Question

use the appropriate standard reduction potentials to determine the equlibrium constant at 200.00 K for the following reaction under acidic conditions:

4H⁺(aq) + MnO₂(s) + 2F²⁺(aq) ------> Mn²⁺(aq) + 2Fe³⁺(aq) + 2H₂O(l)

Answer 1

The half reaction of reduction in the process is: MnO_² + 4H⁺ + 2e^- --> Mn²⁺ + ₂H₂O; Eº=1.23V

The half reaction of oxidation in the process is: Fe³⁺ + 1e^- --> Fe²⁺; Eº= 0.77V

In order to form the full reaction we must subtract twice the latter half reaction, so that the stechiometric coefficients match:

Reaction 1: MnO_² + 4H⁺ + 2e^- --> Mn²⁺ + ₂H₂O. This reaction has a DGº₁

Reaction 2: 2Fe²⁺ --> 2Fe³⁺ + 2e^-. This reaction has a -2DGº₂

Condensated Reaction: MnO_² + 4H⁺ + 2Fe²⁺ --> Mn²⁺ + 2H₂O + 2Fe³⁺. This reaction has a DGº₃.

Therefore, we can express DGº3 as DGº1-2DGº2.

Now remember that DGº= -nFEº, and also DGº=-RT ln(K_eq), so...

(-8.314J/mol)(200K) ln(K_eq) = -2(96485C)(1.23V) - [-1(96485C)(0.77V)]

ln(K_eq)= -163059.65 / (-8.314(200)) = 98.0632.

K_eq= e^98.0632 = 3.876x10⁴².