Problem 50 - Chemistry: An Atoms-Focused Approach

Question

Using the appropriate standard potentials from Appendix \(6,\) determine the equilibrium constant at \(298 \mathrm{K}\) for the following reaction between \(\mathrm{Mn} \mathrm{O}_{2}\) and \(\mathrm{Fe}^{2+}\) in acid solution: \(4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s)+2 \mathrm{Fe}^{2+}(a q) \rightarrow\) $$ \mathrm{Mn}^{2+}(a q)+2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$

Step-by-Step Solution

Answer

The equilibrium constant for the redox reaction between MnO2 and Fe2+ in an acidic solution at 298 K is approximately $$6.71\times10^{22}.$$

Step 1: Write the oxidation and reduction half-reactions

To understand which species are oxidized and reduced in the reaction, balance the number of electrons transferred between the different species for the following reaction: $$ 4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s)+2 \mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ Oxidation half-reaction (Fe2+ is oxidized to Fe3+): $$ \mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+e^{-} $$ Reduction half-reaction (MnO2 is reduced to Mn2+): $$ \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 e^{-}\rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ #region II: Determine the standard reduction potential#

Step 2: Look up standard reduction potentials

Consult Appendix 6 or any standard reduction potential table to find the standard reduction potentials (E˚) for both half-reactions: For Fe3+/Fe2+: E°(Fe3+/Fe2+) = +0.771 V For MnO2/Mn2+: E°(MnO2/Mn2+) = +1.230 V Note that the values provided are already for the reduction reaction, so we do not need to change the sign for either of these. #Region III: Calculate the standard cell potential#

Step 3: Calculate the standard cell potential (E° cell)

By adding the standard reduction potentials of the two half-reactions, we can find the standard cell potential (E° cell) for the overall reaction: E° cell = E° (reduction) + E° (oxidation) E° cell = E° (MnO2/Mn2+) + E° (Fe3+/Fe2+) E° cell = (+1.230 V) + (+0.771 V) E° cell = +2.001 V

Step 4: Calculate the equilibrium constant (K) at 298 K using Nernst equation

Now we use the Nernst equation to find the equilibrium constant at 298 K: $$ \Delta G^\circ = -nFE^\circ_\text{cell} $$ Where ∆G° is the standard Gibbs free energy change, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (96485 C/mol), and E° cell is the standard cell potential. We also know that: $$ \Delta G^\circ = -RT\ln K $$ Where R is the gas constant (8.3145 J/mol*K), T is the temperature (298 K), and K is the equilibrium constant. For this reaction, the number of moles of electrons transferred (n) is 2. Combining the two equations, we get: $$ -RT\ln K = -nFE^\circ_\text{cell} $$ Now we can solve for K: $$ \ln K = \frac{-nFE^\circ_\text{cell}}{-RT} $$ $$ \ln K = \frac{2\times96485\times 2.001}{8.3145\times 298} $$ $$ \ln K = 51.0 $$ Finally, find K by using the exponential function: $$ K = e^{51.0} $$ Now we have all the necessary information to evaluate K: $$ K \approx 6.71\times10^{22} $$ Solution: The equilibrium constant for this reaction between MnO2 and Fe2+ in an acid solution at 298 K is approximately $$6.71\times10^{22}.$$

Key Concepts

Equilibrium Constant

The equilibrium constant ( K ) is a key concept in chemistry, used to determine the extent of a reaction at a specific temperature. For reversible reactions, the equilibrium position indicates how far the reaction has proceeded. When we say that a reaction is at equilibrium, we mean that the rate at which the reactants turn into products is equal to the rate at which the products revert back into reactants.
In simpler terms, it’s a balance between forward and backward reactions. This is why K, the equilibrium constant, becomes important because it tells us how much product can be expected compared to reactants once the system is at equilibrium.
When K is large, as in the reaction between MnO2 and Fe2+, it suggests that the equilibrium greatly favors products. This means that almost all reactants convert into products under standard conditions. When calculating K, the Nernst equation can bridge the gap between cell potential and Gibbs free energy, showing how the potential energy difference in electrochemistry translates into chemical equilibrium.

Standard Reduction Potential

Standard reduction potential ( E° ) measures the tendency of a chemical species to gain electrons, and thus be reduced. These values are important in electrochemistry as they allow us to predict which way a redox reaction will proceed.
To calculate standard cell potential ( E°_{cell} ), you combine the standard reduction potentials of the involved half-reactions. This involves taking the reduction potential of the oxidation reaction and adding it to the reduction potential of the reduction reaction. For example, in our reaction involving Fe2+ and MnO2, the half-reactions have standard potentials of +0.771 V and +1.230 V, respectively.
The value of E°_{cell} indicates the drive behind the movement of electrons from the reducing agent to the oxidizing agent. It's essential to note that a higher positive E° value means a stronger tendency for reduction to occur. Together, these values help chemists determine the feasibility and spontaneousness of a reaction under standard conditions.

Nernst Equation

The Nernst Equation is a fundamental formula in electrochemistry that relates the concentrations of chemical species participating in a redox reaction to the standard cell potential and equilibrium constant.
It is given by:\[E = E^ ext{°} - \frac{RT}{nF} \ln Q\]Here, E is the cell potential under non-standard conditions, E^ ext{°} is the standard potential, R is the gas constant, T is the temperature in Kelvin, F is the Faraday's constant, and Q is the reaction quotient, which reflects the ratio of product concentrations to reactant concentrations at any point away from equilibrium.

• When Q equals the equilibrium constant K, the cell potential E becomes zero, which means the cell is at equilibrium.
• The equation also illustrates how the electrochemical cell potential depends on changes in concentration and temperature.
• By rearranging the Nernst equation, one can solve for K, giving insight into how favorable a reaction is at equilibrium, as was done in the original exercise to find K \approx 6.71 \times 10^{22}.

Understanding the Nernst Equation helps in predicting how cells will behave under non-standard conditions and is critical for tasks involving electrochemical measurements and calculations.