Using the following data
1. CU3+ + 2e- ---> Cu+ E1= 1.28V
2. CU2+ + e- ---> Cu+ E2= 0.15V
3. Cu2+ + 2e- ----> Cu(s) E3= 0.34V
3. Cu+ + e- ----> Cu (s) E4= 0.52V
calculate the standard reduction potential for the reaction of Cu(III) to Cu(II).
Is it .783 V?
Cu³⁺ + 2e⁻ —> Cu⁺ dG°1=-2FE₁ E₁= 1.28 V
Cu⁺ — > Cu²⁺ + e⁻ dG°2=-1FE₂ E₂= 0.15 V
_____________________________________________
Cu³⁺ + e⁻ —> Cu²⁺ dG°final= F(E₂ - 2E₁)
E°final = -dG°final/ (nF)= [F(E₂ - 2E₁)]/(nF)= [0.15 v - 2*1.28v]/1=2.41 v
Cu3+------Cu2+-----Cu+-----Cuo
...........|...0.15.|..0.52..|
|......1.28.........|
I hope the spacing comes out well enough that you can make it out. The idea here is that you have the total of voltage from 3+ to 1+ and you have the single leg of Cu2+ to Cu+. What you need to calculate is the other leg of Cu3+ ==> Cu2+.
It's mostly a little algebra.
Cu3+ +e ==> Cu2+ E = x
Cu2+ + e ==> Cu+ E = 0.15
-------------------------
sum Cu3+ + 2e ==> Cu+ E = 1.28
x + 0.15 = 2*1.28
x = ?
Well, let's put on our thinking caps and do some calculations, shall we?
To find the standard reduction potential for the reaction of Cu(III) to Cu(II), we need to combine the reduction potentials of the reactions involving those ions.
Since Cu(III) isn't directly mentioned in the given data, we need to use some oxidation-reduction trickery. Bear with me!
We know that Cu(III) can be reduced to Cu(II) by either gaining 1 electron (reaction 1) or 2 electrons (reaction 3). So we can write two half-reactions for Cu(III) reduction:
1. Cu(III) + e- -> Cu(II)
2. Cu(III) + 2e- -> Cu(I)
Now, here comes the fun part. We'll use the reduction potentials from the given data to determine the overall potential for reaction 1:
E1 (Cu3+ -> Cu+) = 1.28 V
E2 (Cu2+ -> Cu+) = 0.15 V
To find the overall potential for reaction 1, we add these two potentials:
E1 + E2 = 1.28 V + 0.15 V = 1.43 V
Now, for reaction 2:
E3 (Cu2+ -> Cu) = 0.34 V
E4 (Cu+ -> Cu) = 0.52 V
To find the overall potential for reaction 2, we add these two potentials:
E3 + E4 = 0.34 V + 0.52 V = 0.86 V
But we're not done yet! We need to subtract the overall potential of reaction 2 from the overall potential of reaction 1. This is because we want to know the reduction potential for Cu(III) to Cu(II), not Cu(III) to Cu(I).
So:
1.43 V - 0.86 V = 0.57 V
So, based on my calculations, the standard reduction potential for the reaction of Cu(III) to Cu(II) is 0.57 V. Not quite as exciting as 0.783 V, but still pretty electrifying, don't you think?
To calculate the standard reduction potential for the reaction of Cu(III) to Cu(II), we need to add the reduction potentials for the two individual half-reactions involved.
Given data:
1. Cu3+ + 2e- ---> Cu+ E1 = 1.28V
2. Cu2+ + e- ---> Cu+ E2 = 0.15V
To find the standard reduction potential for the reaction of Cu(III) to Cu(II), we only need to consider the reduction potential for reaction 1, as reaction 2 involves the same species on both sides and does not contribute to the overall cell potential.
Thus, the standard reduction potential for the reaction of Cu(III) to Cu(II) is E1 = 1.28V.
Therefore, the correct answer is 1.28 V, not 0.783 V.
To calculate the standard reduction potential for the reaction of Cu(III) to Cu(II), you can use the Nernst equation and the given reduction potentials for the reactions involving Cu(III), Cu(II), and Cu.
The Nernst equation relates the standard reduction potential (E°) to the actual reduction potential (E) under non-standard conditions:
E = E° - (0.0592 / n) * log(Q)
Where:
E = actual reduction potential
E° = standard reduction potential
n = number of electrons transferred in the reaction
Q = reaction quotient
First, let's identify the reactions involved:
Cu3+ + 2e- → Cu+ (Reaction 1)
Cu2+ + e- → Cu+ (Reaction 2)
Cu2+ + 2e- → Cu(s) (Reaction 3)
Cu+ + e- → Cu(s) (Reaction 4)
Now, let's construct the overall reaction:
Cu3+ + e- → Cu2+ (Reaction 5)
To determine the E° for Reaction 5, we need to use the reduction potentials of Reactions 1 and 2. We can add the individual reactions to obtain the overall reaction and their reduction potentials to obtain the overall reduction potential.
Reaction 1 (-0.15V) + Reaction 2 (-0.15V) = Reaction 5
This simplifies to:
Cu3+ + 3e- → Cu2+
Now, let's calculate the E° for Reaction 5 using the standard reduction potentials for Reactions 1 and 2:
E° = E1 + E2
E° = 1.28V + 0.15V
E° = 1.43V
Therefore, the standard reduction potential for the reaction of Cu(III) to Cu(II) is 1.43V, not 0.783V.