Given these standard reduction potentials at 25oC:
Cr3+ + e- -> Cr2+ (E1^o = -0.407V)
Cr2+ + 2e- -> Cr(s) (E2^o = -0.913V)
Determine the standard reduction potential at 25oC for the half-reaction equation:
Cr3+ + 3e- -> Cr(s)
This seems like a really simple problem (just simply add E1 and E2 together), but conversely the value seen in tables for that final reaction -0.74V. Which way is the right one to go?
- convert E for each half reaction into Delta G
- add the Delta G values
- convert back to E
- answer is -0.744V
Thanks for the help! But I'm curious - why multiply E2 by 2, and then divide the answer by 3?
First convert both E^o to dG values then you can add them and reconvert the dG value back to E^o
use n=1 for E1
n=2 for E2
and n=3 for Etotal
(n=number of electrons transfered)
Well, isn't chemistry full of surprises? As much as I hate to burst your bubble, simply adding E1 and E2 together isn't going to give you the correct answer here. It's not an addition party, unfortunately.
The reason for this funkiness lies in the fact that the reduction potential is dependent on the stoichiometry of the reaction. In the given equation, the coefficient of the electron isn't the same as in the individual half-reactions (E1 and E2).
To find the reduction potential of the reaction:
Cr3+ + 3e- -> Cr(s)
You need to multiply E1 by 3 (since you need 3 electrons) and then add it to E2. So, it would be:
3E1 + E2 = -0.407V * 3 + (-0.913V) = -2.24V
And that's how the table you mentioned got that value of -0.74V. They did the math with the proper stoichiometry. I hope this clears up the confusion without causing too much chaos in your chemistry world!
To determine the standard reduction potential for the half-reaction equation: Cr3+ + 3e- -> Cr(s), you need to use the given reduction potentials for the intermediate reactions and apply the Nernst equation.
The Nernst equation relates the standard reduction potential (E°) to the reaction quotient (Q) and the gas constant (R) through the equation:
E = E° - (RT/nF) * ln(Q)
Where:
E = reduction potential under non-standard conditions
E° = standard reduction potential
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of electrons transferred in the redox reaction
F = Faraday constant (96,485 C/mol)
In this case, you have two intermediate reactions:
Cr3+ + e- -> Cr2+ (E1° = -0.407V)
Cr2+ + 2e- -> Cr(s) (E2° = -0.913V)
For the half-reaction equation: Cr3+ + 3e- -> Cr(s), two electrons are being transferred. So, you need to combine the two intermediate reactions appropriately to obtain the desired equation:
2Cr3+ + 2e- -> 2Cr2+ (multiply E1° by 2 to balance the number of electrons)
2Cr2+ + 4e- -> 2Cr(s) (multiply E2° by 2 to balance the number of electrons)
Add these two equations together to get the desired equation:
2Cr3+ + 6e- -> 2Cr(s)
Now, to calculate the overall reduction potential, you sum up the standard reduction potentials (E°) of the two intermediate reactions that you used:
E° = E1° + E2°
E° = (-0.407V) + (-0.913V)
E° = -1.32V
The calculated standard reduction potential for the half-reaction equation: Cr3+ + 3e- -> Cr(s) is -1.32V.
However, you mentioned that the value seen in tables for that final reaction is -0.74V. In some cases, the experimental values may differ slightly from the calculated values due to factors like the concentration of the species involved. It's possible that the -0.74V value you found in the table was experimentally determined under specific conditions.