- #1
zenterix
- 774
- 84
- Homework Statement
- Using the table below determine the unknown quantity for the following galvanic cell
- Relevant Equations
- $$\mathrm{Pt(s)|H_2(g,1.0bar)|H^+(pH=?)||Cl^-(aq,1.0M)|AgCl(s)|Ag(s)},E=+0.30\text{V}$$
This is the final problem in this problem set from MIT OCW.
Here is what I did to try to solve it
The table cited in the problem is below
We can easily spot the two redox couples that are in the electrochemical cell we are given.
The hydrogen-based electrode has standard potential zero, and the silver-based electrode has standard potential 0.22.
Thus, the hydrogen electrode, with the lower potential, is the reducing agent (ie, it is where oxidation happens) and is the anode.
Electrons flow from the hydrogen electrode to the silver electrode.
Thus, we have ##E^\circ_{\text{cell}}=0.22\text{V}##.
Note that this is the potential of the electrochemical cell under standard conditions (the half-cell potentials in the table are under these standard conditions).
We are given, however, that under the conditions specified in the problem the cell potential is 0.30V.
The conditions specified are concentrations of the ions in solution and partial pressure of hydrogen gas.
We can see how these are relevant when we write out the balanced redox equation for the cell.
We sum the oxidation half-reaction with the reduction half-reaction (which I multiplied by two to get the electrons to cancel)
$$\mathrm{H_2(g)\rightarrow 2H^+(aq)+2e^-}\tag{1}$$
$$\mathrm{2AgCl(s)+2e^-\rightarrow 2Ag(s)+2Cl^-(aq)}\tag{2}$$
and get
$$\mathrm{H_2(g)+2AgCl(s)\rightarrow 2Ag(s)+2Cl^-(aq)+2H^+(aq)}\tag{3}$$
with reaction quotient
$$Q=\mathrm{\frac{[H^+]^2[Cl^-]^2}{P_{H_2}}}=\frac{[H^+]^2[1]^2}{1}\tag{4}$$
$$\implies [H^+]=\sqrt{Q}\tag{5}$$
Using the Nernst equation, we can write
$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{RT}{n_rF}\ln{Q}\tag{6}$$
As far as I can tell at this particular point in time, temperature is not a variable in this problem, but neither is it specified.
If we assume the temperature is 298.15K then the equation above becomes
$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.025693}{n_r}\ln{Q}\tag{7}$$
and so
$$0.30\text{V}=0.22\text{V}-\frac{0.025693}{2}\ln{Q}\tag{8}$$
$$\implies Q=0.001974\tag{9}$$
Thus, using (5) we find
$$\mathrm{[H^+]=0.0445}$$
and thus since
$$pH=-\log{[H^3O^+]}$$
if we can say that ##[H^+]=[H_3O^+]## then
$$pH=1.35$$
The MIT OCW solution to the problem says pH is 1, however.
Here is what I did to try to solve it
The table cited in the problem is below
We can easily spot the two redox couples that are in the electrochemical cell we are given.
The hydrogen-based electrode has standard potential zero, and the silver-based electrode has standard potential 0.22.
Thus, the hydrogen electrode, with the lower potential, is the reducing agent (ie, it is where oxidation happens) and is the anode.
Electrons flow from the hydrogen electrode to the silver electrode.
Thus, we have ##E^\circ_{\text{cell}}=0.22\text{V}##.
Note that this is the potential of the electrochemical cell under standard conditions (the half-cell potentials in the table are under these standard conditions).
We are given, however, that under the conditions specified in the problem the cell potential is 0.30V.
The conditions specified are concentrations of the ions in solution and partial pressure of hydrogen gas.
We can see how these are relevant when we write out the balanced redox equation for the cell.
We sum the oxidation half-reaction with the reduction half-reaction (which I multiplied by two to get the electrons to cancel)
$$\mathrm{H_2(g)\rightarrow 2H^+(aq)+2e^-}\tag{1}$$
$$\mathrm{2AgCl(s)+2e^-\rightarrow 2Ag(s)+2Cl^-(aq)}\tag{2}$$
and get
$$\mathrm{H_2(g)+2AgCl(s)\rightarrow 2Ag(s)+2Cl^-(aq)+2H^+(aq)}\tag{3}$$
with reaction quotient
$$Q=\mathrm{\frac{[H^+]^2[Cl^-]^2}{P_{H_2}}}=\frac{[H^+]^2[1]^2}{1}\tag{4}$$
$$\implies [H^+]=\sqrt{Q}\tag{5}$$
Using the Nernst equation, we can write
$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{RT}{n_rF}\ln{Q}\tag{6}$$
As far as I can tell at this particular point in time, temperature is not a variable in this problem, but neither is it specified.
If we assume the temperature is 298.15K then the equation above becomes
$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.025693}{n_r}\ln{Q}\tag{7}$$
and so
$$0.30\text{V}=0.22\text{V}-\frac{0.025693}{2}\ln{Q}\tag{8}$$
$$\implies Q=0.001974\tag{9}$$
Thus, using (5) we find
$$\mathrm{[H^+]=0.0445}$$
and thus since
$$pH=-\log{[H^3O^+]}$$
if we can say that ##[H^+]=[H_3O^+]## then
$$pH=1.35$$
The MIT OCW solution to the problem says pH is 1, however.
