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(very) nonstandard reduction potential

  1. Apr 26, 2007 #1
    I have a question that I am familiar with, but there is no answer provided and about 1000 ways to make small mistakes. I was hoping someone could take a glance and see if any mistakes jump out. Any tips would really be appreciated.
    What is the reduction potential required for reducing the [tex]{VO}^{2+}[/tex] concentrations to [tex]{10}^{-6}[/tex] M at pH = 2 at room temperature in the presence of [tex]{10}^{-4}[/tex] M [tex]{V}^{3+}[/tex] ?

    [tex]{VO}^{2+}+2{H}^{+}+{e}^{-}[/tex] -> [tex]{V}^3{+} + {H}_{2}O[/tex]


    Relevant Equations

    E={E}^{o}-\frac{RT}{nF}*\ln [\frac{{Products}^{sc}}{{Reactants}^{sc}}]
    R=8.314 VC
    F=96,500 C
    T = 298 K
    n= electrons transferred

    Attempted Solution

    E={0.34}-\frac{(8.314)(298)}{(1)(96500)}*\ln [\frac{{10}^{-4}}{({10}^{-5}){(.01)}^{2}}]

    Which is a result of pH = 2 meaning that the H+ concentration is 0.01. I left out units because LaTex got too confusing with units involved. My answer from the above attempted solution is a reduction potential of -0.0147 V.

    The question then asks what would happen to the above if NaOH was added to solution. I took this to mean that NaOH fully dissolved into Na+ and OH-, increasing pH. But the question asks if the reduction would be made easier or harder... I know that lower H+ concentrations mean a more negative reduction potential, but I'm not sure how that correlates to the easiness or difficulty of reduction.
  2. jcsd
  3. Apr 27, 2007 #2
    I am not familiar with using natural log (ln), so I usually change it to log
    alnb = aln10logb

    I wonder why in the natural log expression, the value for concentration of VO2+ is 10E-05 instead of 10E-06?

    I think you wrote the wrong reaction:
    First, in VO2+ V is +6, in V3+, V is +3
    So you write VO2+ + 3e --> V3+
    Left reaction has O, while right does not, then you add H2O to the right.
    VO2+ + 3e ---> V3+ + H2O
    Then balance the O
    VO2+ + 3e ---> V3+ 2H2O
    Then add H+ to the right
    VO2+ 4H+ + 3e --> V3+ + 2H2O

    Hope that will help you.
  4. Apr 28, 2007 #3
    It is 10^-6! That little typo caused me all kinds of headache. Also, your balanced half reaction is exactly how I would have done it too, but the one above was provided by my professor and is a valid half cell reaction. I'm not exactly sure why it is that way, but I double checked on ChemOffice and found the above reaction and also verified its standard energy of -0.337 V. I'll ask my professor why only 1 e- is involved when V is reduced by 3 and let you know what he says. Thanks for pointing out that typo, I probably would have never noticed it
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