Question

Determine the free energy(ΔG) from the standard cell potential (E_cell⁰ ) for the reaction:

2ClO₂^-(aq)+Cl₂(g)→2ClO₂(g)+ 2Cl^-(aq)

where:

ClO₂+e^-→ClO₂^-E_red⁰ =+0.954 V.
Cl₂+2e^-→2Cl^-E_red⁰ =+1.36 V.

ΔG=+79 kJ

ΔG=-790 kJ

ΔG=-79 kJ

ΔG=-0.79 kJ

Answer 1

                           

Given equation is 2ClO₂^- (aq) + Cl₂(g)→ 2ClO₂(g) + 2Cl^-(aq)

Oxidation (at anode) : 2ClO₂^- (aq) → 2ClO₂(g)

Reduction ( at cathode) : Cl₂(g) → 2Cl^-(aq)

E^o_cell = E^o_cathode - E^o_anode

          = +1.36 - (+0.954) V

          = 0.406 V

We know that G^o = -nFE^o_cell

Where

n = number of moles of electrons transferred = 2 mol

F = Faraday = 96500 Col

Plug the values we get G^o = -79.0x10³ J

                                 G^o = -79.0 kJ