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 = ?
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?
2 answers
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
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
1 answer