- #1
zenterix
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- 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.
This problem is from this problem set from MIT OCW.
Here is how I tried to solve the problem.
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.
Here is how I tried to solve the problem.
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.
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