August 10, 2024, 10:51:18 PM
Forum Rules: Read This Before Posting

### Topic: Activity of a gas, when not to take it into account ? Two phases system  (Read 596 times)

0 Members and 1 Guest are viewing this topic.

#### niobium

• New Member
• Posts: 6
• Mole Snacks: +0/-0
##### Activity of a gas, when not to take it into account ? Two phases system
« on: July 18, 2024, 09:10:29 AM »
"You construct a galvanic cell using the two half cells below:
$$\ce{Au^+}_{(aq)} + 2 e^- => \ce{Au}_{(s)}$$
and
$$\text{chlorine}_{(g)} + 2 e^- => 2 \ce{Cl^-}_{(aq)}$$

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

--------------------------------------------------

The overall oxidation-reduction reaction of this exercise is :
$$2 \ce{Au^+}_{(aq)} + 2 \ce{Cl^-}_{(aq)} => \text{chlorine}_{(g)} + 2 \ce{Au}_{(s)}$$

Here $=>$ stands for a forward reaction.

The given answer given in this exercise is $\Delta E =0.25V$

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.
$Q = \frac{1}{0.10^2 \times 0.50^2} =4.0 \times 10^2$

Then I simply use the Nernst equation (knowing by the tables that $\Delta E^{\circ} = 0.332 V$, with two significant figures)
$$\Delta E = 0.332 - \frac{8.314\times 298 \times \ln(4.0 \times 10^2)}{2\times 96 485} =0.26 V$$

I find $0.26V$ which is very close to the given answer ($0.25V$).

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.

#### Borek

• Mr. pH
• Deity Member
• Posts: 27757
• Mole Snacks: +1804/-411
• Gender:
• I am known to be occasionally wrong.
##### Re: Activity of a gas, when not to take it into account ? Two phases system
« Reply #1 on: July 18, 2024, 09:41:39 AM »
IMHO the answer given is wrong and you are right, that the activity of the chlorine (using given pressure) should be taken into account.

The way the problem is solved suggests changing the partial pressure of the chlorine doesn't change the voltage, which is definitely off.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### niobium

• New Member
• Posts: 6
• Mole Snacks: +0/-0
##### Re: Activity of a gas, when not to take it into account ? Two phases system
« Reply #2 on: July 18, 2024, 09:50:46 AM »
Thank you so much for your answer. I would not be the first time there's an error in the exercises.