Determine the equilibrium constant at 255 K for the following reaction under acidic conditions.

Question

Determine the equilibrium constant at 255 K for the following reaction under acidic conditions.

4H⁺(aq) + MnO₂​(s) + 2Fe²⁺(aq) Mn²⁺(aq) + 2Fe³⁺(aq) + 2H₂O(l)

The two half-reactions are:

MnO₂ ​(s) + 4H⁺ (aq) + 2e^- Mn²⁺ (aq) + 2H₂O (l) E° = 1.23 V

Fe³⁺ (aq) + e^- Fe²⁺(aq) E° = 0.770 V

 

Answer 1

Because the half-reaction involving iron is of a more negative voltage (closer to zero), it implies that ferrous iron Fe²⁺ is going to be the electron donor in a spontaneous reaction. The reactions can be rewritten as:

1. Oxidation reaction, electron donor
2. 2 [Fe²⁺(aq) Fe³⁺ (aq) + e^-] E° = 0.770 V
3. Reduction reaction, electron acceptor
4. MnO₂ ​(s) + 4H⁺ (aq) + 2e^- Mn²⁺ (aq) + 2H₂O (l) E° = 1.23 V
5. Overall reaction: 2Fe²⁺(aq) + MnO₂ ​(s) + 4H⁺ (aq) Mn²⁺ (aq) + 2H₂O (l)+ 2Fe³⁺ (aq)
6. Cell voltage: E_cell = E_reduction - E_oxidation
7. E_cell = 1.23 V - 0.770 V = 0.46 V

To relate cell voltage with equilibrium conditions we want to use the following equation:

1. Nernst equation: E_cell = E⁰_cell - (RT/nF)* ln Q
2. E_cell is the non-standard cell potential, E⁰_cell is the standard cell potential, R is the gas constant, T is temperature, n is the number of electrons, F is Faraday's constant and Q is the reaction quotient
3. At equilibrium, the reaction quotient Q can be expressed as K_eq and E_cell = 0 V. We know E_cell is 0 V because we are at non-standard conditions (T = 255 K) but are at equilibrium so dG (Gibbs free energy) equals zero. The relation dG = -n*F*E_cell helps us make this conclusion.
4. The Nernst equation can then be simplified to: E⁰_cell = (RT/nF)* ln K_eq
5. 0.46 V = ((8.314 J/mol K)*(255K))/((2 mol electrons)*(96485 C/mol electron))*ln K_eq
6. K_eq = 1.526 * 10¹⁸