# (very) nonstandard reduction potential

1. Apr 26, 2007

### 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

$$E={E}^{o}-\frac{RT}{nF}*\ln [\frac{{Products}^{sc}}{{Reactants}^{sc}}]$$
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. Apr 27, 2007

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