"You construct a galvanic cell using the two half cells below:

[tex]\ce{Au^+}_{(aq)} + 2 e^- => \ce{Au}_{(s)} [/tex]

and

[tex] \text{chlorine}_{(g)} + 2 e^- => 2 \ce{Cl^-}_{(aq)} [/tex]

The initial concentration of gold ions is [itex]0.10 M[/itex] and the one of chlorine ions is [itex]0.50 M[/itex]. The initial pressure of chlorine is [itex]1.50[/itex] atm at [itex]25°C[/itex]. Using the table below, calculate the initial voltage of the galvanic cell."

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The overall oxidation-reduction reaction of this exercise is :

[tex]2 \ce{Au^+}_{(aq)} + 2 \ce{Cl^-}_{(aq)} => \text{chlorine}_{(g)} + 2 \ce{Au}_{(s)} [/tex]

Here [itex] => [/itex] stands for a forward reaction.

The given answer given in this exercise is [itex] \Delta E =0.25V [/itex]

The only way I could arrive at this numerical value, is by not taking into account the activity of the chlorine in the reaction quotient.

Here's the detail:

First, I've computed the reaction quotient at the initial state.

[itex] Q = \frac{1}{0.10^2 \times 0.50^2} =4.0 \times 10^2[/itex]

Then I simply use the Nernst equation (knowing by the tables that [itex] \Delta E^{\circ} = 0.332 V[/itex], with two significant figures)

[tex] \Delta E = 0.332 - \frac{8.314\times 298 \times \ln(4.0 \times 10^2)}{2\times 96 485} =0.26 V[/tex]

I find [itex] 0.26V[/itex] which is very close to the given answer ([itex] 0.25V[/itex]).

My question is then, why *does it work* when we don't count the chlorine's activity ?

My modest sketch of an answer to this question is that chlorine is not miscible with water, hence it lives in another phase as the ions. Is this starting clue right ?

My instinct (weren't this answer be given) would have been to include chlorine's activity (i.e. it's partial pressure, 1.5 atm) in the reaction quotient.