}{n} \log{Q}\]
Here, \(E\) is the electromotive force (emf) of the cell under non-standard conditions, \(E^0\) is the standard emf, \(n\) is the number of moles of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient, representing the concentrations of the reactants and products. The 0.05916 value is the result of combining constant factors including the gas constant and Faraday's constant, assuming a temperature of 298 K.

In the exercise's step 5, the Nernst equation was used to calculate the emf given varied concentrations of the reactants and products at 298 K. This adjustment of the standard emf is crucial for real-world applications where conditions can vary significantly from the 'standard' lab conditions, allowing for predictions of cell behavior in a variety of circumstances.