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Calculating the Equilibrium constant of a redox reaction from its half reactions

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the value of the equilibrium constant at 25 degrees celsius for the following reaction:
    5Fe2+ + MnO4- + 8H+ ----) 5Fe3+ + Mn2+ + 4H2O

    2. Relevant equations
    The equation above is the product of these redox reactions:
    1) MnO4- + 8H+ + 5e- ----) Mn2+ + 4H2O E= 1.507 V
    2) Fe3+ + e- -----) Fe2+ E= 0.770 V


    3. The attempt at a solution
    E rxn = (-5)(0.770V) + (1.507 V) = -2.343 V
    log K = 5(-2.343)/0.0592 = -197.88
    10^-197.88 = MATH ERROR

    I really have no clue what I am doing wrong, reaction 2 is reversed and multiplied by 5 to balance out equation 1), so naturally its E value should b multiplied by a coefficient of -5. Any help would be appreciated, the answer given in the back of the textbook is 1.8 x 10^62.
     
  2. jcsd
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