(very) nonstandard reduction potential

[kankerfist]

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.
Question
What is the reduction potential required for reducing the {VO}^{2+} concentrations to {10}^{-6} M at pH = 2 at room temperature in the presence of {10}^{-4} M {V}^{3+} ?

{VO}^{2+}+2{H}^{+}+{e}^{-} -> {V}^3{+} + {H}_{2}O

{E}^{o}=0.34V

Relevant Equations

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

Attempted Solution

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

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.

 

[Shadowz]

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.

 

[kankerfist]

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