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Step-by-Step Solution
Step 1: Write the Cell Representation
The given galvanic cell is:
Pt(s) | H2(g, 1 bar) | HCl(aq) | AgCl(s) | Ag(s) | Pt(s)
Step 2: Identify the Half-Reactions
1) Anode (oxidation) half-reaction:
$ \displaystyle \text{H}_2(g) \longrightarrow 2\,\text{H}^+(aq) + 2\,e^-
$
2) Cathode (reduction) half-reaction:
$ \displaystyle \text{AgCl}(s) + e^- \longrightarrow \text{Ag}(s) + \text{Cl}^-(aq)
$
Step 3: Overall Cell Reaction
By combining the two half-reactions (making sure electrons cancel out), we get:
$ \displaystyle \text{H}_2(g) + 2\,\text{AgCl}(s) \longrightarrow 2\,\text{H}^+(aq) + 2\,\text{Cl}^-(aq) + 2\,\text{Ag}(s)
$
Step 4: Write the Nernst Equation
The cell potential $E_{\text{cell}}$ is given by the standard electrode potential $E_{\text{cell}}^0$ minus a term involving the reaction quotient $Q$. For a two-electron transfer ($n=2$), at 298 K, and using $ \frac{2.303 R T}{F} = 0.06\text{ V}$, the Nernst equation can be written as:
$ \displaystyle E_{\text{cell}} \;=\; E_{\text{cell}}^0 \;-\; \frac{0.06}{n} \,\log Q
$
Here, $E_{\text{cell}}^0$ is the difference between the standard reduction potentials of the cathode and anode. Since the anode is the standard hydrogen electrode (SHE) with $E^0 = 0\,\text{V}$, then
$ \displaystyle E_{\text{cell}}^0 \;=\; E_{\text{AgCl/Ag}}^0 \;-\; 0 \;=\; E_{\text{AgCl/Ag}}^0.
$
Step 5: Express the Reaction Quotient, Q
For the overall reaction:
$ \displaystyle \text{H}_2(g) + 2\,\text{AgCl}(s) \longrightarrow 2\,\text{H}^+(aq) + 2\,\text{Cl}^-(aq) + 2\,\text{Ag}(s),
$
the reaction quotient $Q$ is:
$ \displaystyle Q \;=\; \frac{[\text{H}^+]^2 \,[\text{Cl}^-]^2}{P_{\text{H}_2}}.
$
Step 6: Substitute the Known Values
The cell potential $E_{\text{cell}}$ is given as 0.92 V when the HCl solution is $10^{-6}$ molal (assume activity ~ concentration for simplification). Hence, $[\text{H}^+] \approx 10^{-6}\,\text{M}$ and $[\text{Cl}^-] \approx 10^{-6}\,\text{M}$.
The partial pressure $P_{\text{H}_2}$ is 1 bar. Thus,
$ \displaystyle Q \;=\; \frac{\left(10^{-6}\right)^2 \left(10^{-6}\right)^2}{1} \;=\; 10^{-24}.
$
Step 7: Apply the Nernst Equation
$ \displaystyle E_{\text{cell}} \;=\; E_{\text{AgCl/Ag}}^0 \;-\; \frac{0.06}{2}\,\log(Q).
$
Therefore,
$ \displaystyle 0.92 \;=\; E_{\text{AgCl/Ag}}^0 \;-\; 0.03 \,\log\left(10^{-24}\right).
$
The term $ \log\left(10^{-24}\right) = -24$, so:
$ \displaystyle 0.03 \,\log\left(10^{-24}\right) \;=\; 0.03 \times (-24) \;=\; -0.72.
$
Hence,
$ \displaystyle 0.92 \;=\; E_{\text{AgCl/Ag}}^0 \;-\; \left( -0.72 \right).
$
$ \displaystyle E_{\text{AgCl/Ag}}^0 \;=\; 0.92 \;-\; (-0.72) \;=\; 0.92 + 0.72 \;=\; 1.64 \;\text{V}
$
Double-check your signs carefully: The standard relation in the solution provided is:
$ \displaystyle E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.06}{n}\log Q.
$
But since the cell potential with the given concentrations is 0.92 V, the negative of a negative log term effectively adds:
$ \displaystyle E_{\text{cell}} = E_{\text{AgCl/Ag}}^0 - 0.03 \times (-24).
$
So that term is $ +0.72$. Setting $E_{\text{cell}}$ to 0.92 V gives
$ \displaystyle 0.92 = E_{\text{AgCl/Ag}}^0 + 0.72,
$
which leads to:
$ \displaystyle E_{\text{AgCl/Ag}}^0 = 0.92 - 0.72 = 0.20\,\text{V}.
$
Step 8: Final Answer
The standard electrode potential of the (AgCl/Ag) electrode is therefore
$ \displaystyle 0.20\,\text{V}.
