Homework Statement

In the course of writing out the problem and solving it, I figured out my original question (which turned out to be caused by a mistake I made in literally the final equation in the whole solution). Therefore, this post doesn't really have a question in it at this point. It is simply my whole solution.

You construct a galvanic cell using the two half cells below

$$\mathrm{Au^+(aq)+e^-\rightarrow Au(s)}$$

$$\mathrm{Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)}$$

The initial concentration of ##\mathrm{Au^+}## is 0.10M and of ##\mathrm{Cl^-}## is 0.5M.

The initial pressure of ##\mathrm{Cl_2}## is 1.50atm at ##25^\circ\text{C}##.

Relevant Equations

Using the table below, calculate the initial voltage of the galvanic cell.

Looking at the table above, we see that the top-to-bottom order is from largest to smallest standard cell potential.

Since gold, ie Au, is at the top, every other redox couple on the list has the ability to reduce a ##\mathrm{Au^+(aq)/Au(s)}## redox couple, including the redox couple ##\mathrm{Cl^-/Cl_2}##.

The cell

$$\mathrm{Cl^-(aq)|Cl_2(g)||Au^+(aq)|Au(s)}\tag{1}$$

has emf equal to ##1.69V-1.3583V=0.3317\text{V}##.

It appears we can use the Nernst equation.

$$E_{\text{cell}}=E_{\text{cell}}^\circ-\frac{RT}{n_RF}\ln{Q}\tag{2}$$

At ##\mathrm{298.15K}## we have that ##RT/F=0.025693\text{V}## and so

$$E_{cell}=E^\circ_{cell}-\frac{0.025693\text{V}}{n_r}\ln{Q}\tag{3}$$

So what is ##Q##?

It is the reaction quotient. But what is the reaction?

It seems we have the skeleton equation

$$\mathrm{2Cl^-(aq)+Au^+(aq)\rightarrow Cl_2(g)+Au(s)}\tag{4}$$

and (I am guessing) we need to balance it using a procedure I learned recently which includes the following intermediate steps

Balanced oxidation half-reaction:

$$\mathrm{Au^+(aq)+e^-\rightarrow Au(s)}\tag{5}$$

Balanced reduction half-reaction

$$\mathrm{2Cl^-(aq)\rightarrow Cl_2(g)+2e^-}\tag{6}$$

We multiple (5) by 2 and sum the result with (6) to get

$$\mathrm{2Au^+(aq)+2Cl^-(aq)\rightarrow Cl_2(g)+2Au(s)}\tag{7}$$

and the reaction quotient for this chemical reaction seems to be

$$Q=\mathrm{\frac{P_{Cl_2}}{[Au^+]^2[Cl^-]^2}}$$

Using the values given in the problem we find that

$$Q=\frac{1.52}{0.1^20.5^2}=607.95$$

Therefore

$$E_{cell}=0.3317\text{V}-\frac{0.025693\text{V}}{2}\ln{(607.95)}=0.25\text{V}\tag{8}$$

This matches the answer given on MIT OCW.

The first time I wrote this post, I made a mistake in equation (8). Instead of using ##n_r=2## I used ##n_r=1## and got an incorrect result.

Now that I have made this correct, I don't really have a question. I believe the steps above are all correct now.

EDIT: Oops! See the following post by @Borek for the correction.

This is a case of you can subtract the Standard Action Potential's of the two and get the same result.

Neither Au⁺, Cl^- nor Cl₂ are in the standard state, so the Nernst equation is a must (even if it is just a plug and chug kind of problem).