# 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)

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.

A) -60 kJ

B) -20 kJ

C) -110 kJ