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Problems in relating ∆G with Galvanic Cell (Electrode Potential)
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Cynonian_raj:
For the Reaction Fe^+3 + 3e^- = Fe, E° is -0.036V and E° for Fe^+3 +e^- = Fe^+2 is 0.771V.
Calculate E° for Fe^+2 +2e^- = Fe.
In solution according to Book,
Let Fe^+2 +2e^- = Fe -----------(1)
Firstly, find ∆G for each equation with given E° and then combine equation (and ∆G also) . After combining equation, we get required equation and ∆G for the required equation and then use simply for ∆G = -nFE
And we get required E°.
Here my problem is -
1. How do I get equation (1) after combining those equation (Like mathematics)
How does ot work.
2. How we get ∆G for equation (1) after combining those equation's ∆G.
**Is there any alternate ways for solving this question**
Vidya:
Fe+3 + 3e^- ---->Fe ---- (2) -0.036V
Fe+3 +e^- -----> Fe+2 ----(3) 0.771V.
Reverse equation 3 and add it in equation 2 to get equation 1 .As you reverse equation 3 its cell potential value will become negative.
Now add cell potential values to get cell potential for equation 1
Now ΔG°= nFE°cell
Plug in n = electrons transferred in that reaction
F= 96500C
E°cell calculated above
you will get ΔG° for equation
Similarly you can calculate ΔG° for equation 2 and 3
When you flip equation3, ΔG° sign will also flipped.
Add two ΔG° values to get ΔG° for equation 1
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