Question

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \longrightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ 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 fol- $$ \text { lowing: } \mathrm{pH}=0.0,\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M},\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M} $$ \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

Short answer

(a) In the electrochemical cell, label the anode (Fe) and the cathode (MnO4-). Electrons move from the anode to the cathode through the wire, and ions migrate through the salt bridge. (b) At the anode, Fe2+ ions lose an electron, converting to Fe3+ ions. Electrons are released into the external circuit. (c) Under standard conditions, the cell potential (emf) is calculated as 0.739 V. (d) At 298 K with the given concentrations, the cell potential is approximately 0.783 V.

Step-by-step solution

a) Sketching the electrochemical cell

To sketch the electrochemical cell, first, label the anode (Fe) and the cathode (MnO4-). Electrons move from the anode to the cathode through the external circuit (wire), and ions migrate through the salt bridge. So, in this case, Fe2+ ions move towards the cathode, and Mn2+ ions move towards the anode.

b) Sketch the atomic-level process at the anode

At the anode, the oxidation reaction occurs. The Fe2+ ions lose an electron, converting to Fe3+ ions. Electrons are then released into the external circuit (wire).

c) Calculate the cell potential under standard conditions

First, split the overall redox reaction into its half-reactions: Oxidation half-reaction: \( Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^- \) Reduction half-reaction: \( MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l) \) Under standard conditions, the cell potential (emf) can be calculated using the Nernst equation: \( E_{cell} = E^\circ_{cathode} - E^\circ_{anode} \) Where \(E^\circ_{cathode}\) and \(E^\circ_{anode}\) are the standard reduction potentials for the cathode and anode, respectively, which can be found in a table of standard reduction potentials. For these half-reactions: \(E^\circ_{Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-} = 0.771 \, V \) \(E^\circ_{MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)} = 1.51 \, V \) Now calculate the cell potential: \( E_{cell} = 1.51 \, V - 0.771 \, V = 0.739 \, V \)

d) Calculate the cell potential at 298 K with given concentrations

Using the Nernst equation, the cell potential at 298 K can be calculated as follows: \( E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \times lnQ \) Where: \(E_{cell}\) = cell potential at 298 K \(E^\circ_{cell}\) = standard cell potential \(R\) = gas constant, equal to \(8.314 \, J K^{-1} mol^{-1}\) \(T\) = temperature, equal to 298 K \(n\) = number of electrons transferred, equal to 5 in this case \(F\) = Faraday's constant, equal to \(96485 \, C mol^{-1}\) \(Q\) = reaction quotient For the given concentrations: \( Q = \frac{[Fe^{3+}]^5[Mn^{2+}]}{[Fe^{2+}]^5[MnO_4^-]} \) Use the given concentrations to calculate Q: \( Q = \frac{(2.5 \times 10^{-4})^5 \times 0.001}{(0.10)^5 \times 1.50} \approx 8.33 \times 10^{-16} \) Now calculate the cell potential: \(E_{cell} \approx 0.739 \, V - \frac{8.314 \times 298}{5 \times 96485} \times ln(8.33 \times 10^{-16}) \) \( E_{cell} \approx 0.739 \, V + 0.044 \, V = 0.783 \, V \) The cell potential at 298 K with the given concentrations is approximately 0.783 V.

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two chemical species. These reactions are fundamental in electrochemistry and are essential for the operation of electrochemical cells. Here's how they work:

• Oxidation: This is the process where a species loses electrons. In the redox reaction provided, the oxidation half-reaction is the conversion of \( \mathrm{Fe}^{2+} \) ions into \( \mathrm{Fe}^{3+} \) ions. This occurs at the anode, as represented by the equation \( \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + e^- \).


• Reduction: This is the gain of electrons by a species. The reduction half-reaction here occurs at the cathode, where \( \mathrm{MnO}_{4}^{-} \) ions gain electrons to become \( \mathrm{Mn}^{2+} \) ions, as represented by: \( \mathrm{MnO}_4^-(aq) + 8\mathrm{H}^+(aq) + 5e^- \rightarrow \mathrm{Mn}^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l) \).

When these processes occur together, electrons are transferred from the oxidizing agent to the reducing agent, establishing a flow of electrons, which produces electric current in an electrochemical cell.

Cell Potential

The cell potential, or electromotive force (emf), of an electrochemical cell indicates its ability to produce current and is a direct measure of the energy per electron transferred through the cell. Under standard conditions, this is calculated using the equation:\[E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]Here:

• Standard Reduction Potentials: Every species involved in a redox reaction has a standard reduction potential, which represents the tendency of a substance to gain electrons. In our reaction, the standard reduction potentials are:
  • \( E^{\circ}_{\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}} = 0.771 \, V \)
  • \( E^{\circ}_{\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}} = 1.51 \, V \)


• Calculating Cell Potential: Plugging in these values gives:\[E_{\text{cell}} = 1.51 \, V - 0.771 \, V = 0.739 \, V\]This positive value indicates a spontaneous reaction under standard conditions.

Cell potential helps to gauge the performance and feasibility of electrochemical cells in generating power.

Nernst Equation

The Nernst Equation provides a way to calculate the cell potential under non-standard conditions by considering the concentrations of the reactants and products involved in the cell reaction. It is expressed as:\[E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q\]Where:

• \( E^{\circ}_{\text{cell}} \): Standard cell potential.
• \( R \): Universal gas constant \( (8.314 \, J \, K^{-1} \, mol^{-1}) \).
• \( T \): Temperature in Kelvin. For this exercise, it's \( 298 \, K \).
• \( n \): Number of electrons exchanged, which is 5 here.
• \( F \): Faraday's constant \( (96485 \, C \, mol^{-1}) \).
• \( Q \): Reaction quotient, calculated as:\( Q = \frac{[\mathrm{Fe}^{3+}]^5[\mathrm{Mn}^{2+}]}{[\mathrm{Fe}^{2+}]^5[\mathrm{MnO}_4^-]} \).

Using the given concentrations, \( Q \approx 8.33 \times 10^{-16} \), we find the cell potential:\[E_{\text{cell}} \approx 0.739 \, V - \frac{8.314 \times 298}{5 \times 96485} \ln(8.33 \times 10^{-16}) \approx 0.783 \, V\]The Nernst equation allows us to examine how changes in concentration and temperature affect the electrochemical cell's performance.