How Do You Calculate E°cell Values for Galvanic Reactions?

[salman213]

I needed to find the E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(s)

So E°cell = Ecathode - Eanode

I made the half reactions

Fe3+ (aq) ---> Fe2+(aq)
EØ = +0.77 V (i found this reaction in my table)

IO3-(aq) ---> I2(s)
balanced the 2nd one to

12 H + 2IO3 + 5e- ---> I2 + 6H2O
EØ = I cannot find this value in that table
I FOUND IT ONLINE ON A WEBSITE TO BE EØ = +1.19V, HOW DID THEY GET THIS VALUE! Anyone know?

 

[lightarrow]

I needed to find the E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(s)

So E°cell = Ecathode - Eanode

I made the half reactions

Fe3+ (aq) ---> Fe2+(aq)
EØ = +0.77 V (i found this reaction in my table)

IO3-(aq) ---> I2(s)
balanced the 2nd one to

12 H + 2IO3 + 5e- ---> I2 + 6H2O
EØ = I cannot find this value in that table
I FOUND IT ONLINE ON A WEBSITE TO BE EØ = +1.19V, HOW DID THEY GET THIS VALUE! Anyone know?

2IO3-(aq) + 10Fe2+(aq) + 12H+ --> 10Fe3+(aq) + I2(s) + 6H2O;
E° = +1.19 - 0.77 = +0.42V.

About the reaction:

(1) 2IO3-(aq) + 12 H+ + 10e- ---> I2(s) + 6H2O

You can find its E° if you know the E° of other similar reactions from which you can get that one, for example:

(2) IO3-(aq) + 6H+ +6e- --> I- + 3H2O;_______>E°(2) = +1.08V
(3) I2(s) + 2e- --> 2I- ;____________________>E°(3) = +0.53V

Reaction (1) is given by: 2*Reaction(2) - Reaction(3) and so:

Delta G°(1) = 2*Delta G°(2) - Delta G°(3)

Now you use the equation: Delta G° = -nF*Delta E° where n is the number of electrons of the reaction, so you have:

-10F*Delta E°(1) = 2*[-6F*Delta E°(2)] - [-2F*Delta E°(3)]

10*Delta E°(1) = 12*Delta E°(2) - 2*Delta E°(3)

Delta E°(1) = (1/10)[12*Delta E°(2) - 2*Delta E°(3)] =

= (1/5)[6*Delta E°(2) - Delta E°(3)] =

= (1/5)[6*1.08 - 0.53] = +1.19V.

 

[salman213]

wowi would never have that of that...:S


how is the delta G of the first equation equal to the delta G if the second and third with those operations. Like how do we know to relate delta G?

 

[lightarrow]

wowi would never have that of that...:S


how is the delta G of the first equation equal to the delta G if the second and third with those operations. Like how do we know to relate delta G?

Do you agree on the fact that Reaction (1) is given by 2*Reaction(2) - Reaction(3) ?

2*Reaction(2) = 2IO3-(aq) + 12H+ +12e- --> 2I- + 6H2O

- Reaction(3) = 2I- --> I2(s) + 2e-

Now you sum them:

2IO3-(aq) + 12H+ +12e- + 2I- --> 2I- + 6H2O + I2(s) + 2e-

you simplify canceling 2I- and 2e- on both members:

2IO3-(aq) + 12H+ +10e- --> 6H2O + I2(s).


About summing ΔG: if you have

(1) A --> B you know that ΔG(1) = G(B) - G(A)

(2) C --> D you know that ΔG(2) = G(D) - G(C)

(3) A + C --> B + D you know that ΔG(3) = G(B + D) - G(A + C)

But G(B + D) - G(A + C) = G(B) + G(D) - G(A) - G(C) =

= G(B) - G(A) + G(D) - G(C) = ΔG(1) + ΔG(2)