The percent iron in iron ore can be determined by dissolving the ore in acid, th

Question

The percent iron in iron ore can be determined by dissolving the ore in acid, then reducing the iron to Fe+, and finally titrating the Fe2+ with aqueous KMnO4. The reaction products are Fe3+ and Mn2+. The standard free energy change can be determined from the E of the redox reaction and the number of electrons involved in the reaction. The potential for the oxidation half reaction is - 0.77 volts. The potential for reduction half reactions is 1.51 volts.

Question: There are ______ moles of electrons transferred in the balanced equation and the equilibrium constant for the reaction is ____________. The number of moles of iron is____________moles.

This question can be seen step-by-step in the Textbook Solutions for chemistry, chapter 20, Problem 151 but I can't figure out what the answers are for the questions above.

Explanation / Answer

The balanced reaction

5Fe+2 + MnO4- + 8H+ = 4H2O + 5Fe+3 + Mn2+

then...

there are

Fe+2 to Fe+3 = 1 electron, x 5 coefficeint = 5 electrons

The Kconstant > 1, since this is 1.51V

E°cell = 1.51 - 0.77 = 0.74 V

dG = -n*F*Ecell = -5*96500 * 0.74 = - 357050

dG = -RT*lnK = -8.314*298*ln(K)

-8.314*298*ln(K) = -357050

k = exp(357050/(8.314*298))

K = 3.867*10^62

mol of Iron when balanced, is 5

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