May 30, 2020, 11:56:02 PM
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Topic: Problems in relating ∆G with Galvanic Cell (Electrode Potential)  (Read 279 times)

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Offline Cynonian_raj

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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**

Offline Vidya

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Re: Problems in relating ∆G with Galvanic Cell (Electrode Potential)
« Reply #1 on: August 29, 2019, 12:59:09 AM »
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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