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- $$ \text { lowing: } \mathrm{pH}=0.0,\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M},\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M} $$ \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

Step by step solution

01

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.

02

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).

03

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: \( E_{cell} = 1.51 \, V - 0.771 \, V = 0.739 \, V \)

04

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.

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