Question

# use the appropriate standard reduction potentials to determine the equlibrium constant at 200.00 K for the...

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)

#### Homework Answers

Answer #1

The half reaction of reduction in the process is: MnO2 + 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: MnO2 + 4H+ + 2e- --> Mn2+ + 2H2O. This reaction has a DGº1

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

Condensated Reaction: MnO2 + 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(Keq), so...

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

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

Keq= e^98.0632 = 3.876x1042.

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