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Electrode Potentials
I tried these questions on Redox Equilibria; ELECTRODE POTENTIALS and I just wanna know if the answers are correct
QUESTION:
“Show whether disproportionation will occur in the following cases under standard conditions”
a) VO2+ = VO2+ and V3+
b) MnO2 = MnO4- and Mn2+
MY ANSWERS: (the = signs represent equilibrium arrows)
a) VO2+ = VO2+ and V3+
using the equations:
VO2+ + 2H+ + e- = V3+ + H2O (E = +0.34V)
VO2+ + 2H+ + e- = VO2+ + H2O (E = +1.00V)
Which then makes;
V3+ + H2O + VO2+ + 2H+ + e- = VO2+ + 2H+ + e- + VO2+ + H2O
Simplified down to:
V3+ + VO2+ = 2VO2+
And the E nought value for this is 1.00 – (-0.34) = +1.34V
So the answer for a) is that a disproportionation reaction is feasible, but it is the exact opposite reaction than the one thought to occur (in the question);
Instead of
VO2+ = VO2+ and V3+
We have
V3+ + VO2+ = 2VO2+
b) MnO2 = MnO4- and Mn2+
using the equations
MnO2 + 4H++ 2e- = Mn2+ + 2H2O (E = +1.23V)
MnO4- + 8H+ + 5e- = Mn2+ + 4H2O (E = +1.51V)
MnO4- + 4H+ + 3e- = MnO2 + 2H2O (E = +1.70V)
(I used the first one and the third one, and multiplied them accordingly so the number of electrons were equal)
Which then makes:
3Mn2+ + 6H2O + 2MnO4- + 8H+ + 6e- = 3MnO2 + 12H+ + 6e- + 2MnO2 + 4H2O
Simplified down to:
3Mn2+ + 2MnO4- + 2H2O = 5MnO2 + 4H+
And the E nought value for this is 1.70 – 1.23 = +0.47
So once again the answer for b) is that a disproportionation reaction is feasible, but it is the exact opposite reaction than the one thought to occur (in the question):
Instead of
MnO2 = MnO4- and Mn2+
We have
3Mn2+ + 2MnO4- = 5MnO2
Are these right? (thanx for reading this btw)
QUESTION:
“Show whether disproportionation will occur in the following cases under standard conditions”
a) VO2+ = VO2+ and V3+
b) MnO2 = MnO4- and Mn2+
MY ANSWERS: (the = signs represent equilibrium arrows)
a) VO2+ = VO2+ and V3+
using the equations:
VO2+ + 2H+ + e- = V3+ + H2O (E = +0.34V)
VO2+ + 2H+ + e- = VO2+ + H2O (E = +1.00V)
Which then makes;
V3+ + H2O + VO2+ + 2H+ + e- = VO2+ + 2H+ + e- + VO2+ + H2O
Simplified down to:
V3+ + VO2+ = 2VO2+
And the E nought value for this is 1.00 – (-0.34) = +1.34V
So the answer for a) is that a disproportionation reaction is feasible, but it is the exact opposite reaction than the one thought to occur (in the question);
Instead of
VO2+ = VO2+ and V3+
We have
V3+ + VO2+ = 2VO2+
b) MnO2 = MnO4- and Mn2+
using the equations
MnO2 + 4H++ 2e- = Mn2+ + 2H2O (E = +1.23V)
MnO4- + 8H+ + 5e- = Mn2+ + 4H2O (E = +1.51V)
MnO4- + 4H+ + 3e- = MnO2 + 2H2O (E = +1.70V)
(I used the first one and the third one, and multiplied them accordingly so the number of electrons were equal)
Which then makes:
3Mn2+ + 6H2O + 2MnO4- + 8H+ + 6e- = 3MnO2 + 12H+ + 6e- + 2MnO2 + 4H2O
Simplified down to:
3Mn2+ + 2MnO4- + 2H2O = 5MnO2 + 4H+
And the E nought value for this is 1.70 – 1.23 = +0.47
So once again the answer for b) is that a disproportionation reaction is feasible, but it is the exact opposite reaction than the one thought to occur (in the question):
Instead of
MnO2 = MnO4- and Mn2+
We have
3Mn2+ + 2MnO4- = 5MnO2
Are these right? (thanx for reading this btw)
mik1w
for (b), if the reverse reaction is feasible then the forward reaction will have the negative of the E0 value and not be feasible. I don't think the back reaction is disproportionation either. So for (b), disproportionation is not feasible. I think.. I have only skimmed it
hang on if the backward reaction is +0.47 then it must be feasible (unlike forward reaction)
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