# use tabulated half-cell potentials to calculate deltaGrxn for each of the following reactions at 25c?

use tabulated half-cell potentials to calculate deltaGrxn for each of the following reactions at 25c?

Ok im not sure how to do this, but heres what I thought, you get the half life reactions and then times them by the coeficents?

Heres the questions:

A) 2Fe^3+(aq)+3Sn(s)-->2Fe(s)+3Sn^2+(aq)

B) O2(g)+2H2O(l)+2Cu(s)-->4OH^-(aq)+2Cu^2+(aq)

C) Br2(l)+2I^-(aq)-->2Br^-(aq)+I2(s)

by
selected by

This is a one question at a time deal here. I'll do one of these and that should point you to the method of doing the others.

2 Fe³⁺(aq) + 3 Sn(s) → 2 Fe(s) + 3 Sn²⁺(aq)

First look up the half-cell potentials (I'm using those linked below):

There is no reduction potential in the list for the iron half cell reduction. However, there are two you can use to get it:

Fe³⁺ + e⁻ ⇌ Fe²⁺ . . . . . E⁰ = + 0.77 V . . . . (n = 1, nE⁰ = + 0.77 V)

Fe²⁺ + 2e⁻ ⇌ Fe(s) . . . E⁰ = – 0.44 V . . . . (n = 2, nE⁰ = – 0.88 V)

––––––––––––––––––––––––––––––––––––– . . . ––––––––––––––––––––

Fe³⁺ + 3e⁻ ⇌ Fe(s) . . . E⁰ = – 0.037 V . ⇐ . (n = 3, nE⁰ = – 0.11 V)

What I'm doing here is adding the nE⁰ 's, NOT the E⁰ 's. Remember that –nFE⁰ = ∆G⁰ so I'm adding up quantities that are directly proportional to free energies (I can always add up thermodynamic state function quantities – just like adding up ∆H's using Hess's Law.)

That was the tricky part, the rest is easy:

Sn²⁺ + 2e⁻ ⇌ Sn(s) . . . E⁰ = −0.13 V (The sign switches in the final reaction.)

2 Fe³⁺(aq) + 3 Sn(s) → 2 Fe(s) + 3 Sn²⁺(aq) . . . . E⁰(cell) = 0.13 – 0.037 V = 0.093 V ;

Note: n = 6 (total # of moles of electrons transferred in the reaction as written.)

∆G⁰(rxn) = –nFE⁰(cell) = –6(98.485 kJ/V)(0.093 V) = –54.9 kJ

Good luck on the other two examples.

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.
by
A) -60 kJ

B) -20 kJ

C) -110 kJ
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.