Redox reactions

anteater85

Full Member
10+ Year Member
Jul 2, 2009
84
1
    Can someone help me with this:
    Given that the E0 for Au+3/AU is 1.5 V and the E0 for Li+/Li is -3.05 V. what is Ecell for the reaction: Au3+ +3 Li-->Au + 3 Li

    if [Li+]= 10 M and [Au3+] = 0.01 M

    Thank you
     

    nze82

    Full Member
    May 15, 2009
    1,922
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    CA
    1. Dental Student
      Can someone help me with this:
      Given that the E0 for Au+3/AU is 1.5 V and the E0 for Li+/Li is -3.05 V. what is Ecell for the reaction: Au3+ +3 Li-->Au + 3 Li

      if [Li+]= 10 M and [Au3+] = 0.01 M

      Thank you
      A) Write the two half reactions:

      Au3+ + 3e- --> Au, Eo = +1.5V

      Li+ + 1e- --> Li, Eo = -3.05

      Which half reaction has a higher reduction potential? The first one.
      So, the 2nd reaction must be the one that is oxidized. So, let's flip the reaction and the sign of Eo for this reaction:

      Au3+ + 3e- --> Au, Eo = +1.5V

      Li --> Li+ + 1e-, Eo = +3.05

      B) Multiply the two reaction with proper coefficients, so that you can cancel out the electrons (DO NOT multiply the reduction potential by these coefficients):


      1(Au3+ + 3e- --> Au), Eo = +1.5V

      3(Li --> Li+ + 1e-), Eo = +3.05

      **************************
      Au3+ + 3Li --> Au + 3Li+, Eo = 1.5 + 3.05 = 4.55V

      C) Use the Nernst equation:

      n = 3 (# of electrons)
      Q = [Li+]^3/[Au3+]^1


      E = Eo - (0.0591/n)logQ

      = 4.55 - (0.0591/3)log(10^3/0.01) = +4.45V


       
      Last edited:

      anteater85

      Full Member
      10+ Year Member
      Jul 2, 2009
      84
      1
        A) Write the two half reactions:

        Au3+ + 3e- --> Au, Eo = +1.5V

        Li+ + 1e- --> Li, Eo = -3.05

        Which half reaction has a higher reduction potential? The first one.
        So, the 2nd reaction must be the one that is oxidized. So, let's flip the reaction and the sign of Eo for this reaction:

        Au3+ + 3e- --> Au, Eo = +1.5V

        Li --> Li+ + 1e-, Eo = +3.05

        B) Multiply the two reaction with proper coefficients, so that you can cancel out the electrons (DO NOT multiply the reduction potential by these coefficients):


        1(Au3+ + 3e- --> Au), Eo = +1.5V

        3(Li --> Li+ + 1e-), Eo = +3.05

        **************************
        Au3+ + 3Li --> Au + 3Li+, Eo = 1.5 + 3.05 = 4.55V



        THank you
         
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