What is ΔG°rxn for the cell reaction for each reaction?

Question

What is ΔG°rxn for the cell reaction for each reaction?

Anode: Cr(s) → Cr⁺³(aq) E_red= -0.7237V

Cathode: I₂(aq) → I^-1(aq) E_red = +0.5182V

Anode: NO₃^-1(aq) → HNO₂ (aq) E_red = +0.9513V

Cathode: Au⁺³(aq) → Au(s) E_red = +1.5219V

Anode: ClO₄^-1(aq) → Cl₂(aq) E_red = +1.1928V

Cathode: Cr₂O₇^-2(aq) → Cr⁺³(aq) E_red= +1.2557V

Answer 1

Recall that oxidation takes place at the anode, and reduction takes place at the cathode.

Recall also, that dealing with reduction potentials, the Eº_cell = cathode - anode.

Finally, recall that to find the ∆Gº of the cell, we will use ∆Gº = -nFEº_cell

I'll do the first one, see if you can do the others. If you don't know how to find the number of electrons transferred, that's a separate issue and you should review balancing redox reactions (in acid and base)

Anode: Cr(s) → Cr⁺³(aq) E_red= -0.7237V

Cathode: I₂(aq) → I^-1(aq) E_red = +0.5182V

Eº_cell = cathode - anode = 0.5182 - (-0.7237) = 1.2419 V

∆Gº = -nFE

Balanced redox will transfer 6 mols electrons, so n = 6

F = 96,500 Coulombs / mol = 96.500 J / Vmol

E = 1.2419 V

∆Gº = -(6 mol)(96,500 J / Vmol)(1.2419 V)

∆Gº = 719,060 J = 719.1 kJ

Anode: NO₃^-1(aq) → HNO₂ (aq) E_red = +0.9513V

Cathode: Au⁺³(aq) → Au(s) E_red = +1.5219V

Eº_cell = cathode - anode = 1.5219 - 0.9513 = 0.5706 V

n = 6 mols electrons

Anode: ClO₄^-1(aq) → Cl₂(aq) E_red = +1.1928V

Cathode: Cr₂O₇^-2(aq) → Cr⁺³(aq) E_red= +1.2557V

E_cell = cathode = anode = 1.2557 - 1.1928 = 0.0629