kankerfist
Apr26-07, 12:46 PM
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.
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.